T2-Stresses in Beams solutions 2013.pdf

Solutions To : Tutorial 2 Q1

(a)

For the straight strip of the above dimensions,

4943333 m10512

1m1051040

12

1

12

1 bhI z

Pa10210GPa210 9E

Now, recall that R

EIM z

xz , yI

My

I

My

R

E

z

xzxx

Since m5.2R ,

MPa210Pa10210m5.2

m105.2Pa10210

Nm0.35m5.2

m10512

1Pa10210

639

max

499

xx

xx

zxz

yR

E

R

EIM

y

z

40 mm

5 mm

(b) If the stress is not to exceed 300 MPa,

m75.1

Pa10300

m105.2Pa10210

Pa10300m105.2Pa10210

Pa10300

6

39

639

6

max

R

R

yR

Exx

Hence, the smallest diameter the cask can have if the stress is not to exceed

300 MPa is 3.5 m.

(c) If the material of the strip is brass instead of steel, then

MPa100Pa10100m5.2

m105.2Pa10100

Nm67.16m5.2

m10512

1Pa10100

639

max

499

xx

xx

zxz

yR

E

R

EIM

(d) If the hoop is split at a point by sawing through it, what will its final

shape be?

Straight (elastic unloading).


47

4343

44

m105.664

1020103064

64

dDI z

Since the weight of the national service man is N81.960 , the loading on

the tubular aluminium bar is:

Using singularity functions,

1111

0000

1111

4.13.2941.13.2943.03.2943.294

4.13.2941.13.2943.03.2943.294

4.13.2941.13.2943.03.2943.294

xxxxxM

xxxxxF

xxxxxq

xz

xy

0.3 m 0.3 m 0.8 m

0.3 m 0.3 m 0.8 m

d = 20 mm

D = 30 mm

Since 11114.13.2941.13.2943.03.2943.294 xxxxxM xz

Hence, we have

m4.1m1.1Nm4.13.294

m1.1m3.0Nm29.88

m3.0m0.0Nm3.294

m4.1m1.1Nm4.13.294

m1.1m3.0Nm3.03.294

m3.0m0.0Nm3.294

m4.1m1.1Nm1.13.2943.03.2943.294

m1.1m3.0Nm3.03.2943.294

m3.0m0.0Nm3.294

xx

x

xx

xx

x

xx

xxxx

xxx

xx

xM xz

The bending moment diagram:

(a) Now, recall that yI

M

z

xzxx

, thus

2

max

max

D

I

M

z

xz

xx

MPa5.41Pa10506.41

2

m1030

m105.664

Nm29.88 63

47max

xx

Hence the maximum stress induced in the bar is 41.5 MPa.

Mxz

x

88.29Nm

0.3 1.1 1.4 0.0

To calculate the maximum shear stress:

Shear force,

Shear stress,

We have:

( )

(

)

( )

( )

Hence,

(b) The portion of the bar between his hands will deflect into a circular

shape (because bending moment is constant).

(c) If the National Serviceman releases one arm and continues to do

single-arm chin-ups, the loading on the tubular bar becomes:

1.1 m 0.3 m

R1 = 126.1 N R2 = 462.5 N

Taking moments,

01.181.9604.1

03.081.9604.1

2

1

R

R

N5.462N6.5884.1

1.1

N1.126N6.5884.1

3.0

2

1

R

R


111

000

111

4.15.4621.16.5881.126

4.15.4621.16.5881.126

4.15.4621.16.5881.126

xxxxM

xxxxF

xxxxq

xz

xy

Since 1114.15.4621.16.5881.126 xxxxM xz

Hence, we have

m4.1m1.1Nm5.46246.647

m1.1m0.0Nm1.126

m4.1m1.1Nm1.16.5881.126

m1.1m0.0Nm1.126

xx

xx

xxx

xxxM xz


Mxz

x

138.71Nm

1.1 1.4 0.0

Now, recall that yI

M

z

xzxx

, thus

2

max

max

D

I

M

z

xz

xx

MPa2.65Pa10210.65

2

m1030

m105.664

Nm71.138 63

47max

xx

Hence the maximum stress induced in the bar is increased to 65.2

MPa. (An increment of %1.57%1005.41

5.412.65

.)


Shear force (obtained from vertical equilibrium),

Hence,

Maximum shear stress increases by 57% to 2.27 N/mm

2


Since beam AC is simply-supported, and

is loaded by a concentrated bending

moment of 50 kNm at B, we have:

Taking moments,

010500.3

010500.33

2

31

R

R

kN3

50N

3

1050

kN3

50N

3

1050

3

2

3

1

R

R

75 75

200

50

125

A B

C

2000 mm 1000 mm

50 kNm

A B

C

2.0 m 1.0 m

50 kNm

R1 R2

A B

C

2.0 m 1.0 m

50 kNm


130313

031303

132313

3103

502105010

3

50

3103

502105010

3

50

3103

502105010

3

50

xxxxM

xxxxF

xxxxq

xz

xy

Since 130313 3103

502105010

3

50 xxxxM xz

Hence, we have

m0.3m0.2Nm1050103

50

m0.2m0.0Nm103

50

33

3

xx

xxxM xz


Mxz

x

kNm

2.0 0.0 3.0

kNm

A B

C

2.0 m 1.0 m

50 kNm

To determine the values and

locations of the maximum

tensile and compressive stresses

in the beam, we need to first

locate the position of the

centroid y of the given cross-

section.

221

222

221

mm16250

mm10000mm50200

mm6250mm12550

AA

A

A

To determine y , we have

mm7.58

mm65384615.58

16250

25100005.1126250

255.112

252

12550

21

21

2121

AA

AAy

yAAAA

75 75

200

50

125

75 75

200

50

125

A2

A1

46

4126464

4

42323

4232

3

m107.39

m10107.39mm107.39mm17.39668504

mm1135690012

2500000018090250

12

97656250

mm7.33502005020012

18.531255012550

12

1

mm257.58502005020012

17.58

2

125501255012550

12

1

zI

Consider the beam to the left of B (where the bending moment is negative)

Hence, evidently the top fibres are in tension

and the bottom fibres are in compression.

Using yI

M

z

xzxx

yy

I

M

xx

z

xz3

46

3

46

max

Nm5.839630562m107.39

Nm103

100

m107.39

kNm3

100

For the top fibre, m0587.0mm7.58 y ,

58.7

z

y

ve bending moment

tensileMPa3.49

Nm02.49286314m0587.0Nm5.839630562 23

xx

For the bottom fibre, m1163.0mm3.116 y ,

ecompressivMPa6.97

Nm42.97649034m1163.0Nm5.839630562 23

xx

Consider the beam to the right of B (where the bending moment is positive)

Hence, evidently the top fibres are in

compression and the bottom fibres are in tension.

Using yI

M

z

xzxx

yy

I

M

xx

z

xz3

46

3

46

max

Nm3.419815281m107.39

Nm103

50

m107.39

kNm3

50

For the top fibre, m0587.0mm7.58 y ,

ecompressivMPa6.24

Nm01.24643157m0587.0Nm3.419815281 23

xx

For the bottom fibre, m1163.0mm3.116 y ,

tensileMPa8.48

Nm22.48824517m1163.0Nm3.419815281 23

xx

Hence, the maximum tensile stress is 49.3 MPa at the top fibre left of B and

the maximum compressive stress is –97.0 MPa at the bottom fibre left of B.

Alternatively, depending on accuracy, your answer may be:

Maximum tensile stress is 49 MPa at the top fibre left of B or at the bottom

fibre right of B.

+ve bending moment


Shear force,

We have:

Shear stress,


From symmetry, it is evident

that 2

21

PRR

Due to symmetry, we only

need to perform analysis of

the stress at points B and C.

We can discard point A since

it is a free end and the bending

moment xzM is zero ( stress

due to bending is zero at A).

Bending moments:

At B, Nm10

110

PPM xz

At C, Nm10

31

22

10

PPPM xz

1 m 1 m 1 m 1 m

A B C D E

X

X

0.075 0.075

0.050

0.125

0.050

A B C D E

R1 = R2 =

To determine the values and

locations of the maximum tensile and

compressive stresses in the beam, we

need to first locate the position of the

centroid y of the given cross-

section.

221

222

221

m01625.0

m010.0m050.0200.0

m00625.0m125.0050.0

AA

A

A

To determine y , we have

m0587.0

m50586538461.0

01625.0

025.0010.01125.000625.0

025.01125.0

025.02

125.0050.0

21

21

2121

AA

AAy

yAAAA

0.075 0.075

0.050

0.125

0.050

A1

A2

0.0587

z

y

4545

4

23

23

4

23

23

m1097.3m10966850417.3

m

0337.0050.0200.0050.0200.012

1

0538.0125.0050.0125.0050.012

1

m

025.00587.0050.0200.0050.0200.012

1

0587.02

125.0050.0125.0050.0125.0050.0

12

1

zI

Consider point B, Nm10

PM

Bxz and point C, Nm10

3PM

Cxz

Using yI

M

z

xzxx

At the top fibre, m1163.00587.0125.0050.0 y

At the bottom fibre, m0587.0y

At B (top fibre):

Pa293

Nm9471.292m1163.0m1097.3

Nm10 2

45

P

P

P

xx

At C (top fibre)

Pa879

Nm8413.878m1163.0m1097.3

Nm10

3

2

45

P

P

P

xx

ve bending moment +ve bending moment

At B (bottom fibre)

Pa9.147

Nm8589.147m0587.0m1097.3

Nm10 2

45

P

P

P

xx

At C (bottom fibre)

Pa444

Nm5768.443m0587.0m1097.3

Nm10

3

2

45

P

P

P

xx

The largest tensile stress of Pa444P occurs at the bottom fibre at C.

The largest compressive stress of Pa879P occurs at the top fibre at C.

Since the allowable tensile stress Pa1035MPa35 6T and the

allowable compressive stress Pa10150MPa150 6C , we have

kN8.78N8288.78828

1035444 6

P

P

kN6.170N4641.170648

10150879 6

P

P

Hence, the maximum allowable value for P is kN8.78 .

The tensile stress at C is the controlling factor in determining the maximum

allowable load.

Alternatively, you may also work out the allowable loads P in each case

and take the smallest result to be the limiting load:

Point TOP FIBRE

BOTTOM FIBRE

B

Pa293Pxx Tensile

kN5.119N9249.119453

1035293 6

P

P

Pa879Pxx Compressive

kN6.170N4641.170648

10150879 6

P

P

C

Pa9.147 Pxx Compressive

kN1014N7829.1014198

101509.147 6

P

P

Pa444Pxx Tensile

kN8.78N8288.78828

1035444 6

P

P

As before, the maximum allowable value for P is kN8.78 .

To determine P based on the maximum shear stress allowable:

Shear force,

Shear stress,

We have:

Hence,

But allowable shear stress is 10 MPa. Therefore,

Hence, max. allowable value for P is still 78.8 kN.


From symmetry,

At section n-n,

Pa

rt ( ) ( ) ( ) ( ) ( ) ( )

11250 125 1406.25 50 28.125 21.094

11250 25 281.25 50 28.125 2.344

22500 1687.5 56.25 23.438

(a)

( )( )( )

( )( )

( )( )

(b)

( )( )( )

(( )( )

( )( ))

(c)

Largest shearing stress occurs on the section through the centroid.

( )( )( )

( )( )

( )( )


For rolled steel section W200x46.1, we have:

(a)

[ (

)

]

[ (

)

]

(

)

Horizontal shearing force per meter of weld is:

( )

( )

(b)

[ ( )

]

[ (

)

]

(

)

i.e.,

Alternative Solutions To : Tutorial 2 Q6

For Case (a)

Consider a unit length of the beam

Using the shear stress equation we have

yAIb

F

zzf

xy

xy (1)

A B

Fxy

z z

bf

The horizontal shearing force F induced in the beam is:

yAI

F

byAIb

F

bF

zz

xy

fzzf

xy

fxy

1

1

[ Substituting τxy from Eq. (1) ]

The shearing force is taken up by two weld lines each weld line can

withstand 500 kN.

I.e. 1000 kN for two weld lines

Hence , we have

kN

NFei

FxF

xy

xy

356

10356..

107.211

)108.594(101000

3

6

6

3

Similarly for case (b) Fxy can be calculated.

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T2-Stresses in Beams solutions 2013.pdf