Answered: 6 kN 1.5 kN/m В A

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

The beam is subjected to the load shown. Draw the normal, shear and moment diagrams for the beam AB. What is the absolute maximum bending moment?

Expert Solution

Step 1

Draw the free-body diagram of the beam as shown below,

The reaction at the support and moment at A is,

$R_{A} = 6 + 1.5 \times 2.5 R_{A} = 9.75 k N T h e m o m e n t a t A i s, M_{A} = (1.5 \times 2.5 \times \frac{2.5}{2}) + (6 \times 5) M_{A} = 34.6875 k N \cdot m$

Step 2

Calculation for SFD:

$F o r S e c t i o n A C, F_{x} = R_{A} - 1.5 x F_{x} = 9.75 - 1.5 x A t A, x = 0 F_{A} = 9.75 - 1.5 \times 0 F_{A} = 9.75 k N A t C, F_{C} = 9.75 - 1.5 \times 2.5 F_{C} = 6 k N A t B, F_{B} = F_{C} - 6 F_{B} = 6 - 6 = 0$

The calculation for BMD:

$F o r b e n d i n g m o m e n t l e t u s t a k e s e c t i o n f r o m B, F o r s e c t i o n A C, M_{x} = 6 \times x + 1.5 \times (x - 2.5) \times \frac{(x - 2.5)}{2} M_{x} = 6 x + 0.75 {(x - 2.5)}^{2} W h e r e i s t h e d i s \tan c e f r o m B, A t C, x = 2.5 m M_{C} = 6 \times 2.5 + 0.75 {(2.5 - 2.5)}^{2} M_{C} = 15 k N \cdot m A t A, x = 5 m M_{A} = 6 \times 5 + 0.75 \times {(5 - 2.5)}^{2} M_{A} = 34.6875 k N \cdot m$

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