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
from which it follows that
Recalling that the strain ϵ was defined in Sec. 2.2 as ϵ = δ/L, we have
and, substituting for ϵ from (2.5) into (2.6):
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 P_{i}, L_{i}, A_{i}, and E_{i} the internal force, length, crosssectional area, and modulus of elasticity corresponding to part i, we express the deformation of the entire rod as
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:
Formula (2.9) should be used in place of (2.7), not only when the crosssectional 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.
Determine the deformation of the steel rod shown in Fig. 2.19a under the given loads (E = 29 × 10^{6} psi).
We divide the rod into three component parts shown in Fig. 2.19b and write
To find the internal forces P_{1}, P_{2}, and P_{3}, we must pass sections through each of the component parts, drawing each time the freebody 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 δ_{A} 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 δ_{A} and δ_{B} of points A and B, i.e., by the relative displacement of B with respect to A. Denoting this relative displacement by δ_{B/A}, we write
where A is the crosssectional area of AB and E is its modulus of elasticity.
SAMPLE
The rigid bar BDE is supported by two links AB and CD. Link AB is made of aluminum (E = 70 GPa) and has a crosssectional area of 500 mm^{2}; link CD is made of steel (E = 200 GPa) and has a crosssectional area of 600 mm^{2}. For the 30kN force shown, determine the deflection (a) of B, (b) of D, (c) of E.
SAMPLE
SAMPLE
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:
Since in rod CD, P = 90 kN, we write
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
SAMPLE
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 1.5in.diameter aluminum rod EF. Each bolt is singlethreaded with a pitch of 0.1 in., and after being snugly fitted, the nuts at D and H are both tightened onequarter of a turn. Knowing that E is 29 × 10^{6} psi for steel and 10.6 × 10^{6} psi for aluminum, determine the normal stress in the rod.
SAMPLE
SAMPLE
Tightening the nuts causes tension in the bolts. Because of symmetry, both are subjected to the same internal force P_{b} and undergo the same deformation δ_{b}. We have
The rod is in compression. Denoting by P_{r} the magnitude of the force in the rod and by δ_{r} the deformation of the rod, we write
Tightening the nuts onequarter 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
But δ_{D/B} = δ_{D} − δ_{B}, where δ_{D} and δ_{B} 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,
Substituting from (1), (2), and (3) into (4), we obtain
SAMPLE
2.1  An 80mlong wire of 5mm 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 2kip tensile load is applied to it. Knowing that E = 29 × 10^{6} 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 × 10^{6} 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 18mlong steel wire of 5mm 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 800lb tensile load. Knowing that E = 0.45 × 10^{6} psi, determine (a) the elongation of the rod, (b) the normal stress in the rod. 
2.6  A nylon thread is subjected to a 8.5N 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 12mmdiameter 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 × 10^{6} 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 500lb tensile load is applied to it. Knowing that σ_{all}=22 ksi and E = 10.1 × 10^{6} psi, determine the smallest diameter and shortest length that can be selected for the rod. 
A square yellowbrass 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 4mlong 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 10kN 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 10N 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 4mmdiameter 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.

2.14  The aluminum rod ABC (E = 10.1 × 10^{6} psi), which consists of two cylindrical portions AB and BC, is to be replaced with a cylindrical steel rod DE (E = 29 × 10^{6} 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.

2.15  A 4ft section of aluminum pipe of crosssectional area 1.75 in^{2} 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 × 10^{6} psi for steel and 10.4 × 10^{6} psi for aluminum, determine the deflection of point C when a 15kip force is applied at C.

The brass tube AB (E = 105 GPa) has a crosssectional area of 140 mm^{2} 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 crosssectional area of 250 mm^{2}. 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.


2.17  A 250mmlong aluminum tube (E = 70 GPa) of 36mm outer diameter and 28mm inner diameter can be closed at both ends by means of singlethreaded screwon covers of 1.5mm pitch. With one cover screwed on tight, a solid brass rod (E = 105 GPa) of 25mm 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 onequarter 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.

2.18  The specimen shown is made from a 1in.diameter cylindrical steel rod with two 1.5in.outerdiameter sleeves bonded to the rod as shown. Knowing that E = 29 × 10^{6} psi, determine (a) the load P so that the total deformation is 0.002 in., (b) the corresponding deformation of the central portion BC.

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.

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. 
Members AB and BC are made of steel (E = 29 × 10^{6} psi) with crosssectional areas of 0.80 in^{2} and 0.64 in^{2}, respectively. For the loading shown, determine the elongation of (a) member AB, (b) member BC.


2.22  The steel frame (E = 200 GPa) shown has a diagonal brace BD with an area of 1920 mm^{2}. Determine the largest allowable load P if the change in length of member BD is not to exceed 1.6 mm.

2.23  For the steel truss (E = 200 GPa) and loading shown, determine the deformations of members AB and AD, knowing that their crosssectional areas are 2400 mm^{2} and 1800 mm^{2}, respectively.

2.24  For the steel truss (E = 29 × 10^{6} psi) and loading shown, determine the deformations of members BD and DE, knowing that their crosssectional areas are 2 in^{2} and 3 in^{2}, respectively.

Each of the links AB and CD is made of aluminum (E = 10.9 × 10^{6} psi) and has a crosssectional area of 0.2 in^{2}. Knowing that they support the rigid member BC, determine the deflection of point E.


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 × 10^{6} psi, determine where a 50lb block should be placed on the beam in order to cause contact between B and E.

2.27  Link BD is made of brass (E = 105 GPa) and has a crosssectional area of 240 mm^{2}. Link CE is made of aluminum (E = 72 GPa) and has a crosssectional area of 300 mm^{2}. 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.

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.


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.

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 d_{1}, show that when the diameter is d, the true strain is ϵ_{t} = 2 ln(d_{1}/d). 
2.32  Denoting by ϵ the “engineering strain” in a tensile specimen, show that the true strain is ϵ_{t} = ln(1 + ϵ). 