Question 18.CS.8: Swivel Hook Design Analysis A crane hook for the winch crane...
Swivel Hook Design Analysis
A crane hook fort he winch crane, shown in Figure 18.8a, is rated at P = 3 kN. Determine the tangential stresses at points A and B using Winkler’s formula. Note that, for a large number of manufactured crane hooks, the critical section AB can be closely approximated by a trapezoidal area with half an ellipse at the inner radius and an arc of a circle at the outer radius, as shown in Figure 18.8b. The solution for standardized crane hooks is expedited by readily available computer programs.
Assumptions: The critical section AB is taken to be trapezoidal. The hook is made of A1SI 1020-HR steel with a safety factor of n against yielding.
Given:
\begin{aligned} r_i &=20 mm , \quad b_1=30 mm , \quad b_2=10 mm \\ h &=42 mm , \quad n=5, \\ S_y &=210 MPa , \end{aligned} (from Table B.3)
See Figures 18.7 and 18.8; Section 16.8.
Referring to Figure 18.8b, we obtain the following quantities. The cross-sectional area is
A=\frac{1}{2}\left(b_1+b_2\right) h=\frac{1}{2}(30+10)(42)=840 mm ^2
The distance to the centroid C from the inner edge is
\bar{y}=\frac{h\left(b_1+2 b_2\right)}{3\left(b_1+b_2\right)}=\frac{42(30+2 \times 10)}{3(30+10)}=17.5 mm
Hence,
\bar{r}=r_i+\bar{y}=20+17.5=37.5 mm
By case E of Table 16.1, the radius of the neutral axis, with r_{o} = r_{i} + h = 62 mm, is then
\begin{aligned} R &=\frac{A}{\frac{1}{h}\left[\left(b_1 r_o-b_2 r_i\right) \ln \frac{r_o}{r_i}-h\left(b_1-b_2\right)\right]} \\ &=\frac{840}{\frac{1}{42}\left[(30 \times 62-10 \times 20) \ln \frac{62}{20}-42(30-10)\right]} \\ &=33.9843 \end{aligned}
Equation 16.51 leads to
e=\bar{r}-R=37.5-33.9843=3.5157 mm
The circumferential stresses are determined through the use of Equations 16.55 with a tensile normal load P and bending moment M = –PR. Therefore,
\left(\sigma_\theta\right)_A=-\frac{P}{A}-\frac{P \bar{r}\left(R-r_i\right)}{A e r_i}=-\frac{P}{A}\left[1+\frac{\bar{r}\left(R-r_i\right)}{e r_i}\right] (16.55a)
\left(\sigma_\theta\right)_B=-\frac{P}{A}-\frac{P \bar{r}\left(R-r_o\right)}{A e r_o}=-\frac{P}{A}\left[1+\frac{\bar{r}\left(R-r_o\right)}{e r_o}\right] (16.55b)
\begin{array}{l} \left(\sigma_\theta\right)_A=\frac{P}{A}\left[1+\frac{\bar{r}\left(R-r_i\right)}{e r_i}\right] \\ \left(\sigma_\theta\right)_B=\frac{P}{A}\left[1+\frac{\bar{r}\left(R-r_o\right)}{e r_0}\right] \end{array}
Introducing the required values into the preceding expression, we have
\begin{aligned} \left(\sigma_\theta\right)_A &=\frac{3000}{840\left(10^{-6}\right)}\left[1+\frac{37.5(33.9843-20)}{3.5157(20)}\right] \\ &=30.21 MPa \end{aligned}
\begin{aligned} \left(\sigma_\theta\right)_B &=\frac{3000}{840\left(10^{-6}\right)}\left[1+\frac{37.5(33.9843-62)}{3.5157(62)}\right] \\ &=-13.64 MPa \end{aligned}
where a minus sign means compression.
Comment: The allowable stress \sigma _{all}= 210/5 = 42 MPa is larger than the maximum stress of 30.21 MPa. That is, the crane hook can support a load of 3 kN with a factor of safety of 5 without yielding.
| Table 16.1 Properties for a Variety of Cross-Sectional Shapes |
|
| Cross Section | Radius of Neutral Surface |
 |
\begin{array}{l} R=\frac{h}{\ln \frac{r_o}{r_i}} \\ A=b h \end{array} |
 |
\begin{array}{l} R=\frac{A}{2 \pi\left(\bar{r}-\sqrt{\bar{r}^2-c^2}\right)} \\ A=\pi c^2 \end{array} |
 |
\begin{array}{l} R=\frac{A}{\frac{2 \pi b}{a}\left(\bar{r}-\sqrt{\bar{r}^2-a^2}\right)} \\ A=\pi a b \end{array} |
 |
\begin{array}{l} R=\frac{A}{\frac{b r_o}{h}\left\lgroup \ln \frac{r_0}{r_i} \right\rgroup -b} \\ A=\frac{1}{2} b h \end{array} |
 |
\begin{array}{l} R=\frac{1}{\frac{1}{h}\left[\left(b_1 r_o-b_2 r_i\right) \cdot \ln \frac{r_0}{r_i}-h\left(b_1-b_2\right)\right]} \\ A=\frac{1}{2}\left(b_1+b_2\right) h \end{array} |
