Y=25kN/m, P1=24kN and P2=7kNusing method of section P1 P2ACDВy2 m1.5 m 0.5 m 1) Reactions at A and B2) Shear diagram3) Bending moment diagram4) Maximum absolute bending moment diagram5) Qualitative deflected shape

Answered: Reactions at A and B Shear diagram…

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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Question

Y=25kN/m, P1=24kN and P2=7kN

using method of section

1) Reactions at A and B
2) Shear diagram
3) Bending moment diagram
4) Maximum absolute bending moment diagram
5) Qualitative deflected shape

Expert Solution

Step 1

"As per our guidelines we are supposed to solve first three sub parts, kindly submit other sub parts separately and mention their question number. Thank you".

Draw the free body diagram.

Equilibrium of horizontal forces,

$\begin{array}{rcl} \sum F_{x} & = & 0 \\ A_{x} & = & 0 \end{array}$

Step 2

Equilibrium of vertical forces,

$\sum F_{y} = 0 A_{y} + 25 (2) - 24 - 7 = 0 A_{y} = - 19 kN$

Taking moment at A,

$\begin{array}{rcl} \sum M_{A} & = & 0 \\ M_{A} & = & 25 (2) (\frac{2}{2}) - 24 (3.5) - 7 (4) = 0 \\ M_{A} & = & - 62 kNm \end{array}$

Step 3

It is required to determine the shear force.

$\begin{array}{rcl} S F_{A} & = & - 19 kN \\ S F_{C} & = & - 19 + 25 (2) = 31 kN \\ S F_{D} & = & 31 - 24 = 7 kN \\ S F_{B} & = & 7 - 7 = 0 \end{array}$

It is required to determine the location of point of contraflexure.

Take a section X-X at a distance of x from A.

$\sum F_{X - X} = 0 - 19 + 25 (x) = 0 x = 0.76 m$

The maximum bending moment occurs at a distance of 0.76 m from A.

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