6.15 Numerical Examples
215
To find stresses at C and D.
We have,
M mn = −M AB +
W R
2
(1 − cos θ )
∴ At θ = 90
◦
,
M mn = M C D = −M AB +
W R
2
∴ M C D = −18.17W + W ×
100
2
= 31.83W
Now, stress at
C = σ C =
P
A
+
M C D
AR
1 +
y C
m(R + y C )
= 0 +
31.83W
1500 × 100
1 +
(−25)
0.0217(100 − 25)
= −0.00305W (Compression)
and stress at
D = σ D =
P
A
+
M C D
AR
1 +
y D
m(R + y D )
= 0 +
31.83W
1500 × 100
1 +
25
0.0217(100 + 25)
∴ σ D = 0.00217W (Tensile)
By comparison, the tensile stress is maximum at Point D.
∴ 0.00217W = 120 ∴ W = 55,299.54 N or 55.3 kN
Example 6.12 A ring of mean diameter 100 mm is made of mild steel with 25 mm
diameter. The ring is subjected to four pulls in two directions at right angles to each
other passing through the centre of the ring. Determine the maximum value of the
pulls if the tensile stress should not exceed 80 N/mm
2 (Fig. 6.22).
Solution: Here R = 50 mm.
From Table 6.1, the value of m for circular section is given by,
m = −1 + 2
R
C
2
− 2
R
C
R
C
2
− 1
= −1 + 2
50
12.5
2
− 2
50
12.5
50
12.5
2
− 1
∴m = 0.016