The beam is used to support a dead load of W1 N/m, alive load of W2 kN/m, and a concentrated live load of P kN.Determine (a) the maximum positive (upward) reaction at B, (b)the maximum positive moment at C, and (c) the maximumnegative shear at C. Assume B and D are pins. Influence Lines of Beams (Müller-Breslau's principle)#9|The beam is used to support a dead load of W, N/m, alive load of W2 kN/m, and a concentrated live load of P kN.Determine (a) the maximum positive (upward) reaction at B, (b)the maximum positive moment at C, and (c) the maximumnegative shear at C. Assume B and D are pins.UseW, = Last 2 digits of ID numberW2 = 4th digit of ID numberP = 5th digit of ID numberID # = 18-2363-391W, = 91W2 =P = 6De.g.,IBA|E%3D%3D34 m- 4 m -4 m4 mHint: Dead loads occurs at the whole span unlike live loads

For the above beam, Assume W1=87N/mW2=4kN/mP=3kN

Answered: d live load of P kN. Determine (a) the…

d live load of P kN. Determine (a) the maximum positive (upward) reaction at B, (b) the maximum positive moment at C, and (c) the maximum negative shear at C. Assume B and D are pins.

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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Concept explainers

Beams and Frames

A beam is a horizontal structure or a member which is always subjected to transverse shear loads.

Question

The beam is used to support a dead load of W1 N/m, a
live load of W2 kN/m, and a concentrated live load of P kN.
Determine (a) the maximum positive (upward) reaction at B, (b)
the maximum positive moment at C, and (c) the maximum
negative shear at C. Assume B and D are pins.

Influence Lines of Beams (Müller-Breslau's principle)
#9
|The beam is used to support a dead load of W, N/m, a
live load of W2 kN/m, and a concentrated live load of P kN.
Determine (a) the maximum positive (upward) reaction at B, (b)
the maximum positive moment at C, and (c) the maximum
negative shear at C. Assume B and D are pins.
Use
W, = Last 2 digits of ID number
W2 = 4th digit of ID number
P = 5th digit of ID number
ID # = 18-2363-391
W, = 91
W2 =
P = 6
D
e.g.,
IB
A
|E
%3D
%3D
3
4 m
- 4 m -
4 m
4 m
Hint: Dead loads occurs at the whole span unlike live loads

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