204
6 Two-Dimensional Problems in Elasticity …
σ θ A = σ A =
P
A
+
M
AR
1 +
y A
m(R + y A )
= −
1000
7.06
+
(3.5 × 1000)
7.06 × 3.5
1 +
(−1.5)
0.050(3.5 − 1.5)
or σ A = −2124.65 kg/cm
2 (compressive).
The stress at point B is given by
σ θ B = σ B = +
P
A
+
M
AR
1 +
y B
m(R + y B
=
−1000
7.06
+
3500
7.06 × 3.5
1 +
1.5
0.050(3.5 + 1.5)
∴σ B = 849.85 kg/ cm
2 (Tensile).
Comparison by Straight Beam Formula.
The moment of inertia of the ring cross-section about the centroidal axis is
I =
π d
4
64
=
π(3)
4
64
= 3.976 cm
4
If the link is considered to be a straight beam, the corresponding values are
σ A =
P
A
+
My
I
= −
1000
7.06
+
(+3500)(−1.5)
3.976
∴σ A = −1462.06 kg/cm
2 (compressive).
and σ B =
−1000
7.06
+
3500×1.5
3.976
.
σ B = 1178.8 kg/cm
2 (tensile) (Fig. 6.15).
Example 6.6 An open ring having T-Section as shown in the Fig. 6.16 is subjected
to a compressive load of 10,000 kg. Compute the stresses at A and B by curved beam
formula.
Solution:
Area of the section = A = 2 × 10 + 2 × 14 = 48 cm
2
The value of m can be calculated from Table 6.1 by substituting b 1 = 0 for the
unsymmetric I-section.
From Figure,