Stress in a crane hook section

Data

$b = 30$ mm, $h = 50$ mm, $r_i = 40$ mm, $r_o = 40 + 50 = 90$ mm, $P = 12$ kN.

$$\begin{aligned} A &= bh = 1500\ \text{mm}^2 \\ r_c &= r_i + h/2 = 65\ \text{mm} \end{aligned}$$

Neutral axis radius

For a rectangle:

$$r_n = \frac{h}{\ln(r_o/r_i)} = \frac{50}{\ln(90/40)} = \frac{50}{0.8109} = 61.658\ \text{mm}$$ $$e = r_c - r_n = 65 - 61.658 = 3.342\ \text{mm}$$ $$y_i = r_n - r_i = 21.658\ \text{mm},\qquad y_o = r_o - r_n = 28.342\ \text{mm}$$

Loads on the section

The load acts along the line through the centre of curvature, so the section is subjected to a direct tension and a bending moment about the centroid:

$$M = P\,r_c = 12\times10^3\times65 = 780\times10^3\ \text{N mm} = 780\ \text{N m}$$

The moment tends to straighten (open) the hook, so the inner fibres are in tension and the outer ones in compression.

Stresses

$$\sigma_{direct} = \frac{P}{A} = \frac{12000}{1500} = 8.00\ \text{MPa (tension)}$$ $$\begin{aligned} \sigma_{b,i} &= \frac{M\,y_i}{A\,e\,r_i} = \frac{780\times10^3\times21.658}{1500\times3.342\times40} = 84.24\ \text{MPa (tension)} \\ \sigma_{b,o} &= \frac{M\,y_o}{A\,e\,r_o} = \frac{780\times10^3\times28.342}{1500\times3.342\times90} = -48.99\ \text{MPa (compression)} \end{aligned}$$

Total:

$$\begin{aligned} \sigma_{inner} &= 8 + 84.24 = 92.24\ \text{MPa (tension)} \\ \sigma_{outer} &= 8 - 48.99 = -40.99\ \text{MPa (compression)} \end{aligned}$$

Comparison with straight-beam formula

$$\sigma_{b,straight} = \frac{6M}{bh^2} = \frac{6\times780\times10^3}{30\times50^2} = 62.4\ \text{MPa}$$

The inner fibre bending stress of the curved beam is $84.2/62.4 \approx 1.35$ times the straight-beam value, so the straight beam formula is unsafe for such a sharply curved section.

Answer: Maximum tensile stress $= 92.2$ MPa (inner fibre); maximum compressive stress $= -41.0$ MPa magnitude (outer fibre). The straight-beam formula gives only $62.4$ MPa bending stress.