6.15 Numerical Examples
211
.
R = 2 . 5 c m
B i
A O
B O
A i
2P 1000kg
=
M A
m
n
P
Pcos
Psin
A i A o
P
θ
θ
θ
Fig. 6.20 Loaded closed ring
Hence, the resultant stresses at A and B are,
σ A = 50.6 N/mm
2 (Tension), σ B = −93.02 N/mm
2 (Compression)
Example 6.10 Calculate the circumferential stress on inside and outside fibre of the
ring at A and B, shown in Fig. 6.20. The mean diameter of the ring is 5 cm, and
cross-section is circular with 2 cm diameter. Loading is within elastic limit.
Solution: For circular section, from Table 6.1
m = −1 + 2
R
c
2
− 2
R
c
R
c
2
− 1
= −1 + 2
2.5
1
2
− 2
2.5
1
2.5
1
2
− 1
∴ m = 0.0435
We have,
M A = P R
1 −
2
π
= 0.364 P R = 0.364 × 2.5 P
∴ M A = 0.91 P
Now,
σ A i = −
P
A
+
M A
AR
1 +
y i
m(R + y A )
= −
P
A
+
0.91 P
A × 2.5
1 +
(−1)
0.0435(2.5 − 1)
211
.
R = 2 . 5 c m
B i
A O
B O
A i
2P 1000kg
=
M A
m
n
P
Pcos
Psin
A i A o
P
θ
θ
θ
Fig. 6.20 Loaded closed ring
Hence, the resultant stresses at A and B are,
σ A = 50.6 N/mm
2 (Tension), σ B = −93.02 N/mm
2 (Compression)
Example 6.10 Calculate the circumferential stress on inside and outside fibre of the
ring at A and B, shown in Fig. 6.20. The mean diameter of the ring is 5 cm, and
cross-section is circular with 2 cm diameter. Loading is within elastic limit.
Solution: For circular section, from Table 6.1
m = −1 + 2
R
c
2
− 2
R
c
R
c
2
− 1
= −1 + 2
2.5
1
2
− 2
2.5
1
2.5
1
2
− 1
∴ m = 0.0435
We have,
M A = P R
1 −
2
π
= 0.364 P R = 0.364 × 2.5 P
∴ M A = 0.91 P
Now,
σ A i = −
P
A
+
M A
AR
1 +
y i
m(R + y A )
= −
P
A
+
0.91 P
A × 2.5
1 +
(−1)
0.0435(2.5 − 1)