STRESS AND STRAINS - ikbooks.com

CHAPTER 2STRESS AND STRAINS

StressStress can be classified broadly in three types as described below:

1. Tensile stress: It is illustrated in Fig. 2.1where a tensile load W is applied to a uni-form rod fixed at one end.

l

W

dl

Figure 2.1

Tensile stress, σ =W

Cross-sectional area of rod=

WA

unit is N/mm2 or MN/m2,σ (Greek letter sigma).

2. Compressive stress: As shown in Fig. 2.2when load W tends to compress a rodof cross-section area A, then compressive

stress =WA

.

W

Figure 2.23. Shear stress: If two plates are joined

together with rivet as shown in Fig. 2.3.

The stress in rivet is known as shear stress,it is denoted by τ (Greek letter tau), shear

stress in rivet, τ =FA

.

F

F

Figure 2.3

6 • Strength of Materials

It may be noted that point A is the limit of proportionality and B is the elastic limit. Between pointA and B it is a curve thus not linear relationship. Therefore, actually E is constant within the limitof proportionality, though in Hooke’s law, we had mentioned, within ‘elastic limit’, because A andB are very close to each other. Let us name the important points on the graph:

A: Limit of proportionalityB: Elastic limit: It may be noted that on removal of load up to elastic limits, specimen comes

back to its original dimension.C: Higher yield point: This is the point where yielding of the material begins.

C�: Lower yield point: The stress associated with the lower yield point is known as yield strength.D: Maximum stress: Here the stress is maximum because due to plastic behaviour of the mate-

rial, area of cross section is very low.E: Point of fracture: At this point ‘waisting occurs’ as shown in Fig. 2.5.

WaistingCap

Cone

Figure 2.5

If the material is loaded beyond the elastic limit and then load is removed, a permanent extensionremains, called permanent set.

Proof Stress: For engineering purposes it is desirable to know the stress to which a highlyductile material such as aluminium can be loaded safely before a permanent extension takes place.

This stress is known as the proof stress or offset stress and is defined as the stress at whicha specified permanent extension has taken place in the tensile test. Proof stress is found from thestress-strain curve as given in Fig. 2.6. The extension specified is usually 0.1, 0.2 or 0.5 per cent ofgauge length.

0

0.1%Proofstress

Strain

Stre

ss

0.001

Figure 2.6

Stress and Strains • 7

The proof stress here is found on the basis of 0.1 per cent strain.

Procedure: Draw a line parallel to the initial slope of the curve. The stress at the point where thisline cuts the curve is the 0.1% proo f stress. The 0.2 per cent proof stress is also found in the samemanner.

Note: Though we define Hooke’s law to be taken without elastic limit, but strictly speaking it isapplicable up to the point of proportionality B in Fig. 2.4.

Brittle Materials: Fig. 2.6 shows that the stress-strain graph for brittle materials such as cast iron.The metal is almost elastic and up to fracture but does not obey Hooke’s law. A material such asthis which has little plasticity or ductility and does not neck down before fracture is termed ‘brittle’.The modulus of elasticity for cast iron is not constant but depends on the portion of the curve fromwhich it is calculated.

Stress

Fracture

Strain

Figure 2.7 Cast iron in tension

The following table is very useful for mechanical properties:

Percentage Yield stress 0.1% proof stress Ultimate tensile stressMaterial elongation MN/m2 MN/m2 MN/m2

Copper annealed 60 − 60 220Copper hard 4 − 320 400Aluminium soft 35 − 30 90Aluminium hard 5 − 140 150Black mild steel 25–26Bright mild steel 14–17Structural steel 20 220–250 − 430–500⎧⎪⎪⎪⎨

⎪⎪⎪⎩

Cast ironSpheroidalGraphiteCast iron (annealed)

−

− − − 280–340Stainless steel 60 230 – 600


Now E =stressstrain

Let F, L, A, dl be the force, length, area of cross section, and extension or contraction

respectively,

then E =F.LA.dl

EXAMPLE 2.1: A bar of mild steel has an overall length of 2.1 m. The diameter up to 700 mmlength is 56 mm, the diameter of the remaining 1.4 m is 35 mm. Calculate the extension of the bardue to a tensile load of 55 kN.

E = 200 GN/m2.

SOLUTION:

∴ Remember 1 GN/m2 = 1 kN/mm2

∴ E = 200 kN/mm2

We know E =FlAdl

∴ dl =F.lA.E

Therefore, for portion of 700 mm,

the extension dl1 =55000×700×7×4

200000×22×56×56=

564

mm = 0.0178 mm

Now dl2 for 1400 mm length of 35 mm dia,

dl2 =55000×1400×7×4

200000×22×35×35= 0.4 mm

Total extension = dl1 +dl2 = 0.0178+0.4 = 0.4178 mm

Compound bars: When two or more materials (members) are rigidly fixed together so that theyshare the same load and extend or compress by same amount, the two members form compoundbar. Let us say that in Fig. 2.7 we have to find stress in each material and amount of compression.

l

P Material Aof ES

Material Bof EB

Figure 2.8

Let the outer tube of material A has outsidedia as d1 and inside dia as d2 and inner tube ofmaterial B has outside dia as d3 and inside diaas d4. Both ends are joined rigidly to make com-pound bar of length l.


Hence, d 2 = 5630 mm2

or d = 75 mm

EXAMPLE 2.3: A steel bar of 20 mm diameter and 400 mm long is placed concentrically insidea gunmetal tube (Fig. 2.9). The tube has inside diameter 22 mm and thickness 4 mm. The length ofthe tube exceeds the length of the steel bar by 0.12 mm. Rigid plates are placed on the compoundassembly. Find: a) the load which will just make tube and bar of same length and b) the stresses inthe steel and gunmetal when a load of 50 kN is applied. E for steel= 213 GN/m2, E for gunmetal=100 GN/m2.

SOLUTION:

0.12 mm

P

Figure 2.9

Area of gunmetal tube, Ag =π4(0.032 −0.0222)

= 0.000327 m2

Area of steel bar As =π4(0.02)2 = 0.0003142 m2

a) For tube to compress 0.12 mm:

strain =0.12400

= 0.0003, Let σ1 be the stress in the tube

σ1

Eg= 0.0003,

σ1

100= 0.0003

∴ σ1 = 0.0003×100 = 0.03 GN/m2 = 30000 kN/m2

Hence, load = 30000×0.000327 = 9.81 kN

b) Load available to compress bar and tube as a compound bar is given by, let σ2 be the addi-tional stress produced in the gunmetal tube due to this load and σs be the corresponding stressin the steel bar, then

Load on compound bar = 50−9.81 = 40.19 kN

P = σ2Ag +σsAs

40.19 = σ2 ×0.000327+σ3 ×0.0003142 (i)


Also

σ2

Eg=

σs

Es, ∴ σ2 =

1002100

σs (ii)

From Eqns. (i) and (ii)

σ2 = 40,600 kN/m2 = 40.6 MN/m2

σs = 85300 kN/m2 = 85.3 MN/m2

Final stress in gunmetal = σ1 +σ2

= 40600+30000 = 70,600 kN/m2

= 70.6 MN/m2

Deformation of a Body Due to Self Weight

B

A

lx

dx

Figure 2.10

Let us consider a bar AB which is hanging freely under itsown weight (see Fig. 2.10)

Let w = specific weight of the bar material

Now consider a small section dx at a distance x from A.

Weight of the bar for a length = w× volume

= wAx (A is cross section of the bar)

Now elongation of the elementry length dx due to weight of the bar for length x,(wAx)

=pl

A.E=

(wAx)dxA.E

=wx.dx

E

Total elongation =

l∫

0

wxdxE

=wE

l∫

0

x.dx

=wE

[x2

2

]l

0

∴ elongation, dl =wl2

2E

Because total weight of bar, W = wA.l.

Now elongation dl can be written aswAl.l2AE

Hence, dl =Wl2AE


This result also proves that the extension due to own weight is half if same weight is applied atthe end (of course neglecting extension due to self weight).

EXAMPLE 2.4: A steel bar ABC 18 m long is having cross-sectional area 4 mm2 weighs 22.5 N(Refer Fig. 2.11). If modulus of elasticity of wire is 210 GN/m2, find the deflections at C and B.

A

B

9 m

9 m

C

Figure 2.11

Deflection at C due to self weight of wire AC = dlc

dlc =Wl2AE

=22.5×18000

2×4×210000= 0.241 mm

Deflection at B:

Now deflection at B is due to two reasons: i) due to selfweight of AB and ii) due to weight of BC.

dlB =W/2× l/2

2AE+

W/2× l/2A.E

∴ dlB =W/2× l/2

AE

(12+1

)

=22.5×9000

2×4×210000(1.5) = 0.181 mm

Sometimes a machine member is a acted upon by a number of forces, some acting at outer edgeswhile some are acting inside the body. In such cases in order to find out the total extension orcontraction, the principle of superposition is applied. This has been very well made clear by thefollowing examples:

EXAMPLE 2.5: A steel bar ABC of 400 mm length and 20 mm diameter is subjected to a pointload as shown in Fig. 2.12. Determine the total change in the length of bar. Take E = 200 GPa.

A

60 kN 40 kN

CB

20 kN

200 mm200 mm

Figure 2.12


SOLUTION:For simplification split it into two parts as under:

A C

400 mm

40 kN 40 kN

A CB

200 mm

20 kN 20 kNδ =

PlAE

A =π4(20)2 = 314 mm2

δAC =40×103 ×400314×200000

= 0.255 mm

δAB =20×103 ×200314×200000

= 0.064 mm

Total δ = 0.255+0.064 = 0.319 mm Ans.

EXAMPLE 2.6: A copper rod ABCD of 800 mm2 cross-sectional area and 7.5 m long is subjectedto forces as shown in Fig. 2.13. Find the total elongation of the bar. Take E = 100 GPa

A40 kN 50 kN

C DB20 kN30 kN

2.5 m3.5 m 1.5 m

2.5 m3.5 m 1.5 m

SOLUTION:Splitting into three figures as shown below:

A D

7.5 m

40 kN 40 kN

B C

1.5 m

20 kN 20 kN

DB

4 m

10 kN 10 kN

Figure 2.13


Hence, cross-sectional area at distance x from larger end A� =πd�2

4=

π4(d1 − kx)2

Stress at this section, σ � =PA� =

4Pπ(d1 − kx)2

∴ Strain = ε � =σ �

E=

4PπE(d1 − kx)2

Extension of elementary length dx = ε �dx =4P dx

πE(d1 − kx)2

Total extention of the bar = δ =4PπE

l∫

0

dx(d1 − kx)2 =

4PπE

[(dl − kx)−1

−1×−k

]l

o

=4P

πEk

[1

d1 − kx

]l

0

=4P

πEk

{1

d1 − kl− 1

d1

}but k =

d1 −d2

l

∴ =4P

πE(d1 −d2)

[1

d1 −d1 +d2− 1

d1

]

=4Pl

πE(d1 −d2)

(1d2

− 1d1

)

=4Pl

πE(d1 −d2)· d1 −d2

d1d2

∴ δ =4Pl

πEd1d2

If both the diameters are equal to d.

Then δ =4Pl

πEd2

EXAMPLE 2.7: A round steel rod of different cross-sections is loaded as shown in Fig. 2.15. Findthe maximum stress induced in the rod and its deformations. Take E = 210 GPa.


Extension of Tapered Rectangular Strip

P P b

txx

x

xdx

a

Figure 2.16

Consider any section x− x distant x from the bigger end

Width of the section = t

∴ Area of the section =P

t(a− kx)

∴ Extension of an elemental length dx =Pdx

t(a− kx)E

∴ Total extension of the rod = δ =PtE

l∫

0

dxa− kx

=− PtE

· 1k− loge [(a− kx)]l0

=− PtE

[loge(a− kl)− loge a]

=P

tkE

[a

a− k

]

But k =a−b

l

δ =P.l

Et(a−b)loge

ab


EXAMPLE 2.8: A straight bar of steel rectangular in section is 3 m long and of thickness of 12 mm.The width of rod varies uniformly from 110 mm or one end to 35 mm at the other end. If the rod issubjected to an axial load (tensile) of 25 kN, find the extension of the rod. Take E = 200000 N/mm2.

Extension of the rod, δ =Pl

Et(a−b)loge

ab

P = 25000 N, l = 3000 mm, t = 12 mm

a = 110 mm, ab = 35 mm & E = 200000 N/mm2

∴ δ =25000×3000

2×105 ×12(110−35)loge

11035

=25000×3000

2×105 ×12×75×1.1452

= 0.477 mm Ans

EXAMPLE 2.9: A rigid bar AB is attached to two vertical rods as shown in Fig. 2.17 is horizontalbefore the load is applied. Determine the vertical movement of P if it is of magnitude 60 kN.

Aluminium

Steel

AFor aluminiumL = 3 mA = 500 mm2

E = 75 GPa

For steelL = 4 mA = 300 mm2

E = 210 GPa

C B

60 kN

3.5 m 2.5 m

Figure 2.17


SOLUTION:For Al, ∑MB = 0, 6PAl = 2.5×60

∴ PA =2.5×60

6= 25 kN = 25000 N

δAl =PLA.E

=25000×3000500×75000

= 2 mm

For steel ∑MA = 0 gives

Pst = 3.5×60

Pst =3.5×60

6= 35 kN

σST =35000×4×1000

300×200000= 2.33 mm

C

2.33 mm

2 mm

Y

A

C1

A1

C2

B2

B1

B

Figure 2.18

Now from similar triangles A1 C1 C2 and A1,B1,B2

YA1C1

=B1B2

A1B1;

Y3.5

=2.33−2

6

∴ Y = 0.1925 mm

Now vertical movement of

P =CC2

=CC1 +Y

= 2+0.1925 = 2.1995 mm Ans


Bar of Uniform StrengthAs we have seen earlier that the stress due to self weight is not constant. It increases with theincrease of distance from the lower end.

We wish to find the shape of the bar of which the self weight is considered and is having uniformstress on all sections when subjected to an axial P. Figure 2.19 shows such a bar of uniform stressin which the area increases from the lower end to the upper end.

x

dx

A dA

A

A2

Area A1

Area A1

σ(A+dA)

σA+wAdx

Area A2

P(a) (b)

L

Figure 2.19

Let L be the length of bar, having area A1, and area A2 be cross-sectional areas of the bar at topand bottom, respectively.

Let w be the specific weight of the bar material (1.2. weight per unit volume of the bar).The forces acting on the elementary stripe are:

i) Weight of the strip acting downward and is equal to w× volume of strip.ii) Force on section AB due to uniform stress is equal to σ ×A. This is acting downward. A is

area of elementary stripe.iii) Force on section CD due to uniform (σ) is equal to σ(A+dA). This is acting upwards.

Total force acting upwards = Total force acting downwards

σ(A+dA) = σ ×A+wA dx

σA +σdA = σA+wA.dx

ordAA

=wσ

dx


Using equation,

A1 = A2ewLσ

A1 = 450 e

0.000075×220001777.8 = 450e9.28×10−4

A1 = 450.4 mm2 Ans

EXAMPLE 2.11: A steel rod of 25 mm dia passes centrally through a copper tube of 30 mm insidediameter and 40 mm outside diameter. Copper tube is 850 mm long and is closed by rigid washers ofnegligible thickness, which are fastened by nut threaded on the rod as shown in Fig. 2.20. The nutsare tightened till the load on the assembly is 20 kN. Calculate: i) the initial stresses on the coppertube and steel rod and ii) also calculate increase in the stresses, when one nut is tightened by one-quarter of a turn relative to the other. Take pitch of the thread as 1.5 mm. E for copper = 100 GPa,E for steel = 100 GPa

SOLUTION:

Steel rodWasheron each side

Copper tube

Figure 2.20

Let σs = Stress in steel rod

σc = Stress in copper rod

i)As =

π4(Ds)

2 =π4(25)2 = 156.25π mm2

Ac =π4(D2 −d2) =

π4(402 −302) = 175π mm2

Tensile rod on steel = Compressive load on copper tube

σs =Ac

As×σc =

175π156.25π

×σc


Brass Brass

10 kN

Steel

Figure 2.21SOLUTION:

σs ×100+100σb +100σb (i)

100σs +200σb = 10000

σs +2σb = 100 N/mm2 (ii)

σs

200×103 =σb

100×103

∴ σs = 2σb (iii)

substituting for σs in (ii)

2σb +2σb = 100;

σb =100

4= 25 MPa Ans.

σs = 2×25 = 50 MPa Ans.

EXAMPLE 2.13: Two steel rods and one copper rod each of 20 mm diameter together support aload of 50 kN as shown in Fig. 2.22. Find the stress in each rod. Take Es = 200 GPa,Eb = 100 GPa

50 kN

2 m1.5 m

Brass BrassCopper

Figure 2.22


Ac = As =π4(20)2 = 314 mm2

Total area of steel A�s+314×2 = 628 mm2

σsA�s +σcAc = 50000

628σs +314σc = 50000

2σs +σc = 159.24 (i)

σslsEs

=σclcEc

;

σs ×2000200000

=σc ×1500

100000σs = 1.5σc (ii)

Substituting for σs from Eqn. (ii) in Eqn. (i)

2×1.5σc +σc = 159.24

∴ σc = 39.81 MPa Ans.

σs = 1.5×39.81 = 59.7 MPa Ans.

EXAMPLE 2.14: A uniform bar ABCD has built-in ends A&D. It is subjected to two point loadsP1 and P2 equal to 80 kN and 40 kN at B and C as shown in Fig. 2.23. Find values of reactions atA and D.

A B C D

P1

500 mm 500 mm1000 mm

P2

SOLUTION:

ARA RA

B

500 mm

RBRB

RC RC

500 mm

1000 mm

Figure 2.23


2.10 Figure 2.26 shows a rigid bar ABC hinged at A and suspended at two points B and C by twobars BD and CE, made of aluminium and steel, respectively. The bar carries a load of 20 kNmidway between B and C. The cross-sectional area of aluminium bar BD is 3 mm2 and thatof steel bar CE is 2 mm2. Determine the loads taken by the two bars BD and CE.

1000 mm

500 mm

20 kN

1000 mm1000 mm

D

B C

E

Figure 2.26 [Ans Pa = 3.481 kN,Ps = 13.26 kN]

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