Answered: E Cable 1.5 m A B D -2.0 m- - 2.0 m-…

E Cable 1.5 m A B D -2.0 m- - 2.0 m- –2.0 m- W = 27 kN (a) E Moment release Cable 1.5 m B D -2.0 m 2.0 m- -2.0 m- W = 27 kN (b) PROBLEM 4.3-26

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 11.2.15P: Repeat Problem 11.2-14 using L = 12 ft, ß = 0.25 kips/in., ßRl= 1.5ßL2, and ßR2= 2 ßR1.

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Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.

Question

The simply supported beam ABCD is loaded by a weight W = 27 kN through the arrangement shown in the figure part a. The cable passes over a small frictionless pulley at B and is attached at E to the end of the vertical arm. a. Calculate the axial force JV, shear force V, and bending moment M at section C, which is just to the left of the vertical arm. Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations. b. Repeat part (a) if a roller support is added at C and a moment release is inserted just left of C (see figure part b).

E
Cable
1.5 m
A
B
D
-2.0 m-
- 2.0 m-
–2.0 m-
W = 27 kN
(a)
E
Moment release
Cable
1.5 m
B
D
-2.0 m 2.0 m-
-2.0 m-
W = 27 kN
(b)
PROBLEM 4.3-26

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