2.8^2.8. DEFORMATIONS OF MEMBERS UNDER AXIAL LOADING^67^77^,,^1774^1999%
2.8
DEFORMATIONS OF MEMBERS UNDER AXIAL LOADING

Consider a homogeneous rod BC of length L and uniform cross section of area A subjected to a centric axial load P (Fig. 2.17). If the resulting axial stress σ = P/A does not exceed the proportional limit of the material, we may apply Hooke's law and write

(2.4)

from which it follows that

(2.5)

Recalling that the strain ϵ was defined in Sec. 2.2 as ϵ = δ/L, we have

(2.6)
Fig. 2.17 Deformation of axially loaded rod.

and, substituting for ϵ from (2.5) into (2.6):

(2.7)
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Equation (2.7) may be used only if the rod is homogeneous (constant E), has a uniform cross section of area A, and is loaded at its ends. If the rod is loaded at other points, or if it consists of several portions of various cross sections and possibly of different materials, we must divide it into component parts that satisfy individually the required conditions for the application of formula (2.7). Denoting, respectively, by Pi, Li, Ai, and Ei the internal force, length, cross-sectional area, and modulus of elasticity corresponding to part i, we express the deformation of the entire rod as

(2.8)

We recall from Sec. 2.2 that, in the case of a member of variable cross section (Fig. 2.18), the strain ϵ depends upon the position of the point Q where it is computed and is defined as ϵ = dδ/dx. Solving for dδ and substituting for ϵ from Eq. (2.5), we express the deformation of an element of length dx as

The total deformation δ of the member is obtained by integrating this expression over the length L of the member:

(2.9)

Formula (2.9) should be used in place of (2.7), not only when the cross-sectional area A is a function of x, but also when the internal force P depends upon x, as is the case for a rod hanging under its own weight.

Fig. 2.18 Deformation of axially loaded member of variable cross-sectional area.
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EXAMPLE 2.01

Determine the deformation of the steel rod shown in Fig. 2.19a under the given loads (E = 29 × 106 psi).

Fig. 2.19

We divide the rod into three component parts shown in Fig. 2.19b and write

To find the internal forces P1, P2, and P3, we must pass sections through each of the component parts, drawing each time the free-body diagram of the portion of rod located to the right of the section (Fig. 2.19c). Expressing that each of the free bodies is in equilibrium, we obtain successively

Carrying the values obtained into Eq. (2.8), we have

The rod BC of Fig. 2.17, which was used to derive formula (2.7), and the rod AD of Fig. 2.19, which has just been discussed in Example 2.01, both had one end attached to a fixed support. In each case, therefore, the deformation δ of the rod was equal to the displacement of its free end. When both ends of a rod move, however, the deformation of the rod is measured by the relative displacement of one end of the rod with respect to the other. Consider, for instance, the assembly shown in Fig. 2.20a, which consists of three elastic bars of length L connected by a rigid pin at A. If a load P is applied at B (Fig. 2.20b), each of the three bars will deform. Since the bars AC and AC′ are attached to fixed supports at C and C, their common deformation is measured by the displacement δA of point A. On the other hand, since both ends of bar AB move, the deformation of AB is measured by the difference between the displacements δA and δB of points A and B, i.e., by the relative displacement of B with respect to A. Denoting this relative displacement by δB/A, we write

(2.10)

where A is the cross-sectional area of AB and E is its modulus of elasticity.

Fig. 2.20 Example of relative end displacement, as exhibited by the middle bar.
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SAMPLE

SAMPLE PROBLEM 2.1

The rigid bar BDE is supported by two links AB and CD. Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2; link CD is made of steel (E = 200 GPa) and has a cross-sectional area of 600 mm2. For the 30-kN force shown, determine the deflection (a) of B, (b) of D, (c) of E.

SAMPLE

SAMPLE

SOLUTION
Free Body: Bar BDE
a. Deflection of B.

Since the internal force in link AB is compressive, we have P = −60 kN

The negative sign indicates a contraction of member AB, and, thus, an upward deflection of end B:

b. Deflection of D.

Since in rod CD, P = 90 kN, we write

c. Deflection of E.

We denote by B′ and D′ the displaced positions of points B and D. Since the bar BDE is rigid, points B′, D′, and E′ lie in a straight line and we write

SAMPLE

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SAMPLE

SAMPLE PROBLEM 2.2

The rigid castings A and B are connected by two -in.-diameter steel bolts CD and GH and are in contact with the ends of a 1.5-in.-diameter aluminum rod EF. Each bolt is single-threaded with a pitch of 0.1 in., and after being snugly fitted, the nuts at D and H are both tightened one-quarter of a turn. Knowing that E is 29 × 106 psi for steel and 10.6 × 106 psi for aluminum, determine the normal stress in the rod.

SAMPLE

SAMPLE

SOLUTION
Deformations
Bolts CD and GH.

Tightening the nuts causes tension in the bolts. Because of symmetry, both are subjected to the same internal force Pb and undergo the same deformation δb. We have

(1)
Rod EF.

The rod is in compression. Denoting by Pr the magnitude of the force in the rod and by δr the deformation of the rod, we write

(2)
Displacement of D Relative to B.

Tightening the nuts one-quarter of a turn causes ends D and H of the bolts to undergo a displacement of relative to casting B. Considering end D, we write

(3)

But δD/B = δD − δB, where δD and δB represent the displacements of D and B. If we assume that casting A is held in a fixed position while the nuts at D and H are being tightened, these displacements are equal to the deformations of the bolts and of the rod, respectively. We have, therefore,

(4)

Substituting from (1), (2), and (3) into (4), we obtain

(5)
Free Body: Casting B
(6)
Forces in Bolts and Rod

Substituting for Pr from (6) into (5), we have

Stress in Rod

SAMPLE

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PROBLEMS
2.1 An 80-m-long wire of 5-mm diameter is made of a steel with E = 200 GPa and an ultimate tensile strength of 400 MPa. If a factor of safety of 3.2 is desired, determine (a) the largest allowable tension in the wire, (b) the corresponding elongation of the wire.
2.2 A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2-kip tensile load is applied to it. Knowing that E = 29 × 106 psi, determine (a) the smallest diameter rod that should be used, (b) the corresponding normal stress caused by the load.
2.3 Two gage marks are placed exactly 10 in. apart on a -in.-diameter aluminum rod with E = 10.1 × 106 psi and an ultimate strength of 16 ksi. Knowing that the distance between the gage marks is 10.009 in. after a load is applied, determine (a) the stress in the rod, (b) the factor of safety.
2.4 An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam. It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that E = 200 GPa, determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire.
2.5 A polystyrene rod of length 12 in. and diameter 0.5 in. is subjected to an 800-lb tensile load. Knowing that E = 0.45 × 106 psi, determine (a) the elongation of the rod, (b) the normal stress in the rod.
2.6 A nylon thread is subjected to a 8.5-N tension force. Knowing that E = 3.3 GPa and that the length of the thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.
2.7 Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod. Knowing that, with an axial load of 6000 N acting on the rod, the distance between the gage marks is 250.18 mm, determine the modulus of elasticity of the aluminum used in the rod.
2.8 An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that E = 10.1 × 106 psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips.
2.9 An aluminum control rod must stretch 0.08 in. when a 500-lb tensile load is applied to it. Knowing that σall=22 ksi and E = 10.1 × 106 psi, determine the smallest diameter and shortest length that can be selected for the rod.
 
2.10
A square yellow-brass bar must not stretch more than 2.5 mm when it is subjected to a tensile load. Knowing that E = 105 GPa and that the allowable tensile strength is 180 MPa, determine (a) the maximum allowable length of the bar, (b) the required dimensions of the cross section if the tensile load is 40 kN.
2.11 A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when the rod is subjected to a 10-kN axial load. Knowing that E = 200 GPa, determine the required diameter of the rod.
2.12 A nylon thread is to be subjected to a 10-N tension. Knowing that E = 3.2 GPa, that the maximum allowable normal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine the required diameter of the thread.
2.13 The 4-mm-diameter cable BC is made of a steel with E = 200 GPa. Knowing that the maximum stress in the cable must not exceed 190 MPa and that the elongation of the cable must not exceed 6 mm, find the maximum load P that can be applied as shown.
Fig. P2.13
2.14 The aluminum rod ABC (E = 10.1 × 106 psi), which consists of two cylindrical portions AB and BC, is to be replaced with a cylindrical steel rod DE (E = 29 × 106 psi) of the same overall length. Determine the minimum required diameter d of the steel rod if its vertical deformation is not to exceed the deformation of the aluminum rod under the same load and if the allowable stress in the steel rod is not to exceed 24 ksi.
Fig. P2.14
2.15 A 4-ft section of aluminum pipe of cross-sectional area 1.75 in2 rests on a fixed support at A. The -in.-diameter steel rod BC hangs from a rigid bar that rests on the top of the pipe at B. Knowing that the modulus of elasticity is 29 × 106 psi for steel and 10.4 × 106 psi for aluminum, determine the deflection of point C when a 15-kip force is applied at C.
Fig. P2.15
 
2.16
The brass tube AB (E = 105 GPa) has a cross-sectional area of 140 mm2 and is fitted with a plug at A. The tube is attached at B to a rigid plate that is itself attached at C to the bottom of an aluminum cylinder (E = 72 GPa) with a cross-sectional area of 250 mm2. The cylinder is then hung from a support at D. In order to close the cylinder, the plug must move down through 1 mm. Determine the force P that must be applied to the cylinder.
Fig. P2.16
2.17 A 250-mm-long aluminum tube (E = 70 GPa) of 36-mm outer diameter and 28-mm inner diameter can be closed at both ends by means of single-threaded screw-on covers of 1.5-mm pitch. With one cover screwed on tight, a solid brass rod (E = 105 GPa) of 25-mm diameter is placed inside the tube and the second cover is screwed on. Since the rod is slightly longer than the tube, it is observed that the cover must be forced against the rod by rotating it one-quarter of a turn before it can be tightly closed. Determine (a) the average normal stress in the tube and in the rod, (b) the deformations of the tube and of the rod.
Fig. P2.17
2.18 The specimen shown is made from a 1-in.-diameter cylindrical steel rod with two 1.5-in.-outer-diameter sleeves bonded to the rod as shown. Knowing that E = 29 × 106 psi, determine (a) the load P so that the total deformation is 0.002 in., (b) the corresponding deformation of the central portion BC.
Fig. P2.18
2.19 Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B.
Fig. P2.19 and P2.20
2.20 The rod ABC is made of an aluminum for which E = 70 GPa. Knowing that P = 6 kN and Q = 42 kN, determine the deflection of (a) point A, (b) point B.
 
2.21
Members AB and BC are made of steel (E = 29 × 106 psi) with cross-sectional areas of 0.80 in2 and 0.64 in2, respectively. For the loading shown, determine the elongation of (a) member AB, (b) member BC.
Fig. P2.21
2.22 The steel frame (E = 200 GPa) shown has a diagonal brace BD with an area of 1920 mm2. Determine the largest allowable load P if the change in length of member BD is not to exceed 1.6 mm.
Fig. P2.22
2.23 For the steel truss (E = 200 GPa) and loading shown, determine the deformations of members AB and AD, knowing that their cross-sectional areas are 2400 mm2 and 1800 mm2, respectively.
Fig. P2.23
2.24 For the steel truss (E = 29 × 106 psi) and loading shown, determine the deformations of members BD and DE, knowing that their cross-sectional areas are 2 in2 and 3 in2, respectively.
Fig. P2.24
 
2.25
Each of the links AB and CD is made of aluminum (E = 10.9 × 106 psi) and has a cross-sectional area of 0.2 in2. Knowing that they support the rigid member BC, determine the deflection of point E.
Fig. P2.25
2.26 The length of the -in.-diameter steel wire CD has been adjusted so that with no load applied, a gap of in. exists between the end B of the rigid beam ACB and a contact point E. Knowing that E = 29 × 106 psi, determine where a 50-lb block should be placed on the beam in order to cause contact between B and E.
Fig. P2.26
2.27 Link BD is made of brass (E = 105 GPa) and has a cross-sectional area of 240 mm2. Link CE is made of aluminum (E = 72 GPa) and has a cross-sectional area of 300 mm2. Knowing that they support rigid member ABC, determine the maximum force P that can be applied vertically at point A if the deflection of A is not to exceed 0.35 mm.
Fig. P2.27
 
2.28
Each of the four vertical links connecting the two rigid horizontal members is made of aluminum (E = 70 GPa) and has a uniform rectangular cross section of 10 × 40 mm. For the loading shown, determine the deflection of (a) point E, (b) point F, (c) point G.
Fig. P2.28
2.29 The vertical load P is applied at the center A of the upper section of a homogeneous frustum of a circular cone of height h, minimum radius a, and maximum radius b. Denoting by E the modulus of elasticity of the material and neglecting the effect of its weight, determine the deflection of point A.
Fig. P2.29
2.30 A homogenous cable of length L and uniform cross section is suspended from one end. (a) Denoting by ρ the density (mass per unit volume) of the cable and by E its modulus of elasticity, determine the elongation of the cable due to its own weight. (b) Show that the same elongation would be obtained if the cable were horizontal and if a force equal to half of its weight were applied at each end.
2.31 The volume of a tensile specimen is essentially constant while plastic deformation occurs. If the initial diameter of the specimen is d1, show that when the diameter is d, the true strain is ϵt = 2 ln(d1/d).
2.32 Denoting by ϵ the “engineering strain” in a tensile specimen, show that the true strain is ϵt = ln(1 + ϵ).