6.15 Numerical Examples
217
M C D = (M A − P R) =
−9.085 +
W
2
× 50
∴ M C D = +15.915W
Now,
σ C =
M C D
AR
1 +
y C
m(R − y C )
= +
15.918W
490.87 × 50
1 +
(−12.5)
0.016(50 − 12.5)
∴ σ C = −0.013W (Compression)
and
σ D = +
15.918W
490.87 × 50
1 +
12.5
0.016(50 + 12.5)
= 0.0088 W (Tension)
Stresses at Section CD due to horizontal Loads.
We have, moment at any section MN is given by
M M N = −M A +
P R
2
(1 − cos θ)
At section CD, θ = 0
∴ M C D = −M A +
W
2
R(1 − cos 0)
M C D = −M A = −9.085W
∴ σ C =
P
A
+
M C D
AR
1 +
y C
m(R + y C )
=
W
2 × 490.87
+
(−9.085W)
490.87 × 50
1 +
(−12.5)
0.016(50 − 12.5)
∴ σ C = 0.0083 W (Tensile)
and
σ D =
P
A
+
M C D
AR
1 +
y D
m(R + y D )
=
W
2 × 490.87
+
(−9.085W)
490.87 × 50
1 +
12.5
0.016(50 + 12.5)
∴ σ D = −0.00398W (Compression)