Question

a 2 - in. - diameter round steel rod 10 ft long is subjected to a tensile force of 60 kips. how much will it elongate under the load? the area of the 2 - in. rod is 3.1416 in.². checking to determine whether the stress in the bar is within the elastic limit, we find that (f=\frac{p}{a}=\frac{60}{3.142}=19.1) ksi. this is within the elastic limit of ordinary structural steel (50 ksi), so the formula for finding the deformation is applicable. from data, (p = 60) kips, (l = 120) (length in inches), (a = 3.1416), (e = 29,000,000). substituting these values, we calculate the lengthening of the rod as (delta l=\frac{pl}{ae}=\frac{(60,000)(120)}{(3.142)(29,000,000)} = 0.079) in. problem 2.2.a what force must be applied to a steel bar 1 in. square and 12 in. long to produce an elongation of 0.016 in.? problem 2.2.b how much will a nominal 8×8 - in. douglas fir post 12 ft long shorten under an axial load of 45 kips? problem 2.2.c a routine quality control test is made on a structural steel bar that is 1 in. square and 16 in. long. the data developed during the test show that the bar elongated 0.0111 in. when subjected to a tensile force of 20.5 kips. compute the modulus of elasticity of the steel.

Explanation:

Step1: Identify relevant formula

The formula for elongation of a bar under axial - load is $\Delta L=\frac{PL}{AE}$, where $P$ is the applied load, $L$ is the original length of the bar, $A$ is the cross - sectional area, and $E$ is the modulus of elasticity.

Step2: For Problem 2.2A

We need to find $P$. Rearranging the formula gives $P=\frac{\Delta LAE}{L}$. Given $\Delta L = 0.016$ in, assume $E = 29000000$ psi (modulus of elasticity of steel) and $A=1$ in², $L$ (not given in the part of the problem shown, assume a standard value if needed). But if we assume $L = 12$ in (for example, if it was a standard length in the context), then $P=\frac{0.016\times1\times29000000}{12}\approx386667$ lb or $386.67$ kips.

Step3: For Problem 2.2B

First, find the cross - sectional area of the Douglas fir post $A = 8\times8=64$ in². Given $P = 45000$ lb, assume $E$ for Douglas fir (say $E = 1500000$ psi). Using $\Delta L=\frac{PL}{AE}$, with $L$ (assume $L = 12\times12 = 144$ in if 12 ft long), $\Delta L=\frac{45000\times144}{64\times1500000}=0.0675$ in.

Step4: For Problem 2.2C

We need to find $E$. Rearranging the formula $\Delta L=\frac{PL}{AE}$ gives $E=\frac{PL}{A\Delta L}$. Given $P = 20500$ lb, $A = 1$ in², $L = 16$ in, $\Delta L=0.0111$ in. Then $E=\frac{20500\times16}{1\times0.0111}\approx29279279$ psi.

Answer:

Problem 2.2A: Force value depends on assumed length (example: 386.67 kips with assumed length).
Problem 2.2B: Elongation is 0.0675 in (with assumed values).
Problem 2.2C: Modulus of elasticity is approximately 29279279 psi.