216
6 Two-Dimensional Problems in Elasticity …
.
W
C
A
1 0 0 m
m
B W
W
W
D
M AB
M
N
W
2
W
2
W
2
θ
Fig. 6.22 Closed ring with circular cross-section
Area of cross-section = A = π(12.5)
2
= 490.87 mm
2 .
We have,
M A =
W R
2
1 −
2
π
=
W
2
× 50
1 −
2
π
M A = 9.085W
Now,
σ A =
P
A
+
M A
AR
1 +
y A
m(R + y A )
=
W
2 × 490.87
−
9.085W
490.87 × 50
1 +
(−12.5)
0.016(50 − 12.5)
= 0.0084W (Tensile)
∴ σ B =
P
A
+
M A
−AR
1 +
y B
m(R + y B )
=
W
2 × 490.87
−
9.085W
490.87 × 50
1 +
12.5
0.016(50 + 12.5)
∴ σ B = −0.00398W (Compression)
Also,
6 Two-Dimensional Problems in Elasticity …
.
W
C
A
1 0 0 m
m
B W
W
W
D
M AB
M
N
W
2
W
2
W
2
θ
Fig. 6.22 Closed ring with circular cross-section
Area of cross-section = A = π(12.5)
2
= 490.87 mm
2 .
We have,
M A =
W R
2
1 −
2
π
=
W
2
× 50
1 −
2
π
M A = 9.085W
Now,
σ A =
P
A
+
M A
AR
1 +
y A
m(R + y A )
=
W
2 × 490.87
−
9.085W
490.87 × 50
1 +
(−12.5)
0.016(50 − 12.5)
= 0.0084W (Tensile)
∴ σ B =
P
A
+
M A
−AR
1 +
y B
m(R + y B )
=
W
2 × 490.87
−
9.085W
490.87 × 50
1 +
12.5
0.016(50 + 12.5)
∴ σ B = −0.00398W (Compression)
Also,