ENGR 8
Assignment 11 Examples
Solve this using the left section. Repeat on right section
Solutions Online
7.4 Distributed F = 27 kN @ x=4.5
Mb = 0 -> -Ay(6) + 12(4.5) + 27(1.5)=0
Ay = 15.75
By = 12 + 27 – 15.75 = 23.25
Left section:
Distributed F = 13.5 @ x=3.75
Sum Fy = 0 -> 15.75 – 12 -13.5 – V = 0
V = -9.75
Sum Mc=0 -> -15.75(4.5)+12*3 + 13.5(.75)+M=0
M = 24.75
Right section: Distributed F=13.5@x=5.25
Sum Fy=0-> V – 13.5 + 23.25=0, V = -9.75
Sum Mc=0-> 23.25(1.5)-13.5(.75) –M=0
M = 24.75
7.2 Sum Fy = 0 -> Ay + By = 10
Sum Ma=0 -> 30 + By(3) -10(4.5) = 0
By = 15/3 = 5, Ay = 5
Left section:
Sum Fy = 0 -> 5 – V=0, V = 5
Sum Mc = 0-> 30 – 5(1.5) + M=0
M = 7.5-30 = -22.50
Nc = 0
THEN, determine the shear and bending moment diagram
for the entire beam
Right section:
Sum Fy = 0 -> V + 5 – 10=0, V = 5
Sum Mc = 0-> 5(1.5) – 10(3) –M =0
M = 7.5-30 = -22.50
Nc = 0
7.2 SHEAR AND MOMENT DIAGRAM
0<x<1.5
V=0, M = – 30
1.5< x < 4.5
V = 5,
30 – 5(x-1.5) + M = 0, M=5x – 37.5
4.5<x<6
10 - V = 0, V=10
30 -5*(x-1.5)-5*(x-4.5)+ M=0
M = -30+5x-7.5+5x-22.5 , M=10x-60
7-6 Reactions Ay = 10.5, By = 16.5
Using Freemat and the calculus based approach
(freemat script below)
Then determine shear and moment diagram for beam
% Problem 7.6
% 1) reactions are found to be Ay = 10.5 By = 16.5
% 2) Create a vector of x and w(x)
x = [0:.01:3]
x2 = [3.01:.01:6];
w = -x*2
w2 = -6 + 0*x2;
w=[w w2];
x = [x x2];
plot(x,w)
axis([0 6 0 -7])
title('w(x)')
% 3) Integrage w(x) to find V(x)
for k=1:length(x)
V(k)= trapz(x(1:k),w(1:k)) + 10.5;
end
plot(x,V)
title('V(x)')
% 4) Integrate V(x) to find M(x)
for k=1:length(x)
M(k)= trapz(x(1:k),V(1:k));
end
plot(x,M)
title('M(x)')