Pearson eText for Materials for Civil and Construction Engineers -- Instant Access (Pearson+)
Pearson eText for Materials for Civil and Construction Engineers -- Instant Access (Pearson+)
4th Edition
ISBN: 9780137505586
Author: Michael Mamlouk, John Zaniewski
Publisher: PEARSON+
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Chapter 11, Problem 11.27QP

(a)

To determine

The modulus of elasticity of the reinforced concrete.

(b)

To determine

The load carried by the each of the steel and plain concrete.

(c)

To determine

The minimum required cross sectional area of the column.

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A representative reinforced concrete composite beam section is shown in the figure below. The number of reinforcement is 6 and the diameter of the reinforcement is 16 meters. cross section Determine the neutral axis (x) and calculate the maximum normal stresses that will occur in concrete and steel. It will be assumed that the concrete does not carry tensile stress. The modulus of elasticity of the steel is 200000 Mpa. The modulus of elasticity of concrete is 28000 Mpa. The section width is b= 400 mm. Take the cover as 50 mm. (The maximum bending moment acting on the section is 59 kNm and d= 700 mm.)
A short reinforced concrete column is subjected to a 1000 kN axial compressive load. The moduli of elasticity of plain concrete and steel are 25 and207 GPa, respectively, and the cross-sectional area of steel is 2% of that of thereinforced concrete. Considering the column as a structural member made of a composite material and subjected to load parallel to the steel rebars, calculate the following:a. the modulus of elasticity of the reinforced concreteb. the load carried by each of the steel and plain concretec. the minimum required cross-sectional area of the column given that the allowable compressive stress of plain concrete is 20 MPa and that the allowable compressive stress of plain concrete will be reached before that of steel.
The composite bar shown in the figure is firmly attached to unyielding supports. An Axial force P 52 kips is applied at 65° F. Compute the stress (ksi) in the Aluminum at 116 °F. Assume a = 6.5 x 106/° F for steel and 12.8 x 106°F for aluminum Given: L1 = 17in; L2 = 13in %3D %3D Your final answer should contain two decimal places Aluminum A = 2 in? E = 10 x 10 psi Steel A = 3 in? E = 29 x 10 psi L1 - L2
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