Question 4.19: A crane hook having an approximate trapezoidal cross-section...

\text { Given } S_{y t}=380 N / mm ^{2} \quad(f s)=3.5 .

b_{i}=90 mm \quad b_{o}=30 mm \quad h=120 mm .

R_{o}=170 mm \quad R_{i}=50 mm .

Step I Calculation of permissible tensile stress

\sigma_{\max }=\frac{S_{y t}}{(f s)}=\frac{380}{3.5}=108.57 N / mm ^{2} .

Step II Calculation of eccentricity (e)
For the cross-section XX [Eqs (4.62) and (4.63)],

R_{N}=\frac{\left(\frac{b_{i}+b_{o}}{2}\right) h}{\left(\frac{b_{i} R_{o}-b_{o} R_{i}}{h}\right) \log _{e}\left(\frac{R_{o}}{R_{i}}\right)-\left(b_{i}-b_{o}\right)}             (4.62).

R=R_{i}+\frac{h\left(b_{i}+2 b_{o}\right)}{3\left(b_{i}+b_{o}\right)}             (4.63).

R_{N}=\frac{\left(\frac{b_{i}+b_{o}}{2}\right) h}{\left(\frac{b_{i} R_{o}-b_{o} R_{i}}{h}\right) \log _{e}\left(\frac{R_{o}}{R_{i}}\right)-\left(b_{i}-b_{o}\right)} .

R_{N}=\frac{\left(\frac{90+30}{2}\right)(120)}{\left(\frac{90 \times 170-30 \times 50}{120}\right) \log _{e}\left(\frac{170}{50}\right)-(90-30)} .

= 89.1816 mm.

R=R_{i}+\frac{h\left(b_{i}+2 b_{o}\right)}{3\left(b_{i}+b_{o}\right)} .

=50+\frac{120(90+2 \times 30)}{3(90+30)}=100 mm .

e=R-R_{N}=100-89.1816=10.8184 mm .

Step III Calculation of bending stress

h_{i}=R_{N}-R_{i}=89.1816-50=39.1816 mm .

A=\frac{1}{2}\left[h\left(b_{i}+b_{o}\right)\right]=\frac{1}{2}[(120)(90+30)] .

=7200 mm ^{2} .

M_{b}=P R=(100 P) N – mm .

From Eq. (4.56), the bending stress at the inner fibre is given by,

\sigma_{b i}=\frac{M_{b} h_{i}}{A e R_{i}}             (4.56).

\sigma_{b i}=\frac{M_{b} h_{i}}{A e R_{i}}=\frac{(100 P)(39.1816)}{(7200)(10.8184)(50)} .

=\frac{(7.2435) P}{(7200)} N / mm ^{2}             (i).

Step IV Calculation of direct tensile stress

\sigma_{t}=\frac{P}{A}=\frac{P}{(7200)} N / mm ^{2}               (ii).

Step V Calculation of load carrying capacity
Superimposing the two stresses and equating the resultant to permissible stress, we have

\sigma_{b i}+\sigma_{t}=\sigma_{\max } .

\frac{(7.2435) P}{7200}+\frac{P}{7200}=108.57 .

P = 94 827.95 N.