6.15 Numerical Examples
209
Example 6.8 The dimensions of a 10 tonne crane hook are shown in Fig. 6.18. Find
the circumferential stresses σ A and σ B on the inside and outside fibres respectively
at the section AB.
Solution: Area of the section = A =
9+3
2
× 12 = 72 cm
2 .
Now, y A =
12
3
9+2×3
9+3
= 5 cm.
Therefore y B = (12 − 5) = 7 cm.
Radius of curvature of the centroidal axis = R = 7 + 5 = 12 cm.
For trapezoidal cross-section, m is given by Table 6.1 as,
m = −1 +
12
72 × 12
[(3 × 12) + (12 + 7)(9 − 3)]. ln
12 + 7
12 − 5
− (9 − 3)12
∴ m = 0.080
Moment = M = P R = 10,000 × 12 = 120,000 kg-cm.
Now,
Stress at
A = σ A =
P
A
+
M
AR
1 +
y A
m(R + y A )
=
−10,000
72
+
120,000
72 × 12
1 +
(−5)
0.08(12 − 5)
∴ σ A = −1240 kg/cm
2
(Compression)
Stress at
B = σ B =
P
A
+
M
AR
1 +
y B
m(R + y B )
=
−10,000
72
+
120, 000
72 × 12
1 +
7
0.08(12 + 7)
∴ σ B = 639.62 kg/cm
2
(Tension)
Example 6.9 A circular open steel ring is subjected to a compressive force W = 80
kN as shown in Fig. 6.19. The cross-section of the ring is made up of an unsymmetrical I-section with an inner radius of 150 mm. Estimate the circumferential stresses
developed at points A and B.
Solution:
From Table 6.1, the value of m for the above section is given by
m = −1
+
R
A
[b 1 ln (R + c 1 ) + (t − b 1 ) ln (R + c 3 ) + (b − t) ln (R − c 2 ) − b ln (R − c)]
Hence R = Radius of curvature of the centroidal axis.