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WorksheetsRecitation 4 - Dr. S
Total questions: 17
Worksheet time: 38mins
1a. If the turnbuckle is subjected to an axial force of P = 900 lb, determine the average normal stress developed in section a-a. σ=AN
2.5 ksi
0.9 ksi
1.8 ksi
3.6 ksi
1b. Determine the yield strength of the material that can occur at section a-a without surpassing a Factor of Safety of 3.
FoS=σaaσY
-5.4 ksi
-2.1 ksi
9.8 ksi
3.4 ksi
1c. If the turnbuckle is subjected to an axial force of P = 900 lb, determine the average normal stress developed in each bolt shanks at B and C. Each bolt shank has a diameter of 0.5 in.
2.4 ksi
4.6 ksi
9.0 ksi
6.6 ksi
1d. If each bolt has a yield strength of 60 ksi, determine the Factor of Safety of each bolt.
2
6
3
13
2a. A specimen is originally 1 ft long, has a diameter of 0.5 in, and is subjected to an axial load of 500 lb. When the force is increased from 500 lb to 1800 lb, the specimen elongates 0.009 in.
Determine the longitudinal strain for the material.
ϵ=LoΔL
0.075
0.00075
0.0075
0.75
2b. A specimen is originally 1 ft long, has a diameter of 0.5 in, and is subjected to an axial load of 500 lb. When the force is increased from 500 lb to 1800 lb, the specimen elongates 0.009 in.
Determine the average normal stress present in the specimen.
σ=AN
2214.54 psi
6620.85 psi
3257.97 psi
8791.31 psi
2c. Given the average normal stress found in the previous question, is the situation linear elastic if the yield strength is 22 ksi?
NO
YES
2d. A specimen is originally 1 ft long, has a diameter of 0.5 in, and is subjected to an axial load of 500 lb. When the force is increased from 500 lb to 1800 lb, the specimen elongates 0.009 in.
Determine the modulus of elasticity for the material if it remains linear elastic.
σ=Eϵ
0.96E3 ksi
8.83E3 ksi
4.86E3 ksi
7.21E3 ksi
3. The wires each have a diameter of 0.5 in, length of 2 ft, and are made of 304 Stainless Steel (Yield Strength = 30 ksi, Young's Modulus = 28E3 ksi). The force P is equal to 6 kip.
Draw your FBD and write the info for the problem. Click here when finished.
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3a. Solve for the tensions of each cable.
NAD = -2 kip
NBC = -4 kip
NAD = 4 kip
NBC = 2 kip
NAD = 2 kip
NBC = 4 kip
NAD = -4 kip
NBC = -2kip
3b. Determine the average normal stress in wire AD and wire BC.
σAD=10.19 ksi σBC=20.37 ksi
σAD=−10.19 ksi σBC=−20.37 ksi
σAD=−20.37 ksi σBC=−10.19 ksi
σAD=20.37 ksi σBC=10.19 ksi
3c. Is this situation linear elastic?
σY=30 ksi
YES
NO
3d. Given that the situation is linear elastic, determine the longitudinal strain of wire AD.
E=28×103 ksi
0.659×10−3
0.143×10−3
0.364×10−3
0.289×10−3
3e. Given that the situation is linear elastic, determine the longitudinal strain of wire BC.
E=28×10−3 ksi
0.728×10−3 ksi
0.251×10−3 ksi
0.893×10−3 ksi
0.545×10−3 ksi
3f. Given the initial length of 2 ft for both wires, determine the change in length of wire AD.
ϵ=LoΔL
6.87×10−3 in
5.13×10−3 in
2.11×10−3 in
8.74×10−3 in
3g. Given the initial length of 2 ft for both wires, determine the change in length of wire BC.
ϵ=LoΔL
4.97×10−3 in
17.46×10−3 in
10.96×10−3 in
8.27×10−3 in
3h. Determine the angle of tilt of the rigid beam AB.
1.4 deg
0.14 deg
0.014 deg
0.0014 deg
