UNIT 3 - eGyanKosh [PDF]

UNIT 2 THERMAL STRESSES

Thermal Stresses

Structure 2.1

Introduction Objectives

2.2

Effect of Temperature on Bodies

2.3

Thermal Stresses in Bodies

2.4

Thermal Stresses in Uniform Bars

2.5

2.6

2.4.1

Fully Restrained Bar

2.4.2

Partially Restrained Bar

Thermal Stresses in Stepped Bars 2.5.1

Fully Restrained Stepped Bar

2.5.2

Partially Restrained Stepped Bar

Thermal Stresses in Tapered Bars 2.6.1

Fully Restrained Tapered Bar

2.6.2

Partially Restrained Tapered Bar

2.7

Thermal Stresses in Composite or Compound Bars and Tubes

2.8

Summary

2.9

Answers to SAQs

2.1 INTRODUCTION Thermal stresses are the stresses induced in a body due to change in temperature. Thermal stresses are setup in a body when the temperatures of the body is raised or lowered and the body is not allowed to expand or contract freely. This unit introduces you to the effect of temperature on materials and completely or partially restrained bodies. It introduces the concept of thermal stresses and the method to determine them in simple and compound bars.

Objectives After studying this unit, you should be able to •

explain the effect of temperature on bodies,

•

understand how thermal stresses are developed in materials, and

•

determine thermal stresses in uniform, stepped, tapered and compound bars.

2.2 EFFECT OF TEMPERATURE ON BODIES It is a common observation and experience that the size of a body increases as the temperature increases while it decreases as the temperature decreases. Thus, temperature rise results in expansion, i.e. increase in linear dimensions of the body, and temperature fall results in contraction, i.e. reduction in the linear dimensions of the body. The magnitude of the changes brought in linear dimensions of the body by temperature is defined through an intrinsic property of the material, of which the body is made of, called coefficient of linear thermal expansions or contraction. The coefficient of linear thermal expansion or contraction of a material is defined as the change in length per unit length per degree change in temperature of the body which is made of the material. The values of this coefficient, α of some of the common engineering materials are listed in Table 2.1. 41

Stresses in Solids

From the definition, α can be written as, α=

ΔL 1 × L ΔT

where, ΔL is the change in length over a total length L when the temperature is changed by ΔT. Thus, ΔL = α × L × ΔT Table 2.1 α, Coefficient of Linear Thermal Expansion (K− 1)

Material Low carbon (mild) steel

11.7 × 10− 6 12 × 10− 6

Cast Iron : Gray Malleable

11.7 × 10− 6

Nickel-chrome steel

23 × 10− 6

Aluminium

18.9 × 10− 6

Brass

18 × 10− 6

Bronze

28.8 × 10− 6

Magnesium

2.3 THERMAL STRESSES IN BODIES In the earlier section we had seen that if the temperature of a body is increased by ΔT, there will be an expansion ΔL over a length L. However, this expansion occurs only if the body is free to expand. During this free expansion, the body is stressed. But, if this free expansion is restrained either fully or partially, forces of restraint are generated and the body develops internal strain resulting in internal stresses. These stresses are called thermal stresses. A similar situation arises if the body is restrained against contraction and the temperature is reduced. When expansion is restrained, compressive stresses will be generated because the restraint is equivalent to applying a compressive force. Similarly, when contraction is restrained, tensile stresses will be generated.

2.4 THERMAL STRESSES IN UNIFORM BARS In this section we shall discuss the problem of thermal stresses in bars of uniform crosssection when they are partially or totally restrained against change in length due to temperature changes. Since the area of cross-section is the same over the entire length, the stress will be the same over each cross-section.

2.4.1 Fully Restrained Bar Let us consider a bar of uniform cross-sectional area A and length L and constrained at both ends as shown in Figure 2.1(a) and (b). AREA OF C.S. = A P

(a)

(b) Figure 2.1

42

ΔL

L

L

Let us increase the temperature of the bar by ΔT. If the bar were free to expand, it would have increased to a length L + ΔL as in Figure 2.1(b). However, this increase in length, ΔL, is completely restrained. Hence, the restraining force, P, will be equivalent to a force which would have produced the same change in length. Following these arguments ΔL = αL ΔT = Therefore,

Thermal Strain =

Thermal Stresses

PL or P = AE α ΔT AE

ΔL = α ΔT L

Thermal Stress in the bar = Strain × E = E α ΔT

P . Since the force P is compressive, the stress developed is also A compressive in nature.

Which is the same as

2.4.2 Partially Restrained Bar Now, let us assume that one of the supports of the bar in Figure 2.1(a) can move by a distance ΔL′ (ΔL′ < ΔL) while the bar is expanding. In this case, the bar is free to expand by an amount ΔL′ but is restrained in expanding further by rest of the amount (ΔL − ΔL′). Thus, the force of restraint, P, is now a force which would produce a change in length (ΔL − ΔL′) in the bar. Thus,

P=

Strain =

(ΔL − ΔL′) AE (αL ΔT − ΔL′) AE = L L

(αL ΔT − ΔL′) ⎛ ΔL′ ⎞ = α ΔT − ⎜ ⎟ L ⎝ L ⎠

ΔL′ ⎞ ⎛ Thermal Stress = Strain × E = E ⎜ α ΔT − ⎟ L ⎠ ⎝ Example 2.1 Two parallel walls are stayed together by a steel rod of 5 cm diameter passing through metal plates and nuts at both ends. The nuts are tightened, when the rod is at 150ºC, to keep the walls 10 m apart. Determine the stresses in the rod when the temperature falls down to 50ºC, if (a)

the ends do not yield, and

(b)

the ends yield by 1 cm.

Take E = 2 × 105 N/mm2 and α = 12 × 10− 6 K− 1. Solution Given Length of the rod, L = 10 m =104 mm Diameter of the rod, d = 5 cm = 50 mm Change in temperature, ΔT = 150 − 50 = 100ºC E = 2 × 105 N/mm2 α = 12 × 10− 6 K− 1 (a)

When the ends do not yield (let the stress be σ1) Thermal stress, σ1, in the rod = E α ΔT = 2 ×105 × 12 × 10− 6 × 100 = 240 N/mm2 43

Stresses in Solids

(b)

When the ends yield by 1 cm (let the stress be σ2)

ΔL′ ⎞ ⎛ Thermal Stress, σ2 = ⎜ α ΔT − ⎟ ×E L ⎠ ⎝

⎛ ⎝

= ⎜12 × 10 − 6 × 100 −

10 ⎞ × 2 × 105 4 ⎟ 10 ⎠

= (0.0012 − 0.001) × 2 × 105 = 40 N/mm2 Example 2.2 A composite rod is made by joining a copper rod end to end with a second rod of different material but of same cross-section. At 25oC, the composite rod is 1 m in length of which the length of copper rod is 30 cm. At 125oC, the length of the composite rod increases by 1.91 mm. When the composite rod is not allowed to expand by holding it between two rigid walls, it is found that the length of constituents does not change with rise in temperature. Find the Young’s modulus and coefficient of linear expansion of the second rod. For copper, α = 1.7 × 10– 5 oC– 1, and Y = 1.3 × 1011 Nm– 2. Solution For copper rod α1 = 1.7 × 10– 5 oC– 1, and Y1 = 1.3 × 1011 Nm– 2, l1 = 30 cm = 0.3 m. Let l2 be initial length of the rod of other material at 25oC, and α2 and Y2 be coefficient of linear expansion and Young’s modulus for the material of this rod. Then, at initial temperature θ1 = 25oC, l1 + l2 = 1 m, l2 = 1 − l1 = 1 − 0.3 = 0.7 m

Let the final temperature θ2 = 125oC, the increases in the length of cooper rod and that of other material be Δl1 and Δl2, respectively. Then,

Δl1 + Δl2 = 1.91 mm = 1.91 × 10− 3 m

. . . (A)

Now,

Δl1 = l1 α1 (θ2 − θ1 ) = 0.3 × 1.7 × 10− 5 × (125 − 25) = 0.51 × 10− 3 m

Also,

Δl2 = l2 α 2 (θ 2 − θ1 ) = 0.7 × α 2 × (125 − 25) = 70 α 2

Substituting for Δl1 and Δl2 in Eq. (A), we get 0.51 × 10− 3 + 70 α 2 = 1.91 × 10 − 3

or,

α 2 = 2.0 × 10− 5 o C −1

Now,

F FL Y A Δl ⇒ F= Y = A = Δl A Δl l C

Let A be the cross-section of each of the two rods, therefore, tension produced in Y A Δl1 , and tension produced in the rod of other material, copper rod F1 = 1 l1 Y A Δl2 F2 = 2 . The tension developed in the two rods will be same, i.e. F1 = F2. l2 Therefore,

Y1 A Δl1 Y2 A Δl2 = l1 l2 44

l2 Δl1 0.7 × 0.51 × 10− 3 ⇒ Y2 = 1.3 × 1011 × l1 Δl2 0.3 × 1.40 × 10− 3

or

Y2 = Y1

or

Y2 = 1.105 × 1011 Nm − 2

Thermal Stresses

SAQ 1 Two parallel walls 6 m apart are stayed together by a steel rod 20 mm diameter, passing through metal plates and nuts at each end. The nuts are tightened, when the rod is at a temperature of 100ºC. Determine the stress in the rod, when the temperature falls down to 20ºC, if (a)

the ends do not yield, and

(b)

the ends yield by 1 mm.

Take E = 2 × 105 N/mm2 and α = 12 × 10− 6 K− 1.

2.5 THERMAL STRESSES IN STEPPED BARS In this section, we shall discuss the determination of thermal stresses in bars whose crosssectional area changes in steps over the length.

2.5.1 Fully Restrained Stepped Bar Consider a bar of length L which has a uniform cross-sectional area A1 over length L1 while in the rest of the length L2 the cross-sectional area is A2. Let the Young’s Modulus and the coefficient of linear expansion for the two parts be E1, α1 and E2, α2, respectively. The ends of the bar are fully restrained. The bar is shown in Figure 2.2. Let the temperature of the bar be raised by ΔT. A2, E2, α2 A1, E1, α1

L1

L2 L

Figure 2.2

If the bar were free to expand, it would have extended by a length ΔL given by ΔL = L1 α1 ΔT + L2 α2 ΔT = (L1 α1 + L2 α2) ΔT Since the bar is fully restrained, P, the compressive force developed would have to produce a contraction equal to ΔL. Let this force produce a stress σ1 in part 1 and σ2 in part 2 (these are the thermal stresses).

σ1 =

P P , σ2 = A1 A2 45

Stresses in Solids

σ1 A1 = σ2 A2 or σ2 = σ1 will produce a strain equal to

A1 × σ1 A2

σ1 σ L and a change in length ΔL1 equal to 1 1 . E1 E1

σ2 L2 . Thus, the total change E2 in length due to the application of force P will be equal to (ΔL1 + ΔL2 ) , Similarly, σ2 will produce a change in length ΔL2 equal to

ΔL1 + ΔL2 =

We get,

σ1 L1 σ2 L2 + E1 E2

But this should be equal to ΔL. Therefore,

σ1 L1 σ2 L2 + = ΔL = ( L1 α1 + L2 α 2 ) ΔT E1 E2

But,

σ2 =

∴

⎛L L A ⎞ σ1 ⎜ 1 + 2 1 ⎟ = ΔT ( L1 α1 + L2 α 2 ) ⎝ E1 A2 E2 ⎠

Hence,

A1 σ1 A2

ΔT E1 E2 A2 ( L1 α1 + L2 α 2 ) , and ( A1 E1 L2 + A2 E2 L1 )

∴ σ1 = σ2 =

∴

ΔT E1 E2 A1 ( L1 α1 + L2 α 2 ) ( A1 E1 L2 + A2 E2 L1 )

Herein, we have assumed that the two parts of the bar are made of two different materials. Instead, if the entire bar is of a single material of Young’s Modulus E and coefficient of linear expansion α, E1 = E2 = E and α1 = α2 = α Then σ1 and σ2 reduce to

and

σ1 =

ΔT E A2 α L A1 L2 + A2 L1

σ2 =

ΔT E A1 α L A1 L2 + A2 L1

Further, if the cross-section is uniform throughout (A1 = A2 = A), both the above expressions reduce to σ1 = σ 2 = σ = E α ΔT

a result which we have already derived for uniform bars.

2.5.2 Partially Restrained Stepped Bar Let us now consider the case of a stepped bar as in Figure 2.2, but which is free to extend by an amount ΔL´ (ΔL´ < ΔL) but restrained thereafter. In this case, during the free expansion of ΔL´ the bar remains unstressed but thereafter develops the stresses σ1 and σ2 whose values will now be different than those of Section 2.5.1. By a similar argument as in the previous section, we obtain

σ2 = σ1

A1 A2

σ1 L1 σ2 L2 = = (ΔL − ΔL′) = [ΔT ( L1 α1 − L2 α 2 ) − ΔL′] E1 E2 46

From the above two expressions, the thermal stresses σ1 and σ2 can now be written as :

σ1 =

E1 E2 A2 [ΔT ( L1 α1 + L2 α 2 ) − ΔL′] ( A1 E1 L2 + A2 E2 L1 )

σ2 =

E1 E2 A1 [ΔT ( L1 α1 + L2 α 2 ) − ΔL′] ( A1 E1 L2 + A2 E2 L1 )

Thermal Stresses

In the case of a single material bar

and

σ1 =

EA2 [ΔT α L − ΔL′] , A1 L2 + A2 L1

σ2 =

EA1 [ΔT α L − ΔL′] A1 L2 + A2 L1

For a uniform bar the stress expression reduces to

ΔL′ ⎞ ⎛ σ1 = σ2 = σ = E ⎜ α ΔT − ⎟ L ⎠ ⎝ a result already observed. Example 2.3 A stepped bar made of aluminum is held between two supports. The bar is 900 mm in length at 38ºC, 600 mm of which is having a diameter of 50 mm, while the rest is of 25 mm diameter. Determine the thermal stress in the bar at a temperature of 21ºC, if (a)

the supports are unyielding, and

(b)

when the supports move towards each other by 0.1 mm.

Given

E for aluminum = 74 kN/mm2 α for aluminum = 23.4 × 10−6 K−1

Solution When the Supports are Unyielding We have, L1 = 600 mm; diameter d1 = 50 mm

π(50) 2 mm2 4

Thus,

A1 =

Also,

L2 = 300 mm; diameter d2 = 25 mm

π(25) 2 mm2 Therefore, A2 = 4 and

ΔT = 38 − 21 = 17ºC

Therefore, free contraction, ΔL = L α ΔT = 900 × 23.4 × 10− 6 × 17 = 0.358 mm. Let σ1 be the stress in 50 mm φ part and σ2 be the stress in 25 mm φ part. Then,

σ1 =

900 × 23.4 × 10 −6 × 17 × 74 × 10 3 × π × (25) 2 ΔTEA2 αL = A1L2 + A2 L1 ⎛ π × (50) 2 ⎞ π × (25) 2 4⎜⎜ ⎝

4

× 300 +

4

× 600 ⎟⎟ ⎠

= 14.72 N/mm2 σ2 = σ1

A1 (50) 2 = 14.72 × = 58.88 N/mm2 A2 (25) 2

These stresses are tensile in nature. 47

Stresses in Solids

When the Supports Move towards Each Other by 0.1 mm Here, ΔL´ = 0.1 mm. ∴

σ1 =

EA2 [ΔL − ΔL′] = A1L2 + A2 L1

74 × 10 3 × π × (25) 2 × (0.358 − 0.1) ⎞ ⎛ π × (50) 2 π × (25) 2 4⎜⎜ × 300 + × 600 ⎟⎟ 4 4 ⎠ ⎝

= 10.61 N/mm2 ∴

σ2 = 10.61 ×

(50) 2 = 42.44 N/mm2 2 (25)

Both these stresses are tensile.

SAQ 2 A bi-metallic rod of length 450 mm is mounted horizontally between rigid abutments. The rod has a uniform circular cross-section and is made up of a 150 mm length of steel and a 300 mm length of copper coaxial with each other. If the rod is initially stress-free, determine the stress in the rod caused by a temperature rise of 100 K. Given

E, copper = 105 GNm− 2 E, steel = 210 GNm− 2 α, copper = 18 × 10− 6 K− 1 α, steel = 12 × 10− 6 K− 1

2.6 THERMAL STRESSES IN TAPERED BARS In this section, we shall study the problem of thermal stresses in gradually tapering bars. Because the cross-sectional area is varying along the length of the bar, so does the stress. Herein, we shall derive expression for thermal stress in bars with circular cross-section. For other cross-sections, like square etc., expressions for stresses can be derived in a similar fashion.

2.6.1 Fully Restrained Tapered Bar Consider the bar shown in Figure 2.3, which is circular in cross-section but tapering from a diameter d2 to d1 over the length L. The bar is completely restrained at its ends. Now, let the temperature of the bar increased by ΔT. x

dx

d2

d1

L

Figure 2.3

48

If the bar were not restrained but free to expand it would have extended by an amount ΔL, given by

Thermal Stresses

ΔL = α L ΔT Due to the restraint, a compressive force P would have developed in the bar whose effect is to produce a contraction equal to ΔL. Under this force, a cross-section at a distance x P , where Ax is the area from the larger end would have developed a stress, σx, equal to Ax of that cross-section. ∴

σx =

P = Ax

=

4P ( L − x) ⎛ ⎞ π ⎜ d2 + (d1 − d 2 ) ⎟ L ⎝ ⎠

2

4P x ⎛ ⎞ π ⎜ d1 − (d1 − d 2 ) ⎟ L ⎝ ⎠

2

The strain at that cross-section, εx, can be written as εx =

σx = E

4P x ⎛ ⎞ E π ⎜ d1 − ( d1 − d 2 ) ⎟ L ⎝ ⎠

2

Under this strain, a small element of the bar of length dx would have changed its length by d (ΔL) given by d (ΔL ) = ε x dx =

4 P dx x ⎛ ⎞ πE ⎜ d1 − ( d1 − d 2 ) ⎟ L ⎝ ⎠

2

The total change in length ΔL can then be obtained by integrating the above expression. L

∴

ΔL =

∫

L

d (ΔL) =

0

ΔL =

∫

d (ΔL) =

0

where, a = d1 and b =

=

0

L

Thus,

∫

4P πE

4P x ⎛ ⎞ πE ⎜ d1 − (d1 − d 2 ) ⎟ L ⎝ ⎠ L

∫ 0

1 (a − bx) 2

2

dx

dx

d1 − d 2 L

Now, on writing (a – bx) = t, we get dx = −

dt . b

Substituting these,

ΔL =

4P πE

L

∫ 0

⎛ 1 ⎞ dt 4 P = ⎜− ⎟ ⎝ b ⎠ t 2 πE

L

⎡ L 4P ⎡ 1 ⎤ 4P ⎢ 1 ⎢ = ⎢ ⎥ = d πbE ⎣ a − bx ⎦ 0 πE ⎢ ⎛ 1 − d2 ⎞ ⎢ d1 − ⎜⎝ L ⎟⎠ ⎣

=

L

4P ⎡ 1 ⎤ ⎛ 1 ⎞ ⎡ 1⎤ ⎜ − ⎟ ⎢− ⎥ = ⎢ ⎥ ⎝ b ⎠ ⎣ t ⎦ 0 πE ⎣ bt ⎦ 0 L

⎤ ⎥ ⎡ L ⎤ ⎥ ⎢ ⎥ d − d2 ⎦ x ⎥⎥ ⎣ 1 ⎦0

4PL πE d1 d 2 49

Stresses in Solids

4PL = L α ΔT πE d1 d 2

∴

So the compressive force generated in the bar due to restraining the free expansion for an increase in temperature ΔT is

πE d1 d 2 α ΔT 4

P=

Hence, the thermal stress, σx, at a cross-section with an area of Ax is given as σx =

πE d1 d 2 α ΔT E d1 d 2 α ΔT P = = 2 Ax 4 Ax x ⎛ ⎞ d − d − d ( ) ⎜ 1 1 2 ⎟ L ⎝ ⎠

where x is measured from the end with diameter d1 which is the larger end. The maximum stress, σmax, in the bar occurs at the smaller end with diameter d2.

E d1 α ΔT d2

σmax =

For cross-sections other than circular, the derivation can be proceeded in a similar way.

2.6.2 Partially Restrained Tapered Bar Let us consider the bar of Figure 2.3, but now let it be free to extend by an amount ΔL′( ΔL′ < ΔL ) . By following similar arguments as in the previous section, now the compressive load P developed in the bar will be given by

P=

πE d1 d 2 (ΔL − ΔL′) 4L

= and hence,

σx =

πE d1 d 2 ( L α ΔT − ΔL′) 4L E d1 d 2 ( L α ΔT − ΔL′) (d − d 2 ) ⎛ L ⎜ d1 − 1 L ⎝

⎞ x⎟ ⎠

2

and the maximum stress,

σmax =

E d1 ( L α ΔT − ΔL′) Ld 2

Example 2.4

A circular bar rigidly fixed at both ends is 1 m long and tapers uniformly from 20 cm diameter at one end to 10 cm diameter at the other end. Find the maximum stress in the bar, if its temperature is raised through 50oC. E = 2 × 105 N/mm2 and α = 12 × 10−6 K−1. Solution

Here, d1 = 20 cm = 200 mm d2 = 10 cm = 100 mm L = 1 m = 1000 mm ΔL = 50°C ∴

σ max =

Ed1 α ΔT 2 × 10 5 × 200 × 12 × 10 −6 × 50 = d2 100

= 240 N/mm2. 50

Thermal Stresses

SAQ 3 A straight bar has a circular cross-section, the radius of which varies linearly from 30 mm diameter at one end A to 15 mm at the other end B. The bar is 1 m long and is fixed rigidly at A, but longitudinal movement is possible at B against a spring which opposes movement with a constant stiffness of 20 kN/mm. Initially, there is no longitudinal stress in the bar. The temperature of the bar then falls by 100 K. Determine the change in the bar length if E = 69 GN/m2, and α = 23 × 10−6 K−1.

2.7 THERMAL STRESSES IN COMPOSITE OR COMPOUND BARS AND TUBES In Section 2.5 on stepped bars, we have seen one type of composite bar where bars of different materials are joined serially. In this section, we shall consider another type of compound bar, a bar constructed from two different materials rigidly joined together as shown in Figure 2.4(a), or two bars of different materials but of equal length held between two rigid plates as shown in Figure 2.4(b).

1

E1, A1, α1

2

E2, A2, α2

1

E1, A1, α1

2

E2, A2, α2

L

L

(a)

(b) Figure 2.4

If the individual bars were free to expand (or contract) due to temperature changes, they would do so to different amounts (for the same change in temperature), as shown in Figures 2.4(c) and (d), due to the difference in the coefficients of linear expansion of the two materials. Difference in Free Expansion Lα1ΔT 1 2 Lα2ΔT

L

(c) Lα1ΔT

Difference in Free Expansion

1

2 Lα2ΔT

L

(d) Figure 2.4

51

Stresses in Solids 1 1 2 2 ΔL

L

ΔL

L

(e)

(f) Figure 2.4

However, since the two materials are rigidly joined as a compound bar and subjected to the same temperature rise, each material will attempt to expand to its free length position but each will be affected by the movement of the other. The higher coefficient of expansion material will try to pull the lower expansion material to its free length, but will be held back by the latter to its own free length position. In practice, a compromise will be reached with both extending to a common position in between the individual free length positions. This, in effect, is equivalent to a contraction in bar 2 from its free length position and an expansion of bar 1 from its free position. Thus, the higher coefficient of expansion material develops tensile stresses, when the temperature of the compound bar increases. It will be vice-versa when the temperature decreases. From Figures 2.4(c) to 2.4(f), it is clear that Extension of bar 1 + Contraction of bar 2 = Difference in free lengths Let the stresses in bars 1 and 2 be σ1 and σ2 due to the temperature change. Then the above rule can be written as

σ1L σ 2 L + = (α 2 − α1 ) L ΔT E1 E2 Since there are no external forces acting on the compound bar, for equilibrium, the compressive force in bar 2 should be equal to the tensile force in bar 1. This means that σ1 A1 = σ2 A2 From the above two expressions, σ1 and σ2 can be written as

and

σ1 =

A2 E1 E2 (α 2 − α1 ) ΔT , A1 E1 + A2 E2

σ2 =

A1 E1 E2 (α 2 − α1 ) ΔT A1 E1 + A2 E2

The extension of the compound bar, i.e. ΔL is given as

ΔL = L α1 ΔT +

α1 L E1

ΔL = L α1 ΔT +

A2 E2 (α 2 − α1 ) L ΔT A1 E1 + A2 E2

ΔL =

( A1 E1 α1 + A2 E2 α 2 ) L ΔT A1 E1 + A2 E2

Example 2.5

A compound bar is constructed from three bars 50 mm wide by 12 mm thick fastened together to form a bar 50 mm wide by 36 mm thick. The middle bar is of aluminum alloy for which E = 70 GN/m2 and the outside bars are of brass with E = 100 GN/m2. If the bars are initially fastened at 18°C and the temperature of whole assembly is then raised to 50°C, determine the stresses set up in the brass and the aluminium. 52

Thermal Stresses

For computation purposes, take following values : αbrass = 18 × 10−6 K−1 , and αaluminium = 22 × 10−6 K−1 Solution

Let the stress in aluminium bar be σa and that in each brass bar be σb. Then for equilibrium Force in brass = Force in aluminium σb × 50 × 12 × 2 = σa × 12 × 50 2 σb = σa

or

and from extension considerations, we get

σa σ L + b L = L(α a − α b ) ΔT Ea Eb 2σ b

∴

70 × 10

9

+

σb 100 × 10

9

= (22 − 18) × 10 −6 × (50 − 18)

27σb = 4 × 10-6 × 32 × 700 × 109

or or

σb = 3.32 N/mm2 (T)

and

σa = 6.64 N/mm2 (C)

Example 2.6

A hollow steel cylinder of cross-sectional area 2000 mm2 concentrically surrounds a solid aluminium cylinder of cross-sectional area 6000 mm2. Both cylinders have the same length of 500 mm before a rigid block weighing 200 kN is applied at 20oC as shown in Figure 2.5. Determine (a)

the load carried by each cylinder at 60oC.

(b)

the temperature rise required for the entire load to be carried by the aluminium cylinder alone. 200 kN

AL

ST

500 mm

Figure 2.5

For computation purposes, take following values : Esteel = 210 GN/m2 and Ealuminium = 70 GN/m2 σsteel = 12 × 10– 6 K– 1 and αaluminium = 23 × 10– 6 K– 1 Solution

Figure 2.6 shows the free thermal expansions Δa and Δs together with the common expansion Δ under the load of 200 kN (the subscripts a and s standing for aluminium and steel respectively). For a temperature rise of ΔT K, 53

Stresses in Solids

We have,

Δ a = 500 × 23 × 10− 6 × ΔT = 11.5 × 10− 3 ΔT mm Δ s = 500 × 12 × 10− 6 × ΔT = 6 × 10 − 3 ΔT mm

Δα Δs

Δ

Figure 2.6

Under load, the strains are

and

εα =

Δa − Δ 500

εσ =

Δs − Δ 500

and the corresponding stresses are as follows :

σa =

70 × 103 (Δ a − Δ) = 140 (Δ a − Δ) N mm− 2 500

210 × 103 (Δ s − Δ) = 420 (Δ s − Δ) N mm− 2 500 For equilibrium of vertical forces, σs =

σ a × 6000 + σ s × 2000 = 200 × 103 N

Substituting for σa, σs, Δa and Δs, we get

(11.5 × 10− 3 × ΔT − Δ) + (6 × 10− 3 ΔT − Δ) =

5 21

5 42 The loads taken by the aluminium and the steel are therefore, Hence,

Δ = 8.75 × 10− 3 ΔT −

Pa = σ a × 6000 N

5 ⎞ ⎛ = 840 ⎜ 2.75 × 10− 3 ΔT + ⎟ kN 42 ⎠ ⎝ Ps = σ s × 2000 N

⎛ 5 ⎞ = 840 ⎜ − 2.75 × 10− 3 ΔT ⎟ kN 42 ⎝ ⎠ These equations will be valid as long as Δ is less than Δs. the load will be completely carried by aluminium when Δs becomes equal to Δ. (a)

at 60oC, ΔT = 60 − 20 = 40 K

5 ⎞ ⎛ Pa = 840 ⎜ 2.75 × 40 × 10− 3 + ⎟ 42 ⎠ ⎝ = 192.4 kN Ps = 200 − 192.4 = 7.6 kN

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(b)

Thermal Stresses

The load will be carried completely by aluminium when

6 × 10− 3 × ΔT = 8.75 × 10− 3 × ΔT −

ΔT =

or

5 42

5 × 103 = 43.3o C 2.75 × 42

i.e. at a temperature of (20 + 43.3) = 63.3oC. Example 2.7

A steel bolt of diameter 12 mm and length 175 mm is used to clamp a brass sleeve of length 150 mm to a rigid base plate as in Figure 2.7. The sleeve has an internal diameter of 25 mm and a wall thickness of 3 mm. The thickness of the base plate is 25 mm. Initially, the nut is tightened until there is tensile force of 5 kN in the bolt. The temperature is now increased by 100°C. Determine the final stresses in the bolt and the sleeve. For computation purposes, take following values : Eb = 105 GNm−2 ; Es = 210 GNm−2 αb = 20 × 10 –6 K−1 ; αs = 12 × 10 –6 K−1 25 mm 150 mm

25 mm

12 mm

31 mm

Figure 2.7

Solution

Let the free thermal expansions of steel and brass be Δs and Δb and Δ be the common expansion. Then Δs = (175) (12 × 10−6) (100) = 0.21 mm Δb = (150) (22 × 10−6) (100) = 0.3 mm If the initial stresses in steel and brass due to 5 kN load are σs1 and σb1, then

σ s1 = + σb1 = −

5 × 103 × 4 π (12)

2

= + 44.21 N/mm2

5 × 103 × 4 π (31 − 25 ) 2

2

= − 18.95 N/mm2

(Tensile)

(Compressive)

Equilibrium of thermal stresses σs2 and σb2 requires that σs 2 ×

π (12) 2 π (312 − 252 ) + σb 2 × =0 4 4

3 σs2 + 7 σb2 = 0

or The thermal strains are given by

εs =

(Δ − Δb ) (Δ − Δ s ) and εb = 175 150 55

Stresses in Solids

Thus, the thermal stresses are as follows :

and

σs2 =

210 ( Δ − Δ s ) × 10 3 N/mm2 175

σb 2 =

105 ( Δ − Δ b ) × 10 3 N/mm2 150

Substituting these in the equilibrium equation, 3×

210 105 ( Δ − 0.21) × 103 + 7 × ( Δ − 0.30) × 103 = 0 175 150

Hence,

Δ = 0.262 mm

Thus,

σs 2 =

210 (0.262 − 0.21) × 103 = + 62.26 N/mm2 175

(Tensile)

σb 2 =

105 (0.262 − 0.3) × 103 = − 26.28 N/mm2 150

(Compressive)

Total stresses are therefore as follows : σs = σs1 + σs2 = − 106.5 N/mm2

(Tensile)

σb = σb1 + σb2 = − 4.56 N/mm

(Compressive)

2

SAQ 4 (a)

A steel rod of cross-sectional area 600 mm2 and a coaxial copper tube of cross-sectional area 1000 mm2 are firmly attached at their ends to form a compound bar. Determine the stress in the steel and in copper when the temperature of the bar is raised by 80°C and an axial tensile force of 60 kN is applied. For steel, E = 200 GNm−2 and α = 11 × 10−6 K−1 For copper, E = 100 GNm−2 and α = 16.5 × 10−6 K−1

(b)

A steel rod of 20 mm diameter passes centrally through a tight fitting copper tube of external diameter 40 mm. The tube is closed with the help of rigid washers of negligible thickness and nuts threaded on the rod. The nuts are tightened till the compressive load on the tube is 50 kN. Determine the stresses in the rod and the tube, when the temperature of the assembly falls by 50°C. For steel, E = 200 GNm−2 and α = 12 × 10−6 K−1 For copper, E = 100 GNm−2 and α = 18 × 10−6 K−1

(c)

A rigid slab weighing 600 kN is placed upon two bronze rods and one steel rod each of 60 cm2 cross-sectional area at a temperature of 15°C. The bronze rods are 25 cm while the steel rod is 30 cm long. Before the slab was placed, the top of all the three rods are level. Find the temperature, at which the stress in the steel rod will be zero. E of steel = 200 GNm−2 and E of bronze = 80 GNm−2 α of steel = 12 × 10−6 K−1 and α of bronze = 18 × 10−6 K−1

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2.8 SUMMARY

Thermal Stresses

In this unit, you have studied (a)

the effect of temperature on materials and bodies,

(b)

the concept of thermal stresses, and

(c)

the method of determination of thermal stresses in uniform, stepped, tapered and compound bars.

2.9 ANSWERS TO SAQs SAQ 1

(a)

9.192 N/mm2

(b)

158.7 N/mm2

SAQ 2

201.6 MPa. SAQ 3

− 1.26 mm. SAQ 4

(a)

94.6 N/mm2, 3.3 N/mm2

(b)

σs = 123.15 N/mm2.

(c)

188.6oC σc = 41.05 N/mm2

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Stresses in Solids

STRENGTH OF MATERIALS While on construction site, in day-to-day work, you come across certain materials, i.e. cement, steel girders, angle iron, circular bars etc. which are regularly used in any construction project. While selecting a suitable material, you are always interested to know its strength. The strength of a material may be defined as an ability to resist its failure and behaviour under the action of external forces. A detailed study of forces and their effects, along with some suitable protective measures for the safe working conditions forms the subject matter of Strength of Materials. As a matter of fact, such knowledge is very essential to enable you to analyse and design all types of structures and machines. Therefore, the knowledge and practice of Strength of Materials forms an integral part of learning in all branches of engineering. This course comprises seven units. Units 1 and 2, entitled “Stresses and Strains” and “Principal Stresses” respectively, deal with simple and principal stresses produced in the material on the basis of which an appropriate section of the member (of the given material) is selected for the construction of given structure. Unit 3, entitled “Shear Forces and Bending Moments”, deals with the most important parameters used in designing, namely shear force and bending moment. This unit starts with the classification of supports, beams and loadings. It covers all types of possible combinations of loads and beams for shear force and bending moment computation through a large number of examples. Unit 4, entitled “Stresses in Beams”, covers the details of bending stresses developed across the cross section of a beam under loading. It also discusses the theory of simple or pure bending and concept of moment of resistance along with some practical applications based on these concepts. The unit concludes with the concepts of shear stress distribution in beams. Unit 5, entitled “Deflection of Beams”, presents the concepts of deflection in beams under the action of external forces. The unit covers the slope and deflection computations for cantilever and simply supported beams under standard loadings. Unit 6, entitled “Torsion”, discusses deformations undergone by solid circular shafts while transmitting power. It also discusses the stresses induced in the circular shafts. Unit 7, entitled “Columns and Struts”, deals with different end conditions of columns during buckling process under the action of axial loading. Towards end, the unit also covers the IS specifications for columns. In each unit, you will find a number of SAQs (Self Assessment Questions) for checking your own progress. You are advised to study the text carefully, and then attempt SAQs on your own and verify your answers with those given at the end of the unit. This will develop your confidence in analysing and solving the practical problems. At the end, we wish you all the best for your all educational endeavours.

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