---
title: "Stress in a crane hook section"
subject: "Mechanics of Solids"
chapter: 6
key: "semester_5/mechanics_of_solids:crane-hook-curved-beam-stress"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/crane-hook-curved-beam-stress
site: BCT Engineering
---

# Stress in a crane hook section

Mechanics of Solids · Chapter 6: Curved Beams

- Asked in: year not given, 8 marks
- Video: [PROBLEM ON CRANE HOOK OF A RECTANGULAR SECTION](https://www.youtube.com/watch?v=8ksYSCyySxc) — Machine Design By Bahauddin
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/crane-hook-curved-beam-stress
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-6

## Question

The critical horizontal section of a crane hook is a rectangle 30 mm wide and 50 mm deep. The radius of curvature at the inner fibre is 40 mm, and the line of action of the load passes through the centre of curvature of the section, so its distance from the centroid equals the centroidal radius. For a vertical pull of 12 kN, calculate the maximum tensile and compressive stresses in the section using the Winkler-Bach theory. Compare the inner-fibre stress with the straight-beam value.

## Short answer

Using Winkler-Bach theory, compute neutral axis radius r_n = 61.66 mm, eccentricity e = 3.34 mm. Maximum tensile stress at inner fibre = 92.2 MPa; maximum compressive stress at outer fibre = -41.0 MPa. The straight-beam bending stress is only 62.4 MPa, underestimating the inner stress.

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For the hook section (b = 30 mm, h = 50 mm, ri = 40 mm, ro = 90 mm), what is rn?
   A. 65 mm
   B. 61.66 mm
   C. 57.5 mm
   D. 68.3 mm
   Answer: B. rn = h/ln(ro/ri) = 50/ln(2.25) = 50/0.8109 = 61.66 mm, which is less than rc = 65 mm. (Key point: Neutral axis radius)

2. Because the load line passes through the centre of curvature, the section is treated as carrying
   A. Pure bending only
   B. Pure direct tension only
   C. Direct tension P plus a moment M = P rc about the centroid
   D. Torsion plus shear
   Answer: C. Moving the load to the centroid gives direct tension P/A and a moment P x rc = 780 N m. (Key point: Load as direct force plus moment)

3. What is the maximum tensile stress and where does it occur?
   A. 92.2 MPa at the inner fibre
   B. 84.2 MPa at the outer fibre
   C. 62.4 MPa at the inner fibre
   D. 41.0 MPa at the outer fibre
   Answer: A. Inner fibre: 8 MPa direct plus 84.24 MPa bending = 92.2 MPa tension; the outer fibre is in compression. (Key point: Maximum stress location and value)

4. The inner-fibre bending stress of the curved hook compared with the straight-beam value 6M/(bh^2) = 62.4 MPa is about
   A. The same
   B. 0.74 times
   C. 1.35 times
   D. 2.5 times
   Answer: C. 84.2/62.4 is about 1.35, so the straight-beam formula underestimates the stress in a sharply curved section. (Key point: Comparison with straight beam)

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