206
6 Two-Dimensional Problems in Elasticity …
= −1 +
23.66
48
0 + (2 − 0) ln(23.66 + 10.34)
+ (10 − 2) ln(23.66 − 3.66) − 10 ln(23.66 − 5.66)
Therefore, m = 0.042.
Now, stress at A,
σ A =
P
A
+
M
AR
1 +
y A
m(R + y A )
= −
10,000
48
+
(10,000 × 23.66)
48 × 23.66
1 +
(−5.66)
0.042(23.66 − 5.66)
∴σ A = −1559.74 kg/cm
2 (compressive).
Similarly, stress at B is given by
σ B =
P
A
+
M
AR
1 +
y B
m(R + y B )
= −
10000
48
+
10000 × 23.66
48 × 23.66
1 +
10.34
0.042(23.66 + 10.34)
∴ σ B = 1508.52 kg/cm
2 (tensile).
Example 6.7 A ring shown in the Fig. 6.17, with a rectangular section is 4 cm wide
and 2 cm thick. It is subjected to a load of 2000 kg. Compute the stresses at A and
B and at C and D by curved beam formula.
Solution: Area of the section A = 4 × 2 = 8 cm
2 .
The radius of curvature of the centroidal axis = R = 4 + 2 = 6 cm.
From Table 6.1, the m value for trapezoidal section is given by,
Fig. 6.17 Loaded ring with rectangular cross-section