= 1 m %3D -2 = 0.3 m The column BD, Figure 1, is mnade of aluminum and has a circular cross section of 0.1m diameter. The = diameter of pin A is 10 mm. It's required to calculate: (a) Compressive stress in the link BD (b) Change in the length of the link BD, ALBD (take E=70GPA) (c) Shear stress in pin A. (Pin A is subjected to double shear) A 3 = 0.5 m P = 3 kN d = 0.1 m D .. ..
= 1 m %3D -2 = 0.3 m The column BD, Figure 1, is mnade of aluminum and has a circular cross section of 0.1m diameter. The = diameter of pin A is 10 mm. It's required to calculate: (a) Compressive stress in the link BD (b) Change in the length of the link BD, ALBD (take E=70GPA) (c) Shear stress in pin A. (Pin A is subjected to double shear) A 3 = 0.5 m P = 3 kN d = 0.1 m D .. ..
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter11: Columns
Section: Chapter Questions
Problem 11.6.14P: The wide-flange, pinned-end column shown in the figure carries two loads: a force P1= 450 kN acting...
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