Collection of Examples - Szt courses/3... · Collection of Examples ... Annamária Duliskovich, Eszter Fehér, Bajnok Tamás Nagy, Júlia Ridzi, ... &'ˆyz 11320022 600 22,6 a,

BME, DEPARTMENT OF MECHANICS, MATERIALS & STRUCTURES

Collection of Examples Strength of Materials 1.

Part 1

The Hungarian version was edited by Ágnes O. Csicsely.

BME Department of Mechanics, Materials & Structures Strength of Materials 1.

This book is a revision and extension of Szilárdságtani Példatár by Tibor Kőrössi, Tamás Laki, György Rusznák Dr. Thanks to the following architect students for their contribution in editing and in drawing the figures: Annamária Duliskovich, Eszter Fehér, Bajnok Tamás Nagy, Júlia Ridzi, Dóra Zsódi, Diána Balogh. The book’s text was translated to English by Eszter Fehér and Dániel Tamás Karádi (architect students), and the translation was checked by Krisztina Tóth. The text of the figures was translated by Krisztina Tóth.


Axial tension, compression 1/105

1. AXIAL TENSION, COMPRESSION

We use Hooke’s law during the solution of the introduced problems. According to Hooke’s law, for

materials in elastic state:

� � � � �

where

� � ��

- �is thetensile stress in the cross-section, that is the fraction of the axial tension and the cross-section’s area, and its dimension: N/mm2. „E” is the modulus of elasticity (or Young modulus) and

refers to the material of the investigated element, its dimension is also: N/mm2.

� � �

- � is the strain, the deformation of a 1 mm long part of the bar. It is the fraction of the total elongation ∆ and the initial length l, therefore ε has no unit of measure.

According to the equations above:

� � ��

∆ � � � � ��

When a bar is under tension in the direction of its axis, it expands in this direction and it tends to contract in the direction transverse to the axis of the tension. This lateral contraction can be calculated with the width of the cross-section (diameter of a circle, side length of a rectangle etc.) and the lateral strain. With a formula:

∆� � ��

The relationship between the axial and the lateral strain:

� � ��

Where µ is the Poisson’s ratio, a parameter of the material without unit of measure. The Poisson’s

ratio cannot be more than 0,5. The negative sign refers the contrary nature of the two strains.

If we substitute � into the ∆� formula, and then substitute ∆�� in the place of ε, and �� in the place

of ∆, we obtain a formula for ∆� (similar to ∆):

∆� � ��



The difference between the two formulas is that the elongation depends on the length „l”, and that the contraction depends on the cross-section’s width „d” (furthermore µ points out the fact, that the

strain is lateral).

Take care of using identical units: it is useful to convert all of them to N and mm (mm2, mm3).

In case of axial compression, as long as there is no danger of buckling because of the thickness of the compressed element, the same formula can be used - apart from the sign-, therefore we do not discuss

that case here.



1.) Calculate the maximal tensile force the bar can hold! Calculate the elongation and lateral

contraction if F=FRd!

The solution of these problems are based on Hooke’s law. According to Hooke’s law, for materials in

elastic state:

��,�,� � 270�/��

�� 206000�/��

� � 0,3

� � 12 � $4 � 113��

For maximal tensile force, the stress � is just equals to ��,�,� in the cross-section of the bar. Hence:

&'� � � � ��,�,� � 113 � 270 � 30510� � )*, +,-.

Deformations caused by this force:

∆ � ��,�,� � � � 270 � 6000206000 � /, 0122

∆3 � �� ,�,� � 3� � �0,3 � 270 � 12206000 � �0,0047��456789:58;33<:�;8;9=



2.) Calculate F, if> � *, *1°! (Assume, that the beam is absolutely rigid!)

��,�,� � 300�/��

�� 206000 ��

� � 8 � $4 � 50,3��

Remark: The first task is to convert α

from degrees to radians.

0,06° ⟹ 0,06 � $180 � 1,05 � 10BC9:3 � 0,001059:3

If we know the value of αin radians, we can calculate the vertical displacement of point „A”. (When setting up the equation, consider ifα< 5°40', then α (rad) ≈ sin α ≈ tg α.)

∆ � � α � 5000 � 0,00105 � 5,25mm4elongation= The vertical displacement of point „A” is caused by the elongation of the wire, and this latter is

caused by the force F acting in the wire. According to Hooke’s law, if we know the elongation, we can calculate F:

& � ∆ � � � � � 5,25 � 206000 � 50,34000 � 13599,9� � 13,6��

Check the load-bearing capacity:

� � &� � 13599,950,3 � 270,4 �� O ��,�,� � 300 �� P�&�! Finally we can calculate „F”, as a load which induce known reactions, from moment equation about

point „B”:

2 � & � 5 � �� → & � 52 � �� )S-.



3.) 3 pieces of steel cables with diameter 10 mm are loaded at the same time (as it is shown in the

figure). The cable in the middle has lower ultimate strength, than the rest cable have. TheT � U

diagram is given. Calculate the elongation of the cables if:

a, V � )*-.;

b, V � ++-.!

Calculate the load-bearing capacity of the system!

�� 206000�/��

��,�,W � ��,�,�,W/�� 300/206000 � 1,456 � 10BC � 1,456%o

��,�,� � ��,�,�,�/�� 200/206000 � 0,9709 � 10BC � 0,9709%o

Draw the & � ∆ diagram of the „parallel system”

�W � 2 � 10 � $4 � 157��

�� 10 � $4 � 78,5��

The cable in the middle is in plastic stress state, the other cables are in elastic stress state:

� � ��,�� 0,9709 � 10BC

& � 157 � 206000 � 0,0009709 Y 78,5 � 200 � 47,1 � 10C� � 47,1��

∆ � 0,0009709 � 5000 � 4,854��

All 3 cables are in plastic stress state:

� � ��,W� � 1,456 � 10BC

& � 78,5 � 200 Y 157 � 300 � 62,8 � 10C� � 62,8��

∆ � 0,001456 � 5000 � 7,282��

Load-bearing capacity of the system (in plastic stress state): 62,8kN

side cables

cable in the middle



a ,∆ �? , <�& � 30��

30�� O 47,1��, all of the cables are in elastic stress state.

Since the cables have identical cross section area and elastic modulus, therefore each cable has 10

kN and the elongation is:

∆ � � � � � � � 10000 � 5000206000 � 78,5 � 3,1��4[89;85\=

b, ∆ �? , <�& � 55��

55�� ] 47,1��, the cable in the middle is in plastic state, the other two cables are in elastic state.

& � �W Y �� W Y ��,�,�,� � �� W Y 200 � 78,5 � 55 � 10C�

↓

�W � 39,3 � 10C� � 39,3��

Hence, both of the two cables has severally

&W � 39,32 � 19,65��force

Calculate the stress produced by this force to verify the results:

�W � &� � 1965078,5 � 250,3�/�� O 300�/�� P:�;! The elongation:

∆W � � � � � 250,3 � 5000206000 � 1, ,224stretch= That is nearly twice as much as the previous elongation.

elastic

elastic el.

pl.

plastic

plastic

„a” cables

„b” cable



4.) Draw the T � U diagram of the materials!

4. b) Draw the V � ∆d diagram of the “parallel system”!

4. c)V �?, if ∆d � *, ,*e2.

Steel: ��,�,� � 200�/�� 206000�/�� Timber: �f,� � 15,3�/�� f � 12000�/��

a, Draw the � � � diagram of the materials!

��,� � ��,�,�� 200206000 � 9,71 � 10Bg � *, h/,‰

�f,� � �f,��f � 15,312000 � 1,275 � 10BC � ,, j/+‰

b, Draw the & � ∆ diagram of the parallel system!

The steel is in plastic state, the timber is in elastic state:

� � ��,� � 9,71 � 10Bg

& � 2 � 20 � 200 Y 100 � 9,71 � 10Bg � 12000� 160000 Y 116520 � 276520�� j/1, +j-.

∆ � 0,000971 � 1000 � *, h/,22

The steel is in plastic state, the timber is at the elastic

limit:

steel: 20x20 steel: 20x20

timber: 100x100

steel compression

compression

tension

tension

steel

timber

timber

the timber is

disrupted!

steel elast.

timber plast.

ela

st.

pla

st.

plastic

disrupted!


Axial tension, compression 8/105 � � �f,� � 1,275 � 10BC

& � 160000 Y 100 � 100 � 15,3 � 313000� � ),)-.

∆ � 0,001275 � 1000 � ,, j/+22

The steel is in plastic stress state, the timber is at rupture:

� ] �f,� ] 1,275 � 10BC

& � 160000� � ,1*-.

c, & �?, if ∆ � 1��.

The steel is in plastic stress state, the timber is in elastic stress state:

∆� � ∆f → �� f,because� � f � � � �� f � ∆ � 11000 � 10BC

��,�,� � 200�/��

�f � 10BC � 12000 � 12 �� O �f,� � 15,3�/��

& � &�fmm� Y &fno�mp � ��,�,� � ��fmm� Y �fno�mp � �fno�mp � 160000 Y 12 � 100 � 280000�� j0*-.



5.a) Calculate the elongation of the suspension rods (qr �? ;qt �? =, if V, � *ruvV � jS-. .

Calculate the inclination of the beam!

5.b) Calculate V, ifV � ,*-. and the beam remains horizontal!

Calculate the stresses in the bars (Tw �? ; Tx �?)!

Steel: ��,�,� � 200�/�� 206000�/��

�yz � 12 � $4 � 1135�

&'�yz � 113 � 200 � 22600� � 22,6��

a, Calculate the elongation of the suspension rods(y| �? ;y} �? =, if Fz � 0 andF � 24kN ?

Calculate the inclination of the beam!

&� � &3 O &�yz � 243 � 8�� O 22,6��

&� � 2 � &3 O &�yz � 2 � 243 � 16�� O 22,6��

�� ∆� � & � � � � � 8 � 10C � 2000206000 � 113 � 0,6873��

�� ∆� � & � � � � � 16 � 10C � 4000206000 � 113 � 2,749��

rigid beam



� ≅ tg� � 2,749 � 0,68736000 � 3,436 � 10Bg9:3 � 0,01969° b, Calculate &z, <�& � 10�� and the beam remains horizontal! Calculate the stresses in the bars! (�� ? ; �� ?)

�� é[&� � 10 � 2 Y &z � 76 &� � 10 � 4 � &z � 16

&� � �� &� � �� 567[8:78 → &� � � � &� � �

410 � 2 Y &z � 7= � 2,006 � 410 � 4 � &z � 1= � 4006 → &z � 6, 6� ��

�� 111111113 � 98,33�/�� O �� 200�/��

&� � 10 � 2 Y 6, 6� � 76 � 11, 1� ��

�� 5555113 � 49,16�/�� O �� 200�/��

&� � 10 � 4 � 6, 6�6 � 5, 55� ��

Both of the bars are in elastic stress state.

�� ∆� � & � � � � � 11111 � 2000206000 � 113 � 0,9546��

�� ∆� � & � � � � � 5555 � 4000206000 � 113 � 0,9546��

BME Department of Mechanics, Materials & Structures Strength of materials 1.

Pure shear 11/105

2. PURE SHEAR

If the cross-section is subjected only to pure shear and we assume, that the stress is uniformly

distributed, then the shear stress is:

� � ��

where τ is the average of the shear stress (N/mm2), and Vis the shear force. Note: this formula is not suitable for calculating shear stress in case of problems, when bending is combined with shear. You can

read about this latter case in the section „Bending”.

We should mention shear changes the angles. The shear modulus (G) is a function of the shear stress (τ), the shear strain (γ), elastic modulus and Poisson’s ratio, and the relation between them is

similar to Hooke’s law:

� � � � � → inthiscase:� � �2 � 4 Y �= From this, we can calculate the shear strain:

� � ��

Practically speaking pure shear does not occur in real life, however we assume pure shear in some parts of riveted/bolted joints and in some joints of timber structures. Hence we can calculate the shear

stress in these elements with the formula: � � �� .

Remark: in this chapter, the timber and steel joints, which are used in the discussed examples, are just a few possibilities and they may not be the most recent type of joints. The aim of these examples is to present the method of computation of pure shear, not to introduce the type of joints accurately. Some chapters of the subject called ’Steel and timber structures’ are dealing with these joints.


Pure shear 12/105

Bolted joints, riveted joints

When we check bolted/riveted connections, bolts/rivets have to be checked against shear, against bearing stress we have to check either the bolts/rivets or the plate – the one, that that has lower stress

limit against bearing, and in case of joints under tension we have to check the plate against tension.

Single shearing Double shearing

/The above connection is not recommended to design

because of the eccentricity of the joint./

Firstly it has to be determined whether single or double shearing is acting on the bolt/rivet shank. In the first case one cross-section, in the second case two cross-sections are carrying the shear stress.

The compression causes bearing stress on the surface of the rivet/bolt shank with various directions and intensity. Instead of this we can assume that the stress is uniformly distributed on the plane of the bolt’s diameter, if we calculate the ultimate strength in the same way.

� � 3 � 8 ��,� � &� � &3 � 8

In the formula of „fu,d”, „d” is the diameter of the rivet/bolt shank, at single shear joint, „t” is the thickness of the thinner plate(t1), at double shear joint „t” is the smaller out of2 � 8zand8 .

In the above mentioned formulas we use the nominal size of the diameter of the rivet/bolt shank, that is always an even number.

We have to check the weakened cross-sections of the plates against tension, where we calculate the weakened cross-section by subtracting the holes from the cross-section. We assume that the diameter of

the hole is d+1 mm.

There are rules for the bolt layout. If the bolt layout is corresponding to these rules, it is enough to examine only these introduced type of failures. In other circumstances we should not ignore the shear

stress on the plates, or the possibility of tear in multiple cross-sections.

Timber joints

Timber joints should be examined against all possible effects (tension, compression, shearing). As there are many possible cases, we don’t introduce any general computational method. We present the

detailed steps of the computation in the examples.


Pure shear 13/105

1.) Design the diameter of the bolts if the quality of the materials and the bolt layout is given!

Plates: ��,�,� � 235�/�� ,� � 360�/��

Bolt: ��,� � 240�/�� ,� � 400�/��

Foronebolttheactingforceis:& � &��7 � 3606 � 60�� Shearing:

�pm� � 3 $4

�pm� � &7 � ��,� � 600002 � 240 � 125�� → 3 � 12,6�� → �68�14 Bearing stress:

3pm� � &8on� � ��,� � 6000012 � 360 � 13,9�� → 3 � 13,9�� → �68�14

Accordingly, we use �14 bolts.

Checking the plates:

We check the main plate at the section, which is weakened by the three bolts:

&fz � 235 � 4180 � 15= � 12 � 10BC � S1+, )-. ] 360�� Safe!

&f � 235 � 4180 � 2 � 15= � 12 � 10BC � Sj), *-. ] 300�� Safe!

&fC � 235 � 4180 � 3 � 15= � 12 � 10BC � )0*, /-. ] 180�� Safe!

plate

(kN)

1 piece of

splice plate


Pure shear 14/105

2.) Check the bolt joint of the hinge!

Plates: ��,�,� � 200�/�� ,� � 360�/��

Bolt ��,� � 160�/�� ,� � 360�/��

The force transferred at the hinge is the support reaction of the suspended bar:

&� � 4 � 62 � 12,0��

Force per bolt:

&z � 12 � 5040 � 15,0��

& � 12 Y 15 � 27��

Since the bolts are the same, we should check the most heavily loaded 4& ) bolt.

Shearing:

&�,'� � � � 3 $4 � ��,� � 2 � 12 $4 � 160 � 10BC � 36,19�� ] 27��P:�;! Bearing stress:

&�,'� � 3 � 8on� � ��,� � 12 � 5,7 � 360 � 10BC � 24,62�� O 27�� Unsafe!

So the joint is unsafe.

Calculate the support reactions for a practice!

beam


Pure shear 15/105

3.) Calculate the load-bearing capacity of the marked joint of the suspension roof structure!

Softwood: ��,� � 1,2�/�� ,�,� � 9,7�/�� ,��,� � 1,2�/�� f,�,� � 6,5�/��

Remark: We calculate the maximal force (&'�) for all possible type of failures, and the lowest

value provides the resistance of the joint.

Tension: For tension we check the column at the cross-section weakened by the bolt.

&f,'� � 70 � 4200 � 13= � 6,5 � 85085� � 85,0��

Compression at bolts For compression at bolts, we check the surface at which the grains are perpendicular to the direction

of the stress – therefore we check the beam.

&�,��,'� � 2 � 40 � 200 � 1,2 � 19200� � 19,2��

Compression: The bottom surface of the suspension column is under compression parallel to its grains, we check it as well. (Usually this type of failure is not significant compared to the compression at bolts.) &�,�,'� � 2 � 40 � 200 � 9,7 � 155200� � 155,2��

Shearing: Shear stress acts… a, at the part of the suspension column that is under the beam, and b, at the beams on a spatial surface (see the figure on the right).

Examine both!

a, &�,�,'� � 2 � 200 � 200 � 1,2 � 96000� � 96��

b, &�,�,'� � 2 � 4200 � 150 � 113,1 Y 2 � 40 � 150= � 1,2 � 100529� � 100,5��

fixing bolt


Pure shear 16/105

Load-bearing capacity: The lowest value of load-bearing capacity (&'�) was for compression at bolts.

Therefore the load-bearing capacity of the joint is:

&'� � ,h-.


Pure shear 17/105

4. a) Design the number of bolts!

4. b) Check the joint!

4. c) Calculate the elongation of the bar „CB” !

Plates: ��,�,� � 235�/�� ,� � 360�/��

Bolts: ��,� � 240�/�� ,� � 400�/��

ES= 206000�/��

∑�� 0

�� 115,2 � 4,0 � 86,4 � 34 � 180��

a, Design the number of bolts! For pure shearing:

&�� &'�180 � 10C � 7 � 2 � 14 � $4 � 2407 � 2,436�5[ → 296�[

For bearing stress:

&�� &'�180 � 10C � 7 � 14 � 2 � 6 � 36010 � 400 7 � 3,571�5[ → 2rows

The number of bolts: 2 rows (4 pieces) > 3,571 pieces

b, Check the joint:

Checking of the plate for tension

&'� � 4120 � 2 � 15= � 10 � 235 � 10BC � j,,, +-. ] &��z � 180��Safe!

Checking the splice plates is not necessary as their total thickness exceeds the thickness of the plate.

c, Calculate the elongation of the bar „CB”!

∆ � & � � � � � 180000 � 7000206000 � 4120 � 10= � +, *h/22


Bending 18

3. BENDING

Different special cases can be determined by the position of the plane of the bending or by the stress state of the material. According to the first category, there are symmetrical and unsymmetrical bending (or skew bending), according to the second category there are elastic, elasto-plastic and plastic bending. In the examples, we always examine a combination of these cases. There are two methods to design or check: stress comparison and internal force comparison. Stress comparison method is usually used only

when elastic stress state is assumed.

3.1 Pure symmetrical bending

Pure symmetrical bending in elastic stress state

We consider pure symmetrical bending if one of the principal axis of the cross-section is on the plane of bending. In this case the neutral axis coincides with the other principal axis. Due to the elastic behaviour of the cross-section, the stress is linear on both sides of the neutral axis. The general formula

for the stress:

� � ��¡� � ¢

- where „My” is the bending moment about axis y (acting in the plane perpendicular to the axis y)

- „Iy” is the moment of inertia about axis y;

- „z” is the orthogonal distance of the given point from axis y

Checking by stress comparison:

Maximal stress at thepoint ¢ � ¢oW�

� � ��¡� � ¢oW� � ��£�

- where £� � ¤¥¦§¨© is the section modulus.

The cross-section is safe, if �oW� � ��

- where��,� is the yield stress (bending stress) of the material

Checking by internal force comparison:

Maximal moment-bearing capacity:

�'� � £� � ��

The cross-section is safe, if

�� '�


Symmetrical bending 19/105

Pure symmetrical bending in plastic stress state

In contrast to the previous case, now the cross-section is in plastic stress state, the stress is equal to�� in each point. Since the criteria of the equilibrium is the equality of the area of the tensile and compressive part of the cross-section, the neutral axis halves the area of the cross-section and it is

parallel to the elastic neutral axis. Considering this, the moment-bearing capacity can be calculated as:

�'� � �� ¢� Y �� f � ¢f � �� 4|P�| Y |Pf|= - where |Sc| and |St| are the absolute value of the static moment of the tensile and compressive part

of the cross-section

The cross-section is safe, if

�� '�

A few more words about principal axes and principal inertias. If we know the moment of inertia (Iy

and Iz) of a cross-section about an arbitrary axes pair with an origo in the centroid (the axes are perpendicular to each other)and we also know the product of inertia, then we can calculate the principal

moment of inertias with this formula:

¡z, � ¡� Y ¡¦2 « 12¬¡� � ¡¦® Y ¯�¦

The angle between the principal axes and the original axes isα; and:

tg 2° �� 2 � ¯�¦¡� � ¡¦

(By convention, the positive rotation rotates the positive part of the axis y towards the positive

part of the axis z.)

Further important notes about the position of the principle axes: the principal axis with maximum moment of inertia can be created by rotating the axis with higher moment of inertia from the original axes with α in a direction where the axis passes through quadrants with the opposite sign to the product

of inertia (ebben nem vagyok biztos, hogy jó-e).



for example:

¡� ] ¡¦:73¯�¦<[�6[<8<±; ¡� ] ¡¦:73¯�¦<[7;²:8<±;

C C



1. The given structure is just safe against pure bending. Calculate the value of „b”:

a, assuming elastic stress state and ³´v � S**-.2;

b, assuming plastic stress state and ³´v � +**-.2.¶·,q,v � j**./22j

a, elastic stress state (�oW� � ��,�,� at the most distant fiber from the neutral axis)

�� '�

The necessary (or required) section modulus:

£pm� � ��,�,� � 400 � 10¸200 � 2,0 � 10¸��C

The section modulus expressed with the properties of the structure:

£ � ¡�¢oW�

¡� � 112 � 280C � 200 � 112 � 180C � 4200 � �= � 365,87 � 10¸ � 97,2 � 10¸ Y 486000 � �� 268,67 � 10¸ Y 486000 � �

¢oW� � 1802 Y 50 � 140

£ � ¡�¢oW� � 1,92 � 10¸ Y 3471,43 � �4��C= £pm� � £

2,0 � 10¸ � 1,92 � 10¸ Y 3471,43 � �

�pm� � 23,05�� → �¹pº� � j+22

b, plastic stress state (�oW� � ��,�,� at any point of the cross-section) �� '�

�'�,¹� � ��,�,� � 4|P�| Y |Pf|=

(The neutral axis halves the area of the cross-section, therefore|P�| � |Pf|, i.e. the cross-section is symmetric to the axis of bending)

�'�,¹� � 200 � 2 � |P�| → Ppm� � 500 � 10¸2 � 200 � 1250000��C

With the properties of the structure:

P � 200 � 50 � 490 Y 25= Y � � 90 � 45 � 1150000 Y 4050 � �4��C= Ppm� � P

�pm� � z »��Bzz»��g�»� � 24,69�� → �¹pº� � j+22



2. Determine the radius ‘r’ if the cross-section is just safe for the given ³´v moment with ¶·,q,v, stress

a, assuming elastic stress state;

b, assuming plastic stress state for pure bending!

Calculate the plastic reserve of the cross-section!

a, In elastic stress state:

The necessary section modulus:

£pm� � ��

With the properties of the annulus:

£¹pº� �¼g � ½41,1 � 9=g � 9g¾1,1 � 9 � $4,4 � 9C � 41,1g � 1= � $4,4 � 9C � 41,46 � 1= � 0,328 � 9C

£pm� � £¹pº�

0,328 � 9C � ��,m��,�,�

9 � ¿ ��,m�0,328 � ��,�,�À � ,, S+ � ¿³´v,Ád¶·,q,v)

b, In plastic stress state

„r” appears only in the static moment

P�,pm� � ��,¹�2 � ��,�,�

With the properties of the annulus:

P�,�mon�np��m � 9 � $2 � 4 � 93 � $ � 23 � 9C

P�,¹pº� � 23 � 9C � 41,1C � 1,0C= � 0,221 � 9C

Ppm� � P¹pº�

0,221 � 9C � ��,¹�2 � ��,�,�

9 � ¿ ��,¹�0,442 � ��,�,�À � ,, )j � ¿³´v,Âd¶·,q,v)



Plastic reserve:

�'�,¹� � 0,442 � �� 9C

�'�,m� � 0,328 � �� 9C

�:[8<59;[;9±; � �'�,¹��'�,m� � 1 � )S, /+%



3. Check the cross-section for pure bending


b, assuming plastic stress state!

�� 120��

�� 10�/��

a, In elastic stress state

We can compare the moment or the stress as well.

Implement both for practice!

¡� � 112 � 300 � 480C � 112 � 200 � 240C � 2,534 � 10��g

£� � ¡�¢oW� � 2,534 � 10�240 � 10,56 � 10¸��C

�'�,m� � £� � �� 10,56 � 10¸ � 10 � 10¸ � ,*+, 1-.2 O �� 120,0��Ã7[:�;! Or:

�oW� � ��£� � 120 � 10¸10,56 � 10¸ � ,,, )1./22j ] �� 10�/�� Ã7[:�;! b, In plastic stress state

There is only one way to check the cross-section in plastic stress state: comparing the moments.

�'�,¹� � �� ÄP��Ä Y |PÅ|® � 10 � 2 � 4120 � 300 � 180 Y 120 � 100 � 60= � 10B¸� ,SS-.2 ] �� 120��P:�;!

Stress distribution:

elastic plastic



4. Calculate the moment-bearing capacity of the cross-section for pure bending



Draw the stress diagrams!

��,�,� � 200�/��

a, Assuming elastic stress state

� � 100 � 20 Y 20 � 200 � 6000��

¢Æ � z�� z�Ç �� z��¸�� 136,67��

¡� � 112 � 20C � 100 Y 20 � 100 � 483. 3� � 10= Y 112 � 200C � 20 Y 20 � 200� 4136,67 � 100= � 29,53 � 10¸��g

Or:

¡� � 13 � 20C � 100 Y 13 � 200C � 20 � 6000 � 483, 3� � 20= � 29,53 � 10¸��g

£� � ¡�¢oW� � 29,53 � 10¸136,67 � 216068��C

�'�,m� � £� � ��,�,� � 216068 � 200 � 10B¸ � S), j-.2

b, Assuming plastic stress state

∑�� 0 due to the equality the neutral axis halves the area of the cross section

�2 � 60002 � 3000��

: � �2 � 20 � 300020 � 150��

axis across the

centroid

neutral axis (plastic)

C

z C


Symmetrical bending 26/105 �'�,¹� � ��,�,� � 4|P�| Y |Pf|= �

� 200 � 420 � 100 � 470 � 10= Y 50 � 20 � 25 Y 150 � 20 � 75= � 10B¸ � /S-.2

Or:

�'�,¹� � ��,�,� � 2 � P��m�fpºn� � 200 � 2 � 20 � 150 � 61, 6� ® � /S-.2

Plastic reserve:

� � �'�,¹��'�,m� � 1 � 74 � 43,243,2 � /,%

Normal stress diagrams: Elastic stress distribution Plastic stress distribution

neutral axis

(elastic)

neutral axis

(plastic)

z C



5. Check the structure assuming

a) elastic stress state!

b) plastic stress state!

Draw the stress diagrams!

fs,y,d=235 N/mm2

A � 30 � 4105 Y 120= � 6 750��

zÆ � �105 � 30 � 15 Y 30 � 120 � 606750 � 25��

¡� � 105 � 30C3 Y 30 � 230C3 � 6750 ×25 � 14,0×10¸��g

a., elastic stress state ��oW� � 15 kNm

T¶ � Y15 � 10¸14 � 10¸ � 55=58,93 . 22j⁄ O ��,�,� � 235 � �� ⁄ Ér¶Á! Tr � �15 � 10¸14 � 10¸ � 95= � ,*,,8 . 22j⁄ O ��,�,�� 235 � �� ⁄ Ér¶Á! or:

�'�,m� � 235 � 14 � 10¸95 = � 34,63 � 10¸��O ��=15 k��Ér¶Á! b., plastic stress state

�2 � 67502 � 3375: � 337530 � 112,5��

³Êv,Âd � 235 � 2 � 3 � 112,5 � 38,75 � 10B¸=61,47kNm ] ��=15 ��Ér¶Á!

C y


Unsymmetrical bending 28/105

3. UNSYMMETRICAL BENDING (or in other words: SKEW BENDING)

If none of the principal axes of the cross-section is on the plane of bending, we are talking about unsymmetrical bending. (The moment vector’s line of action is not parallel with either of the principal

planes.)

If the principal planes and the principal moment of inertias are known, unsymmetrical bending can be calculated as a composition of two symmetrical bendings. Hence we decompose the moment into two components lying on the principle planes. (If the cross-section is symmetrical: the axis of symmetry and the axis perpendicular to it are the principle axes (axis y and z) and the corresponding moment of inertias are: „Iy”, „Iz”, in this case Dyz=0.)

� � «��¡� � ¢ «�¦¡¦ � �

4�55Ë9:8;�:� � ��¡� � ¢ Y�¦¡¦ � �= Neutral axis can be defined based on the condition � � 0. Hence

Ì��¡� � ¢Ì � Í�¢¡¢ � �Í 0 � Y��¡� � ¢ Y�¦¡¦ � �

Therefore:

¢ � ��¦ � ¡�� ¡¦ � � � � � �

where

� � ��¦ � ¡�� ¡¦

m is the slope of the neutral axis.

The position of the neutral axis can be defined by construction in many ways. Here is an easy method:

We calculate the stress in two points along an edge of the cross-section. By drawing the edge’s� diagram, we can construct the intersection point of the diagram and the edge. Stress is zero at this point, therefore the point is on the neutral axis. As the neutral axis has to pass through the centroid, the neutral axis is defined by these points and it can be constructed. (This method can be used in any case of unsymmetrical bending appearing in the following examples.)

The sign of the stress is defined by the formula



� � «��¡� � ¢ «�¦¡¦ � �

if we use the correct sign for My, Mz, y and z.

However, the correct sign can be determined by inspection when dealing with the two symmetrical bending separately. (Negative sign is compression; positive sign is tension both in the formula and on the

figures.)

The direction of the positive moment vector is on the figures. (Notice their sign is not identical with the sign of the coordinate system.)

It has to be mentioned that in case of unsymmetrical (or skew) bending the calculation of the maximum is easy only if both terms have extreme value at the same point. For example in case of hexagonal profile or “I”-profile. In these situations, the stress maximum can be calculated with this formula:

��:Î � ��¡� � ¢�:Î Y�¢¡¢ � ��:Î � ��£� Y �¢£¢ Also for elastic unsymmetrical bending the cross-section is safe, if � � �Ï.The internal force

comparison is complicated in case of unsymmetrical bending, thus we don’t use it.



1. a) Determine the maximum of the elastic moment, if ¶v � ,+./22j! b) Calculate the maximum of the load!

a) If we know the principal axes, it is practical to use the

following formula:

�oW� � �� «Ð¥¤¥ � ¢ « ÐÑ¤Ñ � �

Bending moment about the axis y (moment in plane z):

�� M� cos 30° � 0,866M

Bending moment about the axis z (moment in plane y):

�¢ � �� [<730° � 0,5�

Properties of the cross-section:

¡� � 180 � 240312 � 207,36 ∗ 106��4

¡¢ � 1803 � 24012 � 116,64 ∗ 106��4

At points with maximal stress, the two terms have the same sign.

15 � Y 0,866�'�207,36 ∗ 10¸ � 120 Y 0,5�'�116,64 ∗ 10¸ � 90 � �'�40,501 � 10B¸ Y 0,386 � 10B¸=

�Ô3 � 150,887 � 106�� 16,91��

b)

Õ'� � ÐÖ× � 8,455��/�

plane of the bending

C



2. Draw the T2rØ diagram!

¡¢ � ¡� �1204 � $4 � ,1j,01 � ,*122S

The problem seems to be unsymmetrical bending. However, all axes of the circle shaped cross-section is principal axis, therefore the resultant moment acts about a principal axis as well. Accordingly,

the problem is simplified to pure bending.

We should calculate the maximum of the moment, which is apparently in the middle of the beam, since the maximal moment is in the middle in both planes.

�4Õ= � 2,0 � 628 � 9,0��

�4&= � 4,0 � 1,5 � 6,0��

Resultant moment:

� � Ù9 Y 6 � 10,82��

plane of bending (q)

plane of bending (F)



Plane of bending:

8:7° � 69 � 0,667 → ° � 33,69°

Maximal stress:

�oW� � 10,82 � 10¸162,86 � 120 � 7,97�/��



3. Calculate height h, if T2rØ � ,*./22j? 4r··Ú2ÛuÜÁdr·ÝÛe·ÝÞÁ···ÝrÝÁ=

Maximum bending stress occurs in the middle of the beam, at the top and the bottom of the cross-

section, according to the figure.

�� 3 � 42 � 6��

��:Î � 6� 2,4 � 4 � 1,2 � 9,60��

�� 9,60 ∗ 56[ 30° � 8,31��¡� � 180 ∗ \312 £� � 180 ∗ \26 � 30\2

�¢ � 9,60 ∗ [<7 30° � 4,8��¡¢ � 1803 ∗ \12 £¢ � 1802 ∗ \6 � 5400\

The structure is safe, if

��:Î � �3

� � «��¡� � ¢ «�¦¡¦ � � � «��£� «�¦£¦

10 � Y8,31 � 10¸30\ Y 4,8 � 10¸5400\ /� 5400 � \

5400\ � 1495800000 Y 4,8 � 10¸ � \

\2 � 88,89 � \ � 27700 � 0 \z, � 88,89 « Ù88,89 � 4 � 277002

C

plane of the

bending


Unsymmetrical bending 34/105 \z � 216,71

\ � �127,81

The neutral axis, if h=217mm:

¡� � 153,27 � 10¸��g

¡¦ � 105,46 � 10¸��g

tanß � ¡�¡¦ � tan° → ß � 40° The structure is safe, if h = 220 mm.

C



4. Check the structure assuming elastic stress state, and draw stress diagrams!

fs,y,d=235 N/mm2

A � 30 � 4105 Y 120= � 6 750��

zÆ � �105 � 30 � 15 Y 30 � 120 � 606750 � 25��

¡� � 105 � 30C3 Y 30 � 230C3 � 6750 ×25 � 14,0×10¸��g

¡¦ � 30 � 150C12 Y 120 � 30C12 � 3,16 � 10¸��g

MEd=15 kNm

My=15×cos 35°= 12,3kNm

Mx=15×sin 35°=8,6kNm

Neutral axis:

8²ß � 14 � 10¸3,16 � 10¸ � 8²35°à � /j, ,)°

Tj � �12,3 � 10¸14 � 10¸ � 95 � 8,6 � 10¸3,16 � 10¸ ×15= � ,jS,28 . 22j⁄ O ��,�,� � 235 � �� ⁄ Ér¶Á! T, � �12,3 � 10¸14 � 10¸ � 55 Y 8,6 � 10¸3,16 � 10¸ ×52,5=48,34+142,88=191,2 . 22j⁄ O ��,�,� � 235 � �� ⁄

Safe!

plane of

the bending

neutral axis

C


Eccentric tension and compression 36/105

4. ECCENTRIC TENSION AND COMPRESSION

In this section we discuss the following cases:

Eccentric tension and compression of cross sections without tensile strength in elastic and plastic

stress state.

The principles of the calculation of the two cases are the same, but the following notes have to be

mentioned:

In case of eccentric action, the resultant of the stresses and the eccentric force are equal in magnitude, their sign is opposing and their line of action is coincide (ΣV � 0, ΣM � 0. = The stress state and the material properties of the structures are defined in the exercise severally,

but the principles of the calculations are always the same.

Now let us see the details of the different cases.

4.1 Computation assuming elastic stress state

We are looking for the magnitude of the stress, the position of the neutral axis and the stress

diagram.

In case of an ordinary point of action:

� � &� Y & � ;�¡¦ � �; where „F” is the eccentric force, „A” is the area of the cross-section, „Iy” and „Iz” are the moment of

inertias about the axes of the coordinate system: „y” and „z” are the principle axes, „ey” and „ez” are the coordinates of the point of action and „Y” and „Z” are the coordinates of the examined point. The signs of the terms can be determined by mathematical considerations (then the sign of F, „ey”, „ez”, „Y”, „Z” can be determined by their tensile or compressive character and their location in the coordinate system) or by inspection. In the latter method the sign of the first term of the stress formula (axial action) is the same in each point (tension is positive, compression is negative). The sign of the second and third term (bending) is equal to the sign of the first term, if the examined point and the point of action are on the same side of

the neutral axis („z” in the 2nd term, „y” in the 3rd term).

The position of points of the cross-section where the signs of the terms are the same



In the centroid (since here y=0 and z=0)

�6 � &�

Neutral axis can be characterized by its intersection with the axes „y” and „z”. The coordinates of these two points (�0and ¢0) can be computed as:

�6 � � <¢2;� ; ¢6 � <�2;¢

where<� and <¢ are the principle radiuses of gyration of the cross section (iy � ¬IyA ; iz � ¬IzA=.

Negative sign means, that"ey"and"yo" 4or"ez"and"¢è"= are on the opposite side of the axis „y” (or „z”,

respectively).

The baseline of the stress diagram is perpendicular to the neutral axis; it is practical because on lines parallel to the neutral axis the stress is constant (like in case of other kind of stress).

If the point of action is on one of the principle axes, then „ey” or „ez” is zero, therefore one term of the stress formula can be eliminated, and the neutral axis is perpendicular to the principle axis where the

point of action lies.

Further notes: if the axes passing through the centroid, are not principle axes, then the general stress formula has to be used for the bending terms. In this case, the easiest way to give the position of the

neutral axis is to construct points where the stress is zero on two different edges.

It has to be highlighted, that in the previous formulas all data are refers to the principle axes passing

through the centroid (except for „A” and „F”).

4.2 Computation assuming plastic stress state

There are no direct formulas for the solution; we have to set up equilibrium criteria equations ΣM � 0andΣN � 0to calculate the two unknowns (from FRd, from the two coordinates of the point of action, and from the two properties of the neutral axis) in the exercise. However, the order of the equations is relevant, because it defines whether we have to solve a system of equations with one or two

unknowns.

The following exercises reveal the practical way to set up these equations.



1. Check the cross-section, determine the position of the neutral axis and draw stress diagrams a, assuming elastic stress state,


AforceFEd � 800kN force acts at the given point of action (at point “D”).

�3 � 13,0�/��2

a, Elastic stress state

First of all, we calculate the geometrical properties of

the cross-section (area, centroid, moment of inertia).

� � 200 � 400 Y 2 � 100 � 4002 � 120000��

�Æ � 80000 � 200 Y 40000 � 133,3120000 � 177,8��2

¡� � 13� 200 � 4003 Y 2 112 � 100 � 4003 � 120000 � 177,82 � 1540 � 106��4

We do check with the comparison of the stresses. Since the point of action is on one of the principal axes, we use the following formula:

� � &� Y & � ;¦¡� � ¢

;¢ � 177,7 � 100 � 77,7��

The sign of the stresses can be determined by inspection.

(Tension is positive, compression is negative.)

C



σ1 � fd � � FA � F � ezIy � z � �800 � 103120000 � 800 � 103 � 77,81,540 � 109 � 177,8 � �13,85N/mm2

�1 � ,),0+ .22j ] ��3 � 13,00�/��2

So the cross-section is unsafe!

Now let us calculate the stress in the other extreme fiber:

�2 � �&�Y &� ;¢¡� � ¢ � �800 � 103120000 Y 800 � 103 � 77,81,540 � 109 � 222,2 � Yj,), .22j

Determine the position of the neutral axis!

The neutral axis is perpendicular to the axis x, and the neutral axis and the point of acting are on the opposite side of the centroid.

<�2 � ¡�� 1,540 � 109120000 � 12833��2

¢6 � � <�2;¢ � �1283377,8 � �164,9��

b, Plastic stress state:

There are two ways of solution: comparing forces or comparing eccentricity.

In the case of comparing forces, we calculate the load-bearing capacity of the cross-section for a force acting at the point of action and we compare it to the design load (force). The cross-section is safe, if the load-bearing capacity exceeds the design load (force).

In the case of comparing eccentricity, we try to find the most distant position of the point of action that is safe (limit of eccentricity), if the given eccentricity does not exceed it, the cross-section is safe. (In this case FEd � FRd)

Usually comparing eccentricity requires less calculation, since the position of the neutral axis is computed from a linear equation.

C



b.1 Comparing forces:

At first, we compute the position of the neutral axis from the equation ΣMì � 0 (setting up on the

axis through the point of action).

For computational reasons, it is practical to assume, that the compressive stresses are acting on the whole cross section, and then subtract the double of the moment of the tensile stresses.

Σ�� 0

�3 � 120000 � 77,8 � 2 � �3 � 200 � : � í300 � :2î� 2 � �3 � 2 � :4 � :2 � ï300 � 23 � :ð � 0

121,368 � 10¸ � 1560000 Y 2600: � 1950: Y 4,3:C � 0

4,3:C Y 650: � 1560000: Y 121,368 � 10¸ � 0

From this equation: a=82,15 mm

We do the checking by comparing design load &Ed and the load-bearing capacity &Rd

&Ed is calculated from the equation ΣN � 0:

&�3��3 � �Y 2 � �3 � �8;7[<67 � 0

&�3 � 13,0 � 120000 � 2 � 13,0 � 82,15 � 200 Y 241,12 �

13,0 � 4120000 � 36200= � ,*0h-. ] &�� 800��

So the cross-section is safe.

We can verify the result by calculating the moment about another axis (e.g. about an axis through the centroid):

1089 � 77,8 � 2 � 13,0 � ñ200 � 82,15 � 181,125 Y 2 � 20,54 � 82,152 � 167,43ò �

� 84724,2 � 84718 ≅ 0

therefore, the results are correct. (Differences are rounding errors.)

b.2 Comparing eccentricity:

In this method, we compute the position of the neutral axis from the equation ΣN � 0. For computational reasons it is again practical to assume, that the compressive stresses are acting on

the whole cross-section, and then subtract the double of the tension stresses.

&�3 � �3 � �Y 2�3 ��8 � 0

&�� 2 � �� f



800 � 1000 � 13,0 � 120000 � 2 � 13,0 � í200 � : Y 2 � :4 � :2î

6,5 � : Y 5200: � 760000 � 0

From this equation:

: � �5200 « Ù5200 Y 4 � 6,5 � 76000013 � �400 « 526,3 � 126,23��

(negative values are meaningless from the viewpoint of the problem).

Calculate the limit of eccentricity with an equation setting up on the centroid ΣMó � 0!

�'�,¹� � &�� ;'� � 2 � �� f � ��¹

�'�,¹� � 800 � 10C � ;'� � 2 � 13 � 4200 � 126,23 � 159,1 Y 2 � 126,23 � 31,562 � 138,05= �'�,¹� � 118,731 � 10¸�� 118,731��

;'� � �'�,¹�800 � 10C � 148,4�� ] ;�� 77,8��

So the cross-section is safe.



2. The position of the neutral axis is given. Where is the point of acting of the compressive force if the

maximal compressive stress is equal to fd

a. assuming elastic stress state

b. assuming plastic stress state?

Calculate in both cases the compressive force as well!¶v � ,*./22j

Since the neutral axis is perpendicular to the principal axis

„y”, the point of action should be on it.

a. Elastic stress state

� � 80 � 240 Y 150 � 120 � 37200��g

�Æ � �80 � 240 � 40 Y 150 � 120 � 7537200 � 15,6��

¡¢ � 13� 240 � 803 Y 13� 120 � 1503 � 37200 � 15,62 �

� 166910000 � 166,91 � 10¸��g

<¢2 � 16691000037200 � 4486,7��2

86 � 100 � 15,6 � 84,4��

;¢ � � 44984,4 � �53,2��

�� 10 � &'�37200 � &'� � 5,32166,91 � 10¸ � 95,6

�10 ��0,05735 � 10BC � &'�

&Ô3 � �174,2��456��9;[[<67�695;= b. Plastic stress state

In this case we compute FRd from the equation ΣFôõ � 0; then we compute the position of the point of action. Then we set up the ΣM equation on an axis through the centroid (and not on an axis through

the point of action).

ö&n� � 0

�8;7[ � 50 � 120 � 6000��2

�56��9 � 37200 � 6000 � 31200��2

&Ô3 � �10 � 31200 Y 10 � 6000 � �252 � 103� ��252��456��9;[[<67�695;= ΣMÆ � 0

C

n.a.



Firstly, we assume, that the compressive stress acts on the whole cross-section (its static moment is 0 about the centroid), and then we subtract the double of the static moment of that part of the cross-section where the tensile stress acts.

�'�,¹� ��÷ �&'� � ; � 2 � 10 � 120 � 50 � 109,4

�'�,¹� � 13,128 � 10¸�� 13,128��

; � �'�&'� �13,128��252�� 0,0521� � 52,1��



3. Calculate the maximum distance Áqof between the point of action and the centroid, if a tensile

force V´v � S**-. acts in the cross-section and the ultimate strength is ¶·,q,v � ,h* .22j


b. assuming plastic stress state!


¡� �120 � 200C12 � 88 � 168C12 � 45,23 � 10¸��g

� � 120 � 200 � 88 � 168 � 9216��

<�2 � 45,23 � 1069216 � 4907,8��2

The easiest way to compute ;z is substituting it into the stress formula. According to the position of the point of action, the top

extreme fiber is under tension.

190 � Y g�� z¸ Y g��mÑg», C�z�ø � 100

190 � 43,4 Y 0,8844 � ;¦

;¢ � 146,60,8844 � 165,8��

Calculate the stress in the bottom extreme fiber as well.

�: � Y 4000009216 � 400000�165,845,23�106 � 100 � �103,22�/��2

The position of the neutral axis:

¢6 � � <�2;¢ � 4907,8165,8 � 29,6��

b. Plastic stress state

&Ô3 � &�3 � 400��

C

C

C

n.a.

n.a.



ö&n� � 0

400 � 100C � 190 � 9216 � 2 � 190 � 4120 � : � 88 � 4: � 16==

1351040 � 45600: � 33440: Y 535040

816000 � 12160: → : � 67,1��

Σ�ù � 0

�'�,¹� � &�� ;'� � 2 � 190 � 4120 � 67,1 � 66,45 � 88 � 467,1 � 16= � 58,45� 103,44 � 10¸��

;'� � 103,44 � 10¸400 � 10C � 258,6��



4. Compression force.´v � ,**-. and bending moment ³ú,´v � YS*-.2 acts on the cross-

section. Design the value “b”, if the cross-section is just safe


b. assuming plastic stress state! ¶v � ,)./22j


Firstly we should identify the plane of bending. Mz acts on the cross-section, therefore plane „xy” is the plane of bending. It is important to find the correct direction of the moment vector. It is quite simple since the positive moment points to the ascending direction of the axis. These observations are

necessary for the calculations.

The cross-section is just at the limit of its load-bearing capacity (assuming elastic stress state), therefore we know that the stress in the extreme fiber is exactly �Ï.If we set up the stress formula (compressive force = -N, therefore the absolute value of the stress is maximal, if the moment has negative sign), the unknowns are the area of the cross-section and the moment of inertia. They contain the uknown „b”, which we can compute from a simple linear equation.

� � 350�

¡¦ � � � 350C12

��oW� � �� 13 � �� ¦¡¦ � �oW� �

� Bz��C»�� g��ûûü�C»�À�� C»� /� �

13� � 2244,9

� � 172,68�� → � � 180��

Find the position of the neutral axis to draw the stress diagram 4σ=! ;� � �� 40000000100000 � 400��

<¢2 � ¡¢� � 112� 3503 � 180350 � 180 � 10208,33��2

C



�0 � <¢2;� � �10208,33400 � 25,52��

Y�oW� � �� Y�¦¡¦ � �oW� � � 100000180 � 350 Y 40000000zz � 350C � 180 � 175 � 9,3�/��

b. Plastic stress state

If we assume plastic stress state, first we can use the fact, that the cross-section has to be in

equilibrium. Therefore we can equilibrate the stresses considering them either together or separately.

Assuming plastic stress state and eccentric compression we can divide the stress diagram into two parts. The stress acting on the center part of the cross section equilibrate the compression, and the stresses acting on the extreme parts equilibrate the bending moment (see the bottom of the figure on the right). If we calculate the volume of the stress blocks, we obtain the value of the resultant forces of the stresses. The moment arm is the distance between the centroid of the two stress blocks.

¡.� � � � �� 2 � : � � � ��

¡¡.�¦ � : � � � �� 4350 � :=

¡.100000 � 350 � � � 13 � 2 � : � 13 � �

¡¡.40000000 � : � � � 13 � 4350 � :=

¡.� � 10000013 � 4350 � 2 � := ¡¡.� � 4000000013 � 4350 � := � :

10000013 � 4350 � 2� := � 4000000013 � 4350 � := � :

13 � 4350 � 2 � := � 400 � 13 � : � 4350 � :=

1820000 � 10400 � : � 4550 � : � 13 � :

13 � : � 14950 � : Y 1820000 � 0

=



:z, � 14950 « √14950 � 4 � 13 � 182000026

:z � 138,4�� → � � 140��

: � 1011,6��4�:[;9668=



5. Calculate T2rØ and draw the stress diagram, assuming elastic stress state, if V´v � j*, *-.!

Unlike in the previous exercises, now the point of action is not on any principal axis, therefore the acting force has bending moment about both principle axes. Take care, that in this case the horizontal

and vertical axes through the centroid are not principal axes!

The cross-section is symmetric, we know the position of the principal axes:

� � g�√ � 120√2��

� � 240 2 � 28800��

¡¦ � 2 � þ120 � √2®g36 Y 120 � √2® 2 � 40 � √2® �� 138,24 � 10¸��g

¡� � ï 136 � í120 � Ù2î3 � 120 � Ù2ð � 46,08 � 106��4

;¢ � 120�√23 � 40√2��

;� � 120 � Ù2 � 120Ù2��

�oW� � �z � �� ¡� � ¢z ��¦¡¦ � �z � �2000028800 � 20000 � 40 � √246,08 � 10¸ � 40 � √2 �

�20000 � 120 � √2138,24 � 10¸ � 120 � √2 � �0,694 � 1,389 � 4,167 � Y6,25�/��

� � �� ¡� � ¢ ��¦¡¦ � � � �2000028800 � 20000 � 40 � √246,08 � 10¸ � 80 � √2 Y

Y20000 � 120 � √2138,24 � 10¸ � 0 � �0,694 Y 2,778 Y 0 � 2,084�/��

�C � �� ¡� � ¢C ��¦¡¦ � �C � �2000028800 � 20000 � 120 � √246,08 � 10¸ � 40 � √2 Y

Y20000 � 120 � √2138,24 � 10¸ � 120 � √2 � �0,694 � 1,389 Y� 4,167 � 2,084�/��

C

C



<�2 � ¡�� 138,24 � 10628800 � 4800��2

<�2 � ¡¢� � 46,08 � 10628800 � 1600��2

¢0 � � <�2;¢ � � 4800120 � √2 � �28,28��

�0 � � <¢2;� � � 160040 � √2 � �28,28��

In right-angledtriangleswithtwo45°angle, if the point of action is on one of the corners of the

triangle, the neutral axis is parallel to the opposite side of the triangle.

Draw the stress diagram!

neutral axis

C



6.) Check the structure if δδδδ=90°, αααα=0°

a) in elastic stress state,

b) in plastic stress state!

7.) Check the structure in elastic state if δδδδ=90°, αααα=40° !


a) in elastic stress state

b) in plastic stress state by comparing forces and by comparing eccentricity!

9.) Check the structure in elastic state if δδδδ=45°, αααα=40° ! fd= 10,7 N/mm2

The section details:

� � 300 � 100 Y 200 � 100 � 50000��g

¢Æ � 100�200�200Y300�100�5050000 � 110��

¡� � 300 � 10033 Y 100 � 20033 � 50000 � 102 � 361,67 � 106��4

¡¢ � 200 � 100312 Y 100 � 300312 � 241,67 � 106��4

y’

y






a) Elastic stress state:

� � «Ð¥¤¥ � ¢ � «Ð¥�¥

�6�;9 � � 10 � 106361,67 � 106 � 110� �3,041� ��2⁄ � ��:ÎO 10,7� ��2⁄

�Ë��;9 � Y 10 � 106361,67 � 106 � 190� Y5,253� ��2⁄ � Y��:ÎO 10,7� ��2⁄

The cross-section is safe in elastic stress state!

b) Plastic stress state

�Ô3,� � �3 � 4|P5|Y |P8|= The neutral axis halves the area of the cross section (in other words: the neutral axis is a bisector line of the cross section’s area)!

�2 � 500002 � 25000��2

� � 50000��g

¢Æ � 110�� ¡� � 361,67 � 106��4

The section details:

neutral

axis

C



This bisector line of the area will be in the lower rectangular, because its area is bigger than the upper area part. If this hypothesis is wrong, we get a higher value for ’a’ than the height of the lower rectangle. In this case we have to repeat the calculation considering this.

: � 25000300 � 83,33�� O 110��, thehypothesiswasright! Calculating the maximum of the plastic bending moment:

Let’s calculate all the little rectangle’s static moment for the neutral axis. Since the neutral axis halves the lower rectangular, the static moment of it’s compressed side will consist of two parts.

�Ô3,� � 10,7 � 4300 � 83,33 � 41,67 Y 300 � 16,67 � 8,33 Y 100 � 200 � 116,67= �

� 36,56 � 10¸�� 36,56�� ] 10,0��! The neutral axis is a bisector line of the area; therefore we are able to calculate it in another way too: we multiply twice the compressed side’s static moment to the centroid.

�Ô3,� � 2� �3 � ÄP5,5;7896<3Ä �Ô3,� � 2� 10,7 � 300 � 83,33 � 68,33 � 10�6 � 36,56�� ] 10,0��



7.) Check the structure in elastic stress state if δδδδ=90°, αααα=40°! ¶v � ,*, /. 22j⁄

The division of the moment vector: �� 10,0 � cos40° � 7,66�� ¢ � 10,0 � sin40° � 6,428�� Neutral axis: tanß � ¡�¡¢ � tan° � 361,67�106241,67�106 � tan40° → ß � 51,47°

Calculation of stresses: � � «Ð¥¤¥ � ¢ « ÐÑ¤Ñ � �

Rules for signs:

�1 � Y 7,66 � 106361,67 � 106 � 190 Y 6,428 � 106241,67 � 106 � 50 � Y5,354� ��2⁄ � Y��:Î

�2 � � 7,66 � 106361,67 � 106 � 110 � 6,428 � 106241,67 � 106 � 150 � �6,32� ��2 � ��:Î⁄

� � 50000��g

¢Æ � 110�� ¡� � 361,67 � 106��4

¡¢ � 241,67 � 106��4

The cross-section’s details:

plane of

the bending


neutral axis

C



8.) Check the structure if δδδδ=45°, αααα=0


b) in plastic stress state by comparing forces and by comparing eccentricity! fd=10,7 N/mm2

Eccentricity:

;¦ � �� 7,071 � 10¸7071 � 1000��

Neutral axis:

¢6 � ¡�� ;¢ � 361,67 � 10650000 � 1000 � 7,233��

a) Elastic stress state

� � Y�� «��¡� � ¢

�6�;9 � Y 707150000 � 7,071 � 106361,67 � 106 � 110 � �2,01� ��2 O⁄ 10,7� ��2⁄

�Ë��;9 � Y 707150000 Y 7,071 � 106361,67 � 106 � 190 � Y3,856� ��2⁄

The cross section is safe in elastic stress state!

� � 50000��g

¢C � 110�� ¡� � 361,67 � 106��4

The cross-section’s details:

neutral

axis

c



b) plastic stress state

Comparing forces: We know ey, and we are looking for NRd=?

1.) ΣMD=0 → a (the neutral axis is in the base of the T-coss-section)

10,7 � 50000 � 1000 � 2 � 10,7 � 300 � a � í1110 � :2î � 0

:22 � 1110 � : Y 83333,3 � 0O :1 � 77,8�� O 10048\;\��68\;[<[<[9<²8\=:2 � 2142��4�967²[6Ë8<67=

2.) ΣFi,x=0 →NRd

�Ô3 � 10,7 � 50000 � 2� 10,7 � 300 � 77,8 � 35524� � 35,52��] 7,071��! Comparing eccentricity:

1.) ΣFi,x=0 →a (the neutral axis is in the base of the T-coss-section)

7071 � 10,7 � 50000 � 2 � 10,7 � 300 � : → : � 82,23��

2.) ΣMC=0 →eyRd

�'�,¹� � 7071 � ;'� � 2 � 10,7 � 300 � 82,23 � 4110 � 82,232 =

;Ô3 � 5142�� ] ;�3 � 1000�� The cross-section is safe!



9.) Check the structure in elastic state if δδδδ=45°, αααα=40°! fd=10,7 N/mm2

The partition of the moment vector:

�� 7,071 � cos40° � 5,417��

�¢ � 7,071 � sin40° � 4,545��

;� ��¢� � 4,5457,071 � 0,6428�

;¢ � �� 5,4177,071 � 0,766�

Neutral axis:

�6 � ¡¢�� ;� � 241,67 � 10650000 � 642,8 � 7,52��

¢6 � ¡�� ;¢ � 361,67 � 10650000 � 766 � 9,443��

The calculation of the stresses:

� � �� «��¡� � ¢ «�¦¡¦ � ¢

�1 � Y7,071 � 10350000 Y 5,417 � 106361,67 � 106 � 190 Y 4,545 � 106241,67 � 106 � 50 � Y3,928 �� 2⁄


neutral axis

� � 50000��g

¢ù � 110�� ¡� � 361,67 � 106��4

¡¢ � 241,67 � 106��4

The details of the cross-section:

c



�2 �Y7,071 � 10350000 � 5,417 � 106361,67 � 106 � 110 � 4,545 � 106241,67 � 106 � 150 � �4,327�� 2⁄



10.) Check the structure


b) in plastic stress state by comparing forces and by comparing eccentricity! fs,y,d=235 N/mm2

A � 30 � 4105 Y 120= � 6 750��

zC � �105 � 30 � 15 Y 30 � 120 � 606750 � 25��

¡� � 105 � 30C3 Y 30 � 230C3 � 6750 ×25 � 14,0×10¸��g

a, Elastic stress state: MEd=7,5 kNm , NEd=+8,66 kN

; � 7,5 � 10¸8,6 � 10C � 866,05��

T¶ � Y8,66 � 10C6750 Y 7,5 � 10¸14 � 10¸ ×55=+1,28+29,46=30,74 . 22j⁄ O ��,�,� � 235 � �� ⁄ ÝÛ··r¶Á!

TW � 8,66 � 10C6750 � 7,5 � 10¸14 � 10¸ � 95=+1,28-50,89= � Sh, 1,. 22j⁄ O ��,�,� � 235 � �� ⁄ ÝÛ··r¶Á!

ú� � 14 � 10¸ � 8,66 � 10C675 � 7,5 � 10¸ � j, )h22



b, plastic stress state

Comparing forces

ΣMD=0

0 � 235 � 6750 � 866,05 � 2 � 235 � 30 � :4866,05 Y :2= : 2 � 961,05: Y 97430,625 � 0: � 107,38�� O 120�� ΣFix= 0

VÊv � 235 � 6750 � 235 � 2 � 30 � 107,38 � /j, jj-. ] 8,66�� ÝÛ··r¶Á! Comparing eccentricity

ΣFix=0

8,66 � 10C � 235 � 6750 � 2 � 235 � 30 � :: � 11,89�� O 120��

ΣMC=0

8,66 � 10C � ;'� � 2 � 235 � 30 � 11,89 � 39,06

ÁÊv � /,,+, )22 ] 866,05�� ÝÛ··r¶Á!



11. Check the structure in elastic stress state! fs,d,y=235 N/mm2

Draw the normal stress diagram!

A � 30 � 4105 Y 120= � 6 750��

zC � �105 � 30 � 15 Y 30 � 120 � 606750 � 25��

¡� � 105 � 30C3 Y 30 � 230C3 � 6750 ×25 � 14,0×10¸��g

¡¦ � 30 � 150C12 Y 120 � 30C12 � 3,16 � 10¸��g

ez=709 mm

ez=500 mm

My=7,5×cos 35°= 6,14 kNm Mz=7,5×sin 35°= 4,3 kNm

Neutral axis:

¢º ��14 � 10¸ � 8,66 � 10C6750 � 6,14 � 10¸ � �2,925��

�º ��3,16 � 10¸ � 8,66 � 10C6750 � 4,3 � 10¸ � �0,943��

T, � 8,66 � 10C6750 Y 6,14 � 10¸14 � 10¸ � 55+4,3 � 10¸3,16 � 10¸ � 52,5= Y h1, 0S. 22j⁄ O ��,�,�� 235 � �� ⁄ Û··r¶Á!

Tj � 8,66 � 10C6750 Y 6,14 � 10¸14 � 10¸ � 95+4,3 � 10¸3,16 � 10¸ � 15= � 1*, /h. 22j⁄ O ��,�,�� 235 � �� ⁄ ÝÛ··r¶Á!


neutral axis

c



12, a, Draw the structure’s N,V,M diagrams!

b, Check the structure in elastic stress state at the „K” cross-section!

c, Check the structure in plastic stress state at the „K” cross-section!

Draw normal stress diagram for b) and c) too! fs,d,y=235 N/mm2

A= 20×(100+140) =4800 mm2

¢Æ � 100 � 160 � 80 � 80 � 140 � 704800 � 103,33��

¡� � 100 � 20C3 Y 20 � 140C3 � 4800 � 36,66� 12,11 � 10¸��g

NEd= 30kN MEd=22,5 kN eEd=750mm

; � ��

b, Elastic stress state:

T¶ � Y30 � 10C4800 � 22,5 � 10¸12,5 � 10¸ ×56,67=+6,25-105,28= � 99,03 . 22j⁄ O ��,�,�� 235 � �� ⁄ ÝÛ··r¶Á! Tr � Y30 � 10C4800 Y 22,5 � 10¸12,5 � 10¸ ×103,33=+6,25+191,99=198,24 . 22j⁄ O ��,�,�� 235 � �� ⁄ ÝÛ··r¶Á! ú* � �12,11 � 10¸4800 � 750 � �), )1S22

c



c, plastic stress state

Comparing eccentricity

ΣFix=0

30 � 10C � 235 � 4800 � 2 � 235 � 420 � 100 Y 20 � :=: � 16,81��

ΣMC=0

MRd,pl=NEd� ;'� � 2 � 235 � 420 � 100 � 46,6 Y 20 � 16,81 � 28,26= �'�,¹��48,27�� ] 22,5�� ÁÊv,Âd � �'�,¹�� ,1,,, ,,22 ] 750�� ÝÛ··r¶Á! Comparing forces:

ΣMD=0

0 � �2 � 235 � 420 � 100 � 796,66 Y 20 � : � í786,6 � :2î Y 235 � 4800 � 750: � 13,25��

ΣFix=0

NRd,pl= 4800� 235 � 2 � 235420 � 100 Y 20 � 13,25= � 1), S0-. ] 30�� ÝÛ··r¶Á!



13, a, Draw the structure’s N,V,M diagrams!

b, Check the structure in elastic stress state at the „K” cross-section!

c, Check the structure in plastic stress state at the „K” cross-section!

Draw normal stress diagram for b) and c) too! fs,d,y=235 N/mm2

A=2×(20×160+10×120)=8800 mm2

¡� � 160g12 � 140 � 120C12 � 34,45 � 10¸��g

� � 30 � 2 8 � 15��

Mmax= 90+15=105 kNm

b ,Elastic stress state: NEd=-30 kN, MEd= 90 kNm ;�� Ð�×��× � 3000��

T¶ � � C��z�À�� z�øCg,»g�z�ø ×80= � 211,86 . 22j⁄ ] ��,�,�200 � �� ⁄ ��Á·ÝÞÚeÝÚÞÁÛ·.��·r¶Á!

Tr � �30 � 10C8800 Y 90 � 10¸34,54 � 10¸ ×80= � 3,41 Y 208,45=j*+, *S. 22j⁄ ] ��,�,� � 200 � �� ⁄

��Á·ÝÞÚeÝÚÞÁÛ·.��·r¶Á! ú* � �34,54 � 10¸8800 � 3000 � �,, )*022

c



c, plastic stress state

Comparing eccentricity

ΣFix=0

30 � 10C � 200 � 8800 � 2 � 200 � 420 � 160 Y 2 � 10 � :=: � 56,25��

ΣMC=0

MRd,pl=NEd� ;'� � 2 � 200 � 420 � 160 � 70 Y 2 � 10 � 56,25 � 31,875=

�'�,¹��103,94�� ] 90�� &ÁÊv,Âd � �'�,¹�� )S1+22 ] 3000�� ÝÛ··r¶Á! Comparing forces:

ΣMD=0

0 � 200 � 8800 � 3000 � 2 � 200 í20 � 160 � 30 � 70 Y 2 � 10 � : � 43060 � :2=î : � 55,67��

ΣFix=0

NRd,pl= 8800� 200 � 2 � 200420 � 160 Y 2 � 10 � 55,67= � )S, 1S-. ] 30�� ÝÛ··r¶Á!



14, Check the given structure in elastic stress state at the cross-section „K”, and draw fully detailed

normal stress diagram! fd=26,7 N/mm2

A= 3002-1502=67500 mm2

7Æ �C��À � 150 � 22567500 � 125��

¡� � 300g12 � 150g12 � 632,81 � 10¸��g

¡¦ � 632,81 � 10¸ Y 300 � 25 � √2® � 150 � 100 � √2®� 295,31 � 10¸

NK= -20 kN MK=42,43×0,5-20×0,5×√j×0,25=17,68 kNm

³q � ³ú � 17,68√2 � ,j, +-.2; � 12,520 � 0,625�

¢º � �632,81 � 10¸67500 � 625 � �15�� º � �295,31 � 10¸67500 � 625 � �7��

�z � �20 � 10C67500 Y 12,5 � 10¸632,81 � 10¸ ×106,07+12,5 � 10¸295,31 � 10¸ ×141,42

T,�7,785 . 22j⁄ O ��,�,� � 26,7� �� ⁄ ÝÛ··r¶Á! Tj � �20 � 10C67500 � 12,5 � 10¸295,31 � 10¸ ×125×√2 � �/, //h. 22j⁄ O ��,�,� � 26,7 � �� ⁄ ÝÛ··r¶Á!

plane of

the bending

neutral axis

c


Shear with bending 67/105

5. SHEAR WITH BENDING

If not only moment works (which case is called pure bending) on the cross-section but shear forces as well (this latter case is called normal bending, or shear with bending), thus we have to know the shear stresses (�) too. Our calculations below shows only the case of linear elastic bending where the shear forces work in the plane ‘y’.

The average value of shear stresses along stripes that are parallel the axis x:

� � P� � �� ¡�

(It is maximal, where S�/b quotient is giving maximum value.)

where S� is the static moment, prescribed to the axis y in the centroid, over or under the examined

part’s one point of the cross-section; „V” is the shear force applied to the cross-section „b” is the cross-section’s working width at the location of the examination; „I�” is the moment of inertia of the whole

cross-section prescribed to the axis y.

In some cases, depending on the cross-section’s profile, the formula might be reducible. The shear

maximum in the case of rectangle and triangle profile is in the median height.

Its value is:

�oW� � 3 � �2 � �

(Although �om�nW� formula at trapezoid profile is the same, in this case this value is few percent less than the actual stress maximum.)

The maximum shear stress at circle profile is also occurs at the middle of the cross-section.

�oW� � 4 � �3 � �

Combined profiles’ maximum stress is at the height of the centroid if the cross-section width is the smallest in this point. Otherwise we can define properly the situation of the maximum stress by the analysis of the stress diagram. (See the below examples)



1. �´v � j*-. shearforce works at the drawn cross-section. Determine the value of �2rØ and draw

the shear stress diagram!

The profile’s details: � � 100 � 250 Y 120 � 300 � 61000�� The centroid:

�Æ � 300 � 120 � 60 � 100 � 250 � 125300 � 120 Y 100 � 250� �15,82��

¡�Æ � ¡�� Æ �

� 13 � 300 � 120C Y 13 � 100 � 250C �

�61000 � 15,82 � 678,4 � 10¸��g

We are able to calculate the shear stresses with this formula:

� � P� � �� ¡�

at three points: 1. At the neck (in the flange) 2. At the neck (in the web) 3. At the height of the centroid

�z � 300 � 120 � 75,82 � 20 � 10C300 � 678,4 � 10¸ � *, j10./22j

�z � 300 � 120 � 75,82 � 20 � 10C100 � 678,4 � 10¸ � *, 0*+./22j

�oW�,C � 100 � 234,18 � 117,09 � 20 � 10C100 � 678,4 � 10¸ � *, 0*h./22j

c



2. How much �´v shear force might bear the cross-section if the structure is from II. class pine?

(¶�,v � j, +./22j)

Initially draw the diagram of the shear stresses! Since τ�|õ � f�,Ï � 2,5N/mm , we are able to

determine the shear force from the formula:

��,� � P� � �'�� ¡� 8\;7�'� � �� ¡�P�

However, we don’t know yet where the maximum shear

stress will be in the cross-section.

Let’s complete the two half triangle into a profile (see the dashed triangles) which centroid match with the cross-section’s (the rectangle’s) centroid. The shear diagram of these triangles are two congruent parabola which maximum point are in the

half of the triangles’ height (we marked it above and below by Wg,

and CWg distance away from the edges.)

34 � : � 34 � 160√2 � 84,85��

� � 2 � 34 � : � 169,7��

¡� � 112 � 160g � 54,6 � 10¸��g

P� � 84,85 � ñ80 � √2 � 23 � 84,85ò � 407280��C

�� 2,5 � 169,7 � 54,6 � 10¸407280 � 56,875��

c



Let’s define the value of the shear stress in the centroid!

P� � 480 � √2= � 80 � √23 � 482718��C

�Æ � 160 � √2 � 226,3��

� � 56,875��

¡� � 54,6 � 10¸��g

�Æ � 482718 � 56875226,3 � 54,6 � 10¸ � 2,222�/��



3. What maximal force F might be applied on the bent structure (Fmax =?) (do the calculation

assuming elastic state)? What maximal value of the rupture “a” might be tolerated?

��,� � 2,6�/��

�� 13�/��

Since the crack is at the height of the neutral axis, it is not have role in the calculation of the bending

moment because in this height σ � 0.

�oW� � 2& � 2,4 � F � 1,2 � 3,6F

�'�,m� � �� W� � �� b � h 6 � 13 � 120 � 200 6 � 10B¸ � 10,4��

As 3,6 � & � 10,4��, the & � 10,4 3,6� � 2,89��

Determination of ‘a’:

We can determine the value of ‘a’ from the criteria that the rest of the neck-width have to assure the collaboration of the upper and lower structure part during bending. So the neck have to bear the shear stresses.

The shear stress:

� � � � 5,78��,� � 120 � 2::73� � ��

2,6 � 120 � 100 � 50 � 5,78 � 10C4120 � 2:= � z �� Àz

� 43,35120 � 2:

: � 51,7�� 50�



4. Check the structure to bending and shearing assuming linear elastic stress state!

The structure is from second class pine.

The wedge is from first class hardwood.

The static details of the structure:

�oW� � �� 2 � 8 � 16��

�oW� � 8 � 4 8 � 16��

The following examinations have to be executed:

a. Check at place of ³2rØ for bending.

b. Check the compression on the contact surfaces in the

beam at the place of slide force (shear force) ´2rØ.

c. Check the shearing in the beam at ´2rØ .

d. Check for shearing in the wedge at ´2rØ .

a. �� 14�/��

The calculation became unusual here than the usual type. By the scaling standards in the case of the structures joint by two wood we have to take into account a reduction factor, however supposing that

the structure is rigid, we are able to neglect it.

�'�,m� � �� W� � 14 � 200 � 240 6 � 10¸ � 26,88�� ] 16��&! b. The wedges prevent the slip from each other of the two halves structures. So slide forces (shear

forces) will get on the structures. E�|õ works in the terminal wedges. A half wedge is receiving a sliding force as high as the total amount slipping stresses on the 640+100=800mm long and 200 mm wide area to the wedge inner end. Due to the collaboration of the two part we are able to determine the shear

stresses considering the whole cross-section.



The segment which loads the terminal wedges:

�z � 1,5 � �z� � 1,5 � 16000200 � 240 � 0,5�/��

� � 1,5 � � � � 1,5 � 416000 � 0,84 � 8000=200 � 240 � 0,29�/��

�oW� � � � 8 � �W�mpW�m � 200 � 4640 Y 200= � 0,5 Y 0,292 � 66360� � 11, )1-.

We can neglect the shear stresses in the plane of the cross section because of the low value of the τ stresses. The above determined E�|õ handed over to the wedges from the beams by the wedges’ front

side. In the beam the reaction of E�|õ causes shear and the compression on the contact surfaces.

The working surface from the pressure side:

& � 200 � 602 � 6000��

��,�,� � 13,15�/��

�'� � 6000 � 13,15 � 10BC � /0, h-. ] �oW� ¡8<[[:�;! c. The beam section resisting to the shear is:

& � 640 � 200 � 128000��

��,� � 2,5�/��

�oW� � 128000 � 2,5 � 10BC � )j*-. ] �oW�P:�;! d. The wedge section resisting to the shear:

& � 200 � 200 � 40000��

��,� � 2,8�/��

�oW� � 40000 � 2,8 � 10BC � ,,j-. ] �oW� ¡8<[[:�;!



5. a) Design the structure’s web width ‘a’ on bending at elastic stress state! Round the

dimension to 5mm!

b.) Draw highly detailed shear stress diagram! fd=21 N/mm2, fv,d=3,0 N/mm2

a.)

¡� � 2 � ï150 � 50C12 Y 150 � 50 � 100 ð Y : � 150C12� 4153,125 Y 0,28125:= � 10¸

�'�,m� � �� ¡�¢ ¡� � ÐÖ×,!"#× � ¢

¡� � 28 � 10¸21 � 125 � 4153,125 Y 0,28125:= � 10¸: � 48,15��sorrÂÂdÛrtdÁ � +*22

b.)

¡� � 4153,125 Y 0,28125 � 50= � 10¸ � 167,1875 � 10¸��g

�z � 150 � 50 � 100 � 28 � 10C167,2 � 10¸ � 150 � 0,84374�/��

� � 3 � �z � 2,51�/��

�) � í150 � z »ü � 100 � $»ü î � 28 � 10¸167,2 � 10¸ � 50 � j, h0) .22j O ��,� � 3,0 �� Á·ÝÞÚeÝÚÞÁÛ··r¶Á!



6. a) Draw the structure’s N,V,M diagrams!

b) Draw highly detailed normal stress diagram at the place of ±M max!

c) Determine the maximum value of shear stress and draw it’s correct figured diagram!

d) Check the glued joint too!

Do the calculation in elastic stress state! fd= 18,7 N/mm2, fv,d=1,9 N/mm2, Glue: fv,d=3,0 N/mm2

A=50×(160+240+2×80)=28000 mm2

¢Æ � 240 � 50 � 25 Y 80 � 100 � 100 Y 160 � 50 � 17528000 � 89,29��

¡� � 240 � 50C3 Y 160 � 150C3 � 80 � 100C3 � 28000 � 39,29 � 120,11 � 10¸��g

glue

c



b) +Mmax=21 kNm

T¶ � � 21 � 10¸120,11 � 10¸ � 110,71 � �19,36 . 22j⁄ ] �� 18,7 � �� ⁄ ��Á·ÝÞÚeÝÚÞÁÛ·.��·r¶Á! Tr � 21 � 10¸120,11 � 10¸ � 89,29 � ,+, 1,. 22j⁄

-Mmax=20kNm

T¶ � ��z�øz �,zz�z�ø � 110,71 � ,0, S). 22j⁄ s

Tr � � ��z�øz �,zz�z�ø � 89,29 � �14,87 . 22j⁄



c)

�2rØ � j*-.

�2rØ � í160 � 50 � 85,71 Y 80 � ¸�,$zü î � 20 � 10C80 � 120,11 � 10¸ � ,, /). 22j⁄ O 1,9 �� ÝÛ··r¶Á!

d)

�, � 160 � 50 � 85,71 � 20 � 10C80 � 120,11 � 10¸ � ,, Sj/./22j

�j � 240 � 50 � 64,29 � 20 � 10C80 � 120,11 � 10¸ � ,, 1*1. 22j⁄ O 3,0� �� ⁄ ÝÛ··r¶Á!



7., a) Draw highly detailed normal stress diagram at Mmax supposing linear elastic stress state!

b) Check the structure in plastic stress state!

c) Determine the elongation of the hanged beam!

d) Draw highly detailed shear stress diagram and check the glue joint!

beams AD and EC:

fd=16,6 N/mm2

fv,d=1,9 N/mm2

hanged beam DE:

fs,y,d=235 N/mm2

E=210000N/mm2

Glue:

fv,d=3,0 N/mm2 ¡� � 829,56 � 10¸��g

A=65400 mm2

a) +Mmax= 76 kNm

T¶ � � 76829,56×143,59= � 13,15 . 22j⁄

Tr � Y 76829,56×186,41=17,08 . 22j ]⁄ 16,6 . 22j⁄ ��Á·ÝÞÚeÝÚÞÁÛ·.��·r¶Á! -Mmax= 42,5 kNm

T¶ � � 42,5829,56×143,59=7,356 . 22j⁄

Tr � � g ,»� �,»¸ ×186,41= � 9,51 . 22j⁄

glue

c



b) A/2= 65400/2 =32700 mm2

32700=380×80+2×50×a a=23 mm

�'�,¹� � 16,6 � 2 � 4380 � 80 � 103,59 Y 2 � 50 � 23 � 52,09= � 108,53 � 10¸

³Êv,Âd � ,*0, +)-.2 ] 76��&

c)

∆d � 38 � 10C � 2 � 10C210000 � z�ü�¼g

� ,, Sjj22

d)

Vmax= 47 kN

�, � 380 � 80 � 103,59 � 47 � 10C380 � 829,56 � 10¸ � *, S1h+./22j

�j � 380 � 80 � 103,59 � 47 � 10C100 � 829,56 � 10¸ � ,, /0S./22j

�2rØ � 4300 � 186,41 � 200 � 136,41 =100 � 829,56 � 10¸ � ,, 0hh. 22j⁄ O 1,9. 22j⁄ ÝÛ·Ér¶Á! �, � 300 � 50 � 161,41 � 47 � 10C100 � 829,56 � 10¸ � ,, )/j./22j

�, � 300 � 50 � 161,41 � 47 � 10C300 � 829,56 � 10¸ � *, S+/j./22j



8., a) How long and where do we need to reinforce the T shaped cross-section? Suppose elastic stress

state during the calculation! Draw normal stress diagram for the T and I cross-section too!

b) Determine the maximum value of the shear stress at the T and I cross-section!

Draw shear stress diagrams!

c) What reinforcement length do we need, if we suppose elastic stress state?

fs,y,d= 235 N/mm2 fv,d=115 N/mm2

a) In plastic stress state:

„T”

¢Æ � �120 � 20 � 10 Y 20 � 160 � 8020 � 4120 Y 160= � 41,43��

q� � 120 � 20C3 Y 20 � 160C3 � 5600 � 41,43 � ,0, *j � ,*122S

³Êv,Ád� � 235 � 18,02 � 10¸118,57 ×10B¸=)+, /,-.2 O ��=80 k�� ÝuÁÁv·ÞÁÛu¶�ÞeÁ2ÁuÝ! ³Êv,Ád � 235 � 45,86 � 10¸100 ×10B¸=,*/, /h-.2 ] 80 k.2 ÝÛ··r¶Á! q � 120 � 200C12 � 100 � 160C12 � S+, 011 � ,*122S

2-x=35,71/40 x=1,11 m l=2×1,11=2,22 m

reinforcement

cross-section „I” cross-section „T”



b)

�2rØ� � 40 � 10C � 20 � zz�,»$ü 18,02 � 10¸ � 20 � ,+, 1./22j

�2rØ � 4120 � 20 � 90 Y 20 � 80 � 40= � 40 � 10C45,866 � 10¸ � 20 � ,j, j,./22j

c) In elastic stress state

³Êv,Âd � 235 � 2 � 20 � 140 � 48,57 � 10¸ � 1), hj-.2

2-x=63,92/40 x=0,40 m l=2×0,4=0,8 m



9., a) Check the cross-sections K1, K2 and K3 in elastic stress state!

b) Check the cross-sections K1, K2 and K3 in plastic stress state!

c) Check the glued joint on beam AC!

Draw correctly figured stress diagrams for all exercises!

d) Determine the elongation of the beam DE!

Wood:

fd=26 N/mm2

fv,d=3,1 N/mm2

E=14700N/mm2

Glue:

fv,d=3,0 N/mm2

glue

glue

bars AD and BD:

„K1” and „K2”:

bar DE:

c



beam AC:

A=100×(300+400)= 70 000 mm2

¢Æ � �300 � 100 � 50 Y 100 � 400 � 20070000 � 92,86��

¡� � 300 � 100C3 Y 100 � 400C3 � 70000 � 92,86 �� 1629,76 � 10¸��g

a) checking cross-section „K2”

N=-76,36 kN M=162 kNm

T¶ � �76,36 � 10C70000 � 162 � 10¸1629,76 � 10¸ ×192,86=

�� 20,26 . 22j⁄

Tr � �76,36 � 10C70000 Y 162 � 10¸1629,76 � 10¸ ×307,14=

��29,44 . 22j⁄⁄⁄⁄ ] ¶v � 26 . 22j⁄⁄⁄⁄ ÝÛ·.��·r¶Á! checking cross-section „K1”:

N= -25,45 kN M=162 kNm

T¶ � �25,45 � 10C70000 � 162 � 10¸1629,76 � 10¸ ×192,86= � ,9,53 . 22j⁄

Tr � �25,45 � 10C70000 Y 162 � 10¸1629,76 � 10¸ ×307,14=30,17 . 22j⁄ ] ¶v � 26 . 22j⁄ ÝÛ·.��·r¶Á! checking cross-section „K3”:

N= +72 kN M=0

NRd,el=NRd,pl=26×2×50×100×10-3=260kN > 72 kN It is safe!

b) checking cross-sections „K1” and„K2”:

M= 162 kNm NRd,pl=?

ΣMC=0

162 � 10¸ � 2 � 26 � 100 � : � í307,14 � W î : � 128,18��



ΣFix=0

.Êv,Âd � 426 � 70000 � 2 � 26 � 100 � 128,18= � 10BC � ,,+), +-. ] 76,36�� ÝÛ··r¶Á!

c)

�ÜdÚÁ � 50,91 � 10C � 4300 � 100 � 142,86=100 � 1629,76 � 10¸ � ,, ))h. 22j⁄ O 3,0. 22j⁄ ÝÛ··r¶Á!

d)

∆d � 72 � 10C � 6 � 10C14700 � 2 � 50 � 100 � j, h)h22


Eccentric compression testing of materials without tension 85/105

6. ECCENTRIC COMPRESSION TESTING OF MATERIALS WITHOUT TENSION

Equilibrium occurs in only those cases, where the eccentric force (compression) doesn’t affect out of the cross-section’s convex contour. Other words, the point of action is inside the cross-section’s convex

contour.

Before the stress examinations, let’s have a look at the core of section.

The core of section

The boundary of the core of section is that locus of points of actions, where the force effect, and its neutral axis “touches” the convex contour of the cross-section (in other words: the area of bearing). We can determine the corner points of the core of section, like the points of actions, which neutral axis are the cross-section’s convex contour boundary lines. The following formulas are the basis of the calculations:

�¦,o � � ¡�� ¢o � � <� ¢o

��,o � � ¡¦� � �o � � <¦ �o

Since in these formulas the �¦,o and z�, and the ��,o and �o might be commutable respectively, we

are able to proceed like that we consider that the neutral axis are the sides of the core of section: In this case the cross-section’s corner points will be known as point of actions.

Determination in elastic stress state:

First of all, we have to examine whether there is a tensile strength in the cross-section or not. This will only be interesting if the compression force’s point of action is outside the core of section. In the case when the point of action is in the core, the whole cross-section has compression stress, there is not any difference between the cross sections with tensile strength, and the cross sections without tensile

strength. Consequently, the known formulas will be the same in this case.

If the point of action is out of the core, a part of the cross-section does not work, and the other part of

it equilibrates the external force with only compression stresses. The general formulas are the following:

� � &P� � ¢

�oW� � &P� � ¢oW�

Here the F is the force, P� is the static moment of the compressed cross-section part prescribed to the border line of the two different areas (working and not working area), z is the perpendicular distance of

the examined point from the mentioned border line.

� � ¡�P�



Here the p is the (perpendicular) distance of point of action and the border line, ¡� is the moment of inertia of the compressed area prescribed to the given point, P� is the static moment of the mentioned

area prescribed to the border line.

We would like to draw your attention that in these formulas, all of the data (except the force F) are

given to the same axis, which is called now the border line.

In the case of rectangular and triangular cross-section segment, the formulas simplify. If “c” is the distance of the point of action and the cross-section’s working border line (or in triangular case it is one of the peaks), then:

9;58:7²Ë:9[;²�;78:� � 25�oW� � 2&��

89<:7²Ë:9[;²�;78:� � 5�oW� � 3&��

The stress diagram is situated on the one side of the border line, with the base line perpendicular to the border line.

Determination (calculation) in plastic stress state:

The solution is very simple. Since: � � �� (and it is compression stress everywhere.), we face with axial compression. We have to determine A% that the given point of action shouldbe the centroid of the compressed cross-section. (Sometimes this might be determined in numerous ways.). After all &�|õ can be determined from the following formula:

&oW� � ��



1. Determine the core of section!

First of all, calculate the geometrical properties:

¢Æ � 900 � 300 � 150 � 300 � 300 � 150900 � 300 Y 300 � 300 � 40500000 � 13500000270000 Y 90000 � 27000000360000 � 75��

¡� � 900 � 300C3 Y 300 � 300C3 � 360000 � 75 � 8775 � 10¸��g

¡¦ � 900 � 300C12 Y 300 � 300C12 � 18900 � 10¸��g

<� � 8775 � 10¸360000 � 24375��

<¦ � 18900 � 10¸360000 � 52500��

Considering the lines that are covering the cross-section as neutral axis we determine the regarding point of actions.

��,z � *�¦,z � �24375225 � �,*0, )22

��, � �52500450 � �,,1, /22�¦, � *

��,C � �52500525 � �,**22�¦,C � �24375525 � �S1, S22

��,g � *�¦,g � �24375375 � �1+22

C



Then we get the core of section by joining all of the obtained point of action.

C



2. What might be the value of V2rØ at point ‘D’

a. in elastic stress state,

b. in plastic stress state?

Draw stress diagrams too! ¶v � ,, +./22j

a. In elastic stress state:

In case of the right-angled square cross-section the distance of the core’s corner points are the one-sixth of the given side from the centroid.

36 � 6006 � 100�� O 150��

So in this example the point of action is outside the core.

5 � 300 � 150 � 150��

35 � 450��

&oW� � �� b � 3c

&oW� � 3 � 150 � 600 � 1,52 � 202,5 � 10C�� 202,5��

b. In plastic stress state

&oW� � �� b � 2c

&oW� � 2 � 150 � 600 � 1,5 � 270 � 10C�� j/*-.

C



3. What might be the value of V2rØ which works in the point ‘D’

a. in elastic stress state,

b. in plastic stress state?

Draw stress diagrams too! ¶v � ,, +./22j

a. In elastic stress state

50 � ¸��¸ ,sothepointofactionisinsidethecore. � � 600 � 600 � 360000��

¡¦ � 600 � 600C12 � 10800 � 10¸��g

£¦ � z��z�øC�� 36 � 10¸��g

�1,5 � � &oW�360000 � &oW� � 5036 � 10¸

54 � 10¸ � 100 � &oW� Y 50 � &oW�� 150 � &oW�

&oW� � 54 � 10¸150 � 360 � 10C� � 360��

b. In plastic stress state

&oW� � 2 � 250 � 600 � 1,5 � 450 � 10C�� 450��

5 � 300 � 50 � 250��

C



4. Check the brick column at cross-sections 1-1 and 2-2 a. in elastic stress stat,.

b. in plastic stress state! �& � 16,5��/�C

�' � 1,35

�� 1,2�/��

Cross-section 1-1

�zÐ � 1,35 � 0,51 � 0,51 � 1,5 � 16,5 � 8,69��

&zÐ � 100��

�zÐ � 100 Y 8,69 � 108,69��

5 � 510 � 255 � 140 � 115��

a, In elastic stress state:

35 � 3 � 115 � 345�� O 370��

or

3706 � 61,7 O 185 � 115 � 70��

so the point of action is outside the core

�oW� � 2 � 108,69 � 10C3 � 115 � 510 � 1,24 �� ] 1,2 �� ¡8<[�()[:�;! b, In plastic stress state

25 � 2 � 115 � 230��

�¤oW� � 230 � 510 � 1,2 � 140,8 � 1000� � 140,8�� ]108,69��¡8<[P�&�! Cross-section 2-2

� Ð � 1,35 � 0,51 � 0,51 � 2,0 � 16,5 � 11,59��

&zÐ � 100��

& Ð � 25��

� Ð � 100 Y 25 Y 11,59 � 136,59��

C C’

C



;� � 25 � 10C � 175136,59 � 10C � 32,03��

5 � 255 � 32,03 � 222,97��

a, In elastic stress state

35 � 3 � 222,97 � 668,91�� ] 510��[6�6<78"D" <[<7[<3;8\;569;! £ �510C6 � 22110 � 10C��C

�oW� � �136,59 � 10C510 � 510 �136,59 � 10C � 32,0322,11 � 10¸ � �0,723 �� O �1,2 �� ¡8<[P�&�! b, In plastic stress state

25 � 2 � 222,97 � 445,94��

� � � 445,94 � 510 � 1,2 � 273 � 1000� � 273�� ] 136,59��¡8<[P�&�!



5. What forces �´v and .´v are acting on the column, if the stress diagram is like the drawing

below? �� 1,0�/��

¡� � 900 � 650C3 � 400g3 � 73854,7 � 10¸��g

P� � 900 � 650 � 325 � 400 � 200 � 155,2 � 10¸��g

1,0 � �Ð155,2 � 10¸ � 650

�Ð � 155,2 � 10¸ � 1,0650 � 238,8 � 10C� � 238,8��

� � 73854,7 � 10¸155,2 � 10¸ � 475,9��

The moment of the resultant force is equal to the total

moment of the components.

Let’s write the equation to the whole cross-section’s centroid:

�� 5000 � 238,8 � 10C � 4475,9 � 200= �� 238,9 � 10C � 275,95000 � 13182� � 13,2��



6. a) Determine the distance “a” for bending in elastic stress state! Round the measurement to 5

mm!

b) Draw highly detailed shear stress diagram!

�� 21� �� ⁄

��,� � 2,6� �� ⁄

In elastic stress state the stress at the terminal points:

�� ¡� � ¢

For this the unknown inertia is:

¡� � �� ¢ � 29,25 � 10¸21 � 130 � 181,07 � 10¸��g

The inertia depends on the wanted “a”. The inertia prescribed based on the “a”:

¡� � 2 � ï: � 50C12 Y : � 50 � 105 ð Y 50 � 160C12 � 1,123 � 10¸ � : Y 17,07 � 10¸

The wanted “a” is able to determined from the two equations. The higher the inertia the lower the stress, so we have to round “a” up.

: � 181,07 � 10¸ � 17,07 � 10¸1,123 � 10¸

: � 146�� → r � ,+*22

If : � 150mm, then the inertia is:

¡� � 1,123 � 10¸ � 150 Y 17,07 � 10¸ � 185,52 � 10¸��g



�z � 150 � 50 � 105 � 15 � 10C185,52 � 10¸ � 150 � *, SjS+. 22j⁄

� � 3 � �z � 3 � 0,4245 � ,, j/). 22j⁄

�oW� � 4150 � 50 � 105 Y 80 � 50 � 40= � 15 � 10C185,56 � 10¸ � 50 � ,, +)j. 22j⁄



7. Determine the core of section! Based on the stress diagrams determine the measure and the place

of the normal force!

�� 1,0� �� ⁄

The details of the cross-section:

� � 4 � 510 � 1040400��

�Æ �B»z�À Y C�»z�À 4 � 510 � 127,5��

¡¦ � 510g3 Y 3 � 510g3 � 4 � 510 � 127,5 � 73289,7 � 10¸��g

¡� �510 � 1530C12 Y 510 � 510C12 � 157854,7 � 10¸��g

<¦ � ¡¦� � 73289,7 � 10¸1040400 � 70443,8��

<� � ¡�� 157854,7 � 10¸1040400 � 151725��

Core of section determination:

��,z � � <¦ �z � � 73289,7 � 10¸1040400 � 382,5 � �184,17��

��, � � <¦ � � � 73289,7 � 10¸1040400 � 4637,5 Y 255= � �78,93��

�¦, � � <� ¢ � � 157854,7 � 10¸1040400 � 4765 Y 127,5= � �170,0��

��,C � � <¦ �C � � 73289,7 � 10¸1040400 � 637,5 � �110,5��

�¦,g � � <� ¢g � �157854,7 � 10¸1040400 � 765 � �198,33��

C



a) By the stress diagram the cross-section is in elastic stress state. The whole cross-section is being pressed, so the point

of action is inside the core.

The srtess in the centroid:

�Æ � B�,�z� � � 637,5 � 0,1 � �0,6625� �� ⁄

Since the stress in the centroid is the quotient of the normal force and the whole area, we are able to determine the

normal force like:

� � �Æ � � � �0,6625 � 1040400 � 10BC � 10h, j1+-.

The stress in the external line based on eccentric

compression:

�� ;�¡¦ � ¢

�1,0 � �0,6625 �689,265 � 10C � ;�73289,7 � 10¸ � 382,5

Áq � h), 0j22

b) When the point of action is outside the core, the cross-section is cracked. The normal force can be determined by the stresses’ resultant force. The volume of the stress block

is:

� � 1530 � 420 � 4�1,0= � 10C2 � �)j,, )-.

The normal force acts in the centroid of the stress block. It’s

centroid distance from the edge of the cross-section is:

5 � 13 � 420 � 140��

;� � 382,5 � 140 � jSj, +22

C



c) The point of action is outside the core and the cross-section is cracked. The difference from the above is that the base of the stress block is not a rectangular. To determine the normal force’s point of action we

have to use the following formula:

� � ¡�P�

Where the ¡� is the inertia of the compressed cross-section part to the border line, P� is the static moment of the compressed cross-section part to the border line, � is the distance of the point of action and the

border line.

¡� � 2 � 510 � 90C3 Y 510 � 600C3 � 36967,86 � 10¸��g

P� � 2 � 510 � 90 2 Y 510 � 600 2 � 95,931 � 10¸��C

� � ¡�P� � 36967,8695,931 � 385,36��

;� � 637,5 � 600 Y 385,86 � Sjj, 0122

� � P� � �� 95,931 � 10¸ � 4�1,0=600 � 10BC � �,+h, 00+-.

d) The cross-section is in plastic stress state: Because of the stress block is a prism, to the determination of its centroid we have to make use of the know formulas from the determination of the centroid with

compound cross-sections. The normal force is equal to the volume of the stress block.

� � �� 4�1,0= � 397800 � 10C � �)h/, 0-.

�� 510 � 510 Y 4510 � 420= � 1530 � 397800��

� � P�� 95,931 � 10¸397800 � 241,15��

;� � 637,5 � 4600 � 241,15= � j/0, 1+22 ] ��,z The point of action is outside the core of

section.

e) The cross-section is in plastic stress state. We can determine the normal force from the determination

of the stress block’s volume.

� � �� 420 � 1530 � 4�1,0= � 10C � �1Sj, 1-.

The normal force acts at the cross-section’s centroid. While the stress block is a cuboid, its distance from

the cross-section edge is:

4202 � 210��

;� � 382,5 � 210 � ,/j, +22 ] ��,C The point of action is outside the core of section.



8. a) Where and how long do we have to reinforce the T shaped cross-section? Do the determination

assuming elastic stress state! Draw normal stress diagram to the cross-sections T and I!

b) Determine the maximum shear stress at cross-section T and I! Draw fully detailed shear stress

diagrams!

c) Where and how long do we have to reinforce the cross section assuming plastic stress state?

�� 21� �� ⁄

��,� � 2,6� �� ⁄

Cross-section’s details:

�* � 50 � 4150 Y 160=� 15500��

¢Æ* � �50 � 150 � 25 Y 50 � 160 � 8015500 � 29,19��

¡�* � 150 � 50C3 Y 50 � 160C3 � 15500 � 29,19 � 61,31 � 10¸��g

¡�¤ � 150 � 260C12 � 100 � 160C12 � 185,57 � 10¸��g

a) The load bearing capacity of the cross-section without reinforcement:

�'�,m�* � 21 � 61,31 � 10¸ � 10B¸130,81 � h, 0S-.2 O 15�� → thestructurehastobereinforced!

reinforcement



The reinforced cross-section’s load bearing capacity:

�'�,m�¤ � 21 � 185,57 � 10¸130 � jh, h0-.2 ] 15�� → (+! We have to reinforce the structure at that section where the moment is higher than 9,84 kNm.

Î � 2,0 � 9,847,5 � 0,688� → /*e2

b)

�oW�* � 50 � zC�,�zü � 7,5 � 10C61,31 � 10¸ � 50 � ,, *S/. 22j⁄

�oW�¤ � 4150 � zC�ü � 100 � ��ü = � 7,5 � 10C185,57 � 10¸ � 50 � 1,5322 � *, /11. 22j⁄

c)

The border line is a bisector line of the cross section’s area.

: � z»»�� »� � 155�� → the neutral axis is at the edge of the cross-section.

The load-bearing capacity of the cross-section without reinforcement:

�'�,¹�* � 21 � 2 � 50 � 155 � 53,31 � 10B¸ � ,/, )+-.2 ] 15��→ Itdoesnotneedreinforcement.

plastic

elastic elastic



9., Determine the value of qmax acting on the brick column



Draw fully detailed stress diagram to a) and b) too!

fd=1,1 N/mm2

a) At elastic stress state:

Assume that “D” is inside the core.

�1,1 � �281,28 � 10C510 � 1000 � �z��»z�ü¸

M=23,7762×106Nmm

Á � �� 23,7762 � 10C281,28 � 0S, +)22O 5106 � 85��

The assumption was right!

,2rØ � �0,72 � 23,770,72 � )), *j-. 2⁄

b) In elastic stress state:

281,28×103=100×2c×1,1

c=127,85 mm

M=281,28×103×(510/2 -127,85)

M=35,76×106Nmm

,2rØ � Ð�,$ � C»,$¸�,$ � Sh, 1/-./2



10.,

a) Determine the normal force (its place and value) based on the normal stress diagrams!

b) Draw the approximate shape of the core of the cross-section (the coordinates of the core’s points

do not have to be calculated)!

fd=1,2 N/mm2

a)

¡¦ � »��g��Àz Y 2 � z �� »�Àz � 2763,5 � 10¸��g

A=250×490+2×120×250=182500 mm2

σcentroid=-0,7 N/mm2

N=-0,7×182500×10-3=-127,75 kN

�1,2 � �0,7 � 127,75 � 10C � ;�2763,5 � 10¸ � 245

e=44,15 mm

b)

. � �1,2 � 120 � 2502 � 10BC � �,0-.

e=245-40=205 mm

c)

¡� � 250 � 370C3 Y 2 � 120 � 250C3 � 5471,1 � 10¸��g

P� � 250 � 370 2 Y 2 � 120 � 250 2 � 24,6125 � 10¸��C

�� ¡��P�� 5471,1 � 10¸24,61 � 10¸ � 222,29��

e=222,29-125=97,29 mm

. � �� P��oW� � �1,2 � 24,6125 � 10¸370 � 10BC � �/h, 0j-.

d)

N=-1,2×(250×370+2×120×250)×10-3=-183 kN

� � P�� 24,6125 � 10¸152500 � 161,4��

e=161,4-125=36,4 mm

e)

N=-1,2×120×250×10-3=-36 kN

e=125+60=185 mm

C



11., Determine the compression force acting at the centroid of the brick column

a) at point „D1”, in elastic stress state,

b) at point „D2” in plastic stress state!

Draw fully detailed stress diagram to a) and b) too!

fd=1,2 N/mm2

a) elastic stress state

D1 is outside the core

.2rØ � �1,2 � 400 � 3 � 1002 � �72000� � �/j-.

b) plastic stress state

D2 is the centroid of the stress block

600 � 300 � 150 � 600 � 100 � 50 Y 400 � : � í100 Y :2î

Wü Y 100: � 60000 � 0 260,56 mm

:z, � �100 « √100 Y 2 � 60000 � -460,56 (False solution)

.2rØ � �1,2 � 4600 � 400 Y 400 � 260,56= � �413066� � �S,), ,-.



12., Determine the value of Nmax acting on the brick column

a) assuming elastic stress state,

b) assuming plastic stress state!

fd=1,1 N/mm2

Mmax=4,5×1×0,5=2,25kNm

a) Elastic stress state:

�1,1 � �� 10C510 � 2,25 � 10¸»z�À¸

.2rØ � �j+h, 1S-.

c) Plastic stress state

�� 10C � �1,1 � 510 � 2 � ï255 � 2,25 � 10¸� � 10C ð

Y z � � 143,055� Y 1262,25 � 0 277 kN = Nmax

�z, � 143,055 « Ù143,055 � 2 � 1262,25 � 9,11 kN

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