Bending Moment Diagram for Different Load Cases

THE COURSEWORK
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Author’s Note
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Note:
i = 8 ; j = 6 ; k = 1 ; l = 2
Span = 3 + (0.2 i) = 3 + (0.2 * 8)
Span = 4.6 m
Load Case – 1
The distance of the load from the left-support is = 1 + (0.1 * j) = 1 + (0.1 * 6) = 1.6 m
Writing free-body-diagram of the given load case,
Writing the equilibrium conditions for the given problem,
ƩFx = 0
ƩFy = RA + RB = 10
ƩMA = 10 * 1.6 = RB * 4.6
Therefore, RB = 3.4782 kN
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That gives,
RA = 10 - RB = 10 – 3.4782 = 6.5218 kN
MATLAB Code to find the Bending Moment Diagram:
clear all;
load = 10; %load in kN
act = 1.6; %distance of the load from the left support in m
RB = load * act / 4.6; %reaction at support-A
RA = 10 - RB; %reaction at support-B
x1 = [0:0.1:4.6]; %increment of 0.1 m
m1 = RA * x1 - 10 * (x1 - 1.6).*(x>1.6);
plot(x1,m1)
grid on;
xlabel('x in meters')
ylabel('Moment in kNm')
title('Bending Moment Diagram')
Load Case - 2
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The value of 1 + k = 1 + 1 = 2
The resulting case will be:
Writing free-body-diagram of the given load case, (and converting uniformly distributive
load to point-load)
Writing the equilibrium conditions for the given problem,
ƩFx = 0
ƩFy = RA + RB = 4.6
ƩMA = 4.6 * 3.45 = RB * 4.6
Therefore, RB = 3.45 kN
That gives,
RA = 4.6 - RB = 4.6 – 3.45 = 1.15 kN
MATLAB Code to find the Bending Moment Diagram:
clear all;
load = 4.6; %load in kN
act = 3.45; %distance of the load from the left support in m
RB = load * act / 4.6; %reaction at support-A
RA = 4.6 - RB; %reaction at support-B
x1 = [0:0.1:4.6]; %increment of 0.1 m
m1 = RA * x1 - 4.6 * (x1 - 3.45).*(x1>3.45);
plot(x1,m1)
grid on;
xlabel('x in meters')
ylabel('Moment in kNm')
title('Bending Moment Diagram')
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