6.15 Numerical Examples
213
.
W
C
A
1 0 0 m
m
B
W
D
50mm
.
50mm
Y A
Y B
30mm
R
Section AB
Fig. 6.21 Closed ring with rectangular cross-section
σ B i = 9.11 ×
500
π
= 1450 kg/cm
2
σ B 0 = −4.81 ×
500
π
= −765.54 kg/cm
2
Example 6.11 A ring of 200 mm mean diameter has a rectangular cross-section
with 50 mm in the radial direction and 30 mm perpendicular to the radial direction as
shown in Fig. 6.21. If the maximum tensile stress is limited to 120 N/mm
2
, determine
the tensile load that can be applied on the ring.
Solution: R = 100 mm, Area of cross-section = A = 30 × 50 = 1500 mm
2 .
From Table 6.1, the value of m for the rectangular section is given by
m = −1 +
100
1500×50
[30 × 50 + 0] ln
100+25
100−25
− 0
∴ m = 0.0217
To find M AB .
213
.
W
C
A
1 0 0 m
m
B
W
D
50mm
.
50mm
Y A
Y B
30mm
R
Section AB
Fig. 6.21 Closed ring with rectangular cross-section
σ B i = 9.11 ×
500
π
= 1450 kg/cm
2
σ B 0 = −4.81 ×
500
π
= −765.54 kg/cm
2
Example 6.11 A ring of 200 mm mean diameter has a rectangular cross-section
with 50 mm in the radial direction and 30 mm perpendicular to the radial direction as
shown in Fig. 6.21. If the maximum tensile stress is limited to 120 N/mm
2
, determine
the tensile load that can be applied on the ring.
Solution: R = 100 mm, Area of cross-section = A = 30 × 50 = 1500 mm
2 .
From Table 6.1, the value of m for the rectangular section is given by
m = −1 +
100
1500×50
[30 × 50 + 0] ln
100+25
100−25
− 0
∴ m = 0.0217
To find M AB .