1. A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of steel is 7850 kg/m3 and E 200 x 103 MN/m2, find the total elongation of the rod. Answer: 8 = 54.33 mm
1. A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of steel is 7850 kg/m3 and E 200 x 103 MN/m2, find the total elongation of the rod. Answer: 8 = 54.33 mm
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![Exercise No. 2: Simple Strain
1. A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended
vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of
steel is 7850 kg/m3 and E = 200 × 103 MN/m2, find the total elongation of the rod.
Answer: 8 = 54.33 mm
2. A steel wire 30 ft long, hanging vertically, supports a load of 500 lb. Neglecting the weight of the
wire, determine the required diameter if the stress is not to exceed 20 ksi and the total
elongation is not to exceed 0.20 in. Assume E = 29 x 106 psi.
Answer: d = 0. 1988 in
II
()](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4410631b-69d0-4886-ac35-d6b82d6689e9%2Fe2ee8654-5342-4284-9861-1f3f47c6a4d7%2Fo757im_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercise No. 2: Simple Strain
1. A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended
vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of
steel is 7850 kg/m3 and E = 200 × 103 MN/m2, find the total elongation of the rod.
Answer: 8 = 54.33 mm
2. A steel wire 30 ft long, hanging vertically, supports a load of 500 lb. Neglecting the weight of the
wire, determine the required diameter if the stress is not to exceed 20 ksi and the total
elongation is not to exceed 0.20 in. Assume E = 29 x 106 psi.
Answer: d = 0. 1988 in
II
()
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