It's urgent please solve as soon as possible Figure 1 shows a statically indeterminate structure supporting a gradient distributed load.Determine the following:-a) The reactions at Points A and Busing the integration method.using the integration method.b) Slope and deflection at mid-span of the structure,c) Sketch and label the bending moment diagram35 (N/m)15 (N/m)15 (m)Figure 1

Answered: Figure 1 shows a statically…

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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It's urgent please solve as soon as possible

Figure 1 shows a statically indeterminate structure supporting a gradient distributed load.
Determine the following:-
a) The reactions at Points A and B
using the integration method.
using the integration method.
b) Slope and deflection at mid-span of the structure,
c) Sketch and label the bending moment diagram
35 (N/m)
15 (N/m)
15 (m)
Figure 1

Expert Solution

Step 1

FBD:

Step 2

The structure is made up of these three components:

Step 3

The deflection at point B due to the specific loading respectively is as follows:

$δ_{B} = \frac{15 \times 15^{4}}{8 E I} δ_{B} = \frac{11 \times 20 \times 15^{4}}{120 E I} δ_{B} = \frac{R_{B} \times 15^{3}}{3 E I}$

B is pin support so the net deflection at B is going to be zero:

$\frac{15 \times 15^{4}}{8 E I} + \frac{11 \times 20 \times 15^{4}}{120 E I} - \frac{R_{B} \times 15^{3}}{3 E I} = 0 R_{B} = 166.875 N (u p w a r d)$

Step 4

Resolving vertical forces:

$\sum_{}^{} F_{y} = 0 R_{A} + R_{B} = (15 \times 15) + (\frac{1}{2} \times 15 \times 20) R_{A} = 375 - 166.875 = 208.125 N$

Take moment about point A:

$\sum_{}^{} M_{A} = 0 M_{A} + (166.875 \times 15) - [15 \times 15 \times 7.5] - [\frac{1}{2} \times 20 \times 15 \times \frac{2}{3} \times 15] = 0 M_{A} = 684.375 N - m (a n t i c l o c k w i s e)$

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