6.15 Numerical Examples
203
Fig. 6.14 Loaded circular
link
.
. A
B
P
R
Therefore, b = 0.5693.
If t is the thickness of the cylinder, then.
b = 0.5 + t = 0.5693.
∴ t = 0.0693 m or 69.3 mm.
Example 6.5 The circular link shown in Fig. 6.14 has a circular cross-section 3 cm
in diameter. The inside diameter of the ring is 4 cm. The load P is 1000 kg. Calculate
the stress at A and B. Compare the values with those found by the straight beam
formula. Assume that the material is not stressed above its elastic strength.
Solution:
Cross - sectional area = A =
π
4
(3)
2
= 7.06 cm
2
.
For circular cross-section m is given by
m = −1 + 2
R
c
2
− 2
R
c
R
c
2
− 1
Here R = 2 + 1.5 = 3.5 cm.
c = 1.5 cm. (Refer Table 6.1).
Therefore,
m = −1 + 2
3.5
1.5
2
− 2
3.5
1.5
3.5
1.5
2
− 1
m = 0.050
At section AB, the load is resolved into a load P and a bending couple whose
moment is positive. The stress at A and B is considered to be the sum of the stress
due to axial load P, and the stress due to the bending moment M.
Therefore, stress at point A is