Determine the displacement of points B and C of the beam shown in Fig. a. EI is constant. MBC MoEI(a)хtan Btan AMoГвA — VвEIBIC/A = Vc4(b)(c)tan C

Answered: M B C Mo EI (a) х tan B tan A Mo ГвA —…

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

Determine the displacement of points B and C of the beam shown in Fig. a. EI is constant.

M
B
C Mo
EI
(a)
х
tan B
tan A
Mo
ГвA — Vв
EI
B
IC/A = Vc
4
(b)
(c)
tan C

Expert Solution

Step 1

Given : A bending moment $M_{o}$ is applied at point C of the cantilever beam of length $L$ and $E I$ is constant where $E$ is the young's modulus and $I$ is the moment of inertia.

To determine the deflection at point B and C we apply the application of area moment method which can be calculated as

$t_{\frac{B}{A}} = m o m e n t a r e a o f \frac{M}{E I} d i a g r a m b e t w e e n A a n d B$ and $t_{\frac{C}{A}} = m o m e n t a r e a o f \frac{M}{E I} d i a g r a m b e t w e e n C a n d A$

The above method is valid when the point of zero slope is known that is the reference point and another point is the origin which is the point of non-zero slope.

Step 2

The area moment diagram and point of deflection of the given cantilever beam is shown below

Step 3

For displacement of point C-

The reference point that is point of zero slope is A and origin that is point of non zero slope is C

$t_{\frac{C}{A}} = v_{C} - v_{A} = \frac{1}{E I} [A_{1} x_{1} + A_{2} x_{2}]$

where $A_{1}$ and $A_{2}$ is the area of bending moment diagram while $x_{1}$ and $x_{2}$ is measured from the origin

Therefore $t_{\frac{C}{A}} = v_{C} - v_{A} = \frac{1}{E I} [M_{0} \frac{L}{2} \times \frac{L}{2} + M_{0} \frac{L}{2} \times \frac{L}{2}] t_{\frac{C}{A}} = v_{C} = \frac{M_{0}}{E I} [\frac{L^{2}}{4} + \frac{L^{2}}{4}] = \frac{M_{0}}{2 E I} L^{2}$

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