8. Using the value obtained in the previous problem, which of the following mostnearly gives the deformation caused by the applied load?c. 34.04 mma. 31.59 mmb. 29.40 mmd. 35.00 mm Situation 2: A steel rod having a length of 50 m is suspended vertically fromone end. It supports a tensile load of 250 kN at the lower end. The unit weight ofthe steel is 7,850 kg/m³. The allowable tensile stress for steel is 140 MPa. Use E= 200 GPa.

Answered: Image /qna-images/answer/65b9c6b8-afc0-4dfb-8692-6d119850a711.jpg

Answered: Situation 2: A steel rod having a…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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A bar made from aluminum (E = 72 GPa) has a diameter of 12 mm. When the bar is subjected to a tensile axial force P, its diameter is decreased by 0.018 mm. Assume the Poisson’s ratio to be 0.33. Determine the value of loads P and its axial deformation. P = ? N δ = ? mm
The rigid bar of negligible weight is supported as shown in the figure. If W = 104 kN, compute thetemperature change of the assembly that will cause a tensile stress of 50 MPa in the steel rod. Use the following data:
The rigid bar of negligible weight is supported as shown in the figure. If W = 80 kN, compute the 2 temperature change of the assembly that will cause a tensile stress of 50 MPa in the steel rod. Use the following data: A (in) a (PF) E (GPa) Steel Rod 300 11.7 x 104 200 Bronze Rod 1400 18 x 106 83 Bronze 3 m Steel 1.5 m 1.0 m 2.5 m 1.5 m W
Situation 4: An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in Figure. Axial loads are applied at the positions indicated. Steel A = 550 mm² 10. Compute the axial stress at the bronze 11. Compute the axial stress at the aluminum 12. Compute the axial stress at the steel 60 KN 3.0 m Aluminum A = 400 mm² 2.0 m Bronze A = 220 mm² 20 KN 1.0 m 30 kN
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SITUATION. As shown in the figure below, a rigid bar with negligible mass is pinned at O and attached to two vertical rods. Assume that the rods were initially stress-free. Allowable stress in steel is 120 MPa and in bronze is 60 MPa. For this problem, a = 1.8m; b = 1.3m; c= 1.7m. a b C P Steel: A-900 mm² Bronze: A=1200 mm² E = 83 GPa L=2.0 m E = 200 GPa L=1.5m What is the value of P without exceeding the allowable stress of bronze, in kN? O 161 O 174 O 145 124 4
two rods attached to a rigid bar AB with negligible mass . A 101 kN load was applied at the end of this bar. The computed stresses for steel and bronze are Blank 1 mPa and Blank 2 mPa respectively.
MN=85
A rigid bar of negligible weight is supported as shown below. Compute the value of the load W that will cause the stress in the steel rod to be 55 MPa. Bronze L = 3 m A = 1300 mm? E = 83 GPa Steel L = 1.5 m A = 320 mm? E = 200 GPa - 1 m %3D 2.5 m 1.5 m w/
The assembly shown in figure below consists of a light rigid bar AB, pinned at O, that is attached to the steel and aluminum rods. In the position shown, bar AB is horizontal and there is a gap, A = 5 mm, between the lower end of the steel rod and its pin support at C. Compute the stress in the aluminum rod when the lower end of the steel rod is touch to its support. A 0.75 m Steel A = 250 mm E = 200 GPa *i* O 1.5 m Aluminum L = 2m A = 300 mm' E = 70 GPa B D
1. The column in the figure shown has a strain on its compressive edge equal to -0.003 and has a tensile strain of +0.00185 on its other edge. Detemine the nominal values of Pn and Mn that cause this strain distribution if f'c = 21 MPa, fy = 414 MPa. Draw the column interaction diagram. en 600 Pn mm 8 - 6 32mm 75 225 225 75 600

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Situation 2: A steel rod having a length of 50 m is suspended vertically from one end. It supports a tensile load of 250 kN at the lower end. The unit weight of the steel is 7,850 kg/m³. The allowable tensile stress for steel is 140 MPa. Use E = 200 GPa.