determine MaВ, МВА, МвС, Мсв, Мср, Mрс Using Slope Deflection Method for Figure 03 and all reactions at support. Use I = 1/12 b(h3) and E = 2.10x107 KPa Figure 03 1090 KN/m B 4.25m 5.15m b2= 0.55m b3= 0.75m 545 KN h2= 0.75m h3= 0.75m 9.95m L4, m b1= 0.75m 3.22m h1= 0.75m 3.28m

determine MaВ, МВА, МвС, Мсв, Мср, Mрс Using Slope Deflection Method for Figure 03 and all reactions at support. Use I = 1/12 b(h3) and E = 2.10x107 KPa Figure 03 1090 KN/m B 4.25m 5.15m b2= 0.55m b3= 0.75m 545 KN h2= 0.75m h3= 0.75m 9.95m L4, m b1= 0.75m 3.22m h1= 0.75m 3.28m

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

Laws of Equilibrium

A body is said to be in equilibrium if the system of forces acting on the body tends to keep the body at rest, in other words, there is no translation motion or net torque on the body. Under such conditions, the equilibrium conditions are to be applied to analyze the condition. The branch of mechanics that deals with this condition is known as 'statics'. Under equilibrium, there are no net forces on the body.

Question

determine MAАВ, МВА, МВС, Мсв, Мср, Мрс
Using Slope Deflection Method for Figure 03
and all reactions at support. Use I = 1/12 b(h3) and E = 2.10x107 KPa
Figure 03
1090 KN/m
C
4.25m
5.15m
b2= 0.55m
b3= 0.75m
545 KN
h2= 0.75m
h3= 0.75m
9.95m
L4, m
b1= 0.75m
3.22m h1= 0.75m
3.28m
國D

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