212
6 Two-Dimensional Problems in Elasticity …
= −
P
A
− 5.21
P
A
∴ σ A i = −6.21
P
A
(Compressive)
σ A 0 = −
P
A
+
M
AR
1 +
y o
m(R + y B )
= −
P
A
+
0.91 P
A × 2.5
1 +
1
0.0435(2.5 + 1)
∴ σ A o = 1.755
P
A
(Tension)
Similarly,
M B = (M A − P R)
= (0.364 P R − P R) = −0.636 P R
= −0.636 × 2.5 P
∴ M B = −1.59P
Now,
σ Bi =
M B
AR
1 +
y i
m(R + y i )
= −
1.59 P
A × 2.5
1 +
(−1)
0.0435(2.5 − 1)
∴ σ Bi = 9.11
P
A
(Tension)
and
σ Bo = −
1.59 P
A × 2.5
1 +
(+1)
0.0435(2.5 + 1)
= −4.81
P
A
(Compression)
Now, substituting the values of P = 500 kg,
A = π (1)
2
= 3.14159 cm
2
, above stresses can be calculated as below.
σ A i = −6.21 ×
500
π
= −988 kg/cm
2
σ A 0 = 1.755 ×
500
π
= 279.32 kg/cm
2