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Question: Determine the load carrying capacity of the hook.
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Fig 2.2 shows a crane hook of approximate trapezoidal cross-section. It is made of plain carbon steel 45C8 (Syt = 380 N/mm2) and the factor of safety is 3.5. Determine the load carrying capacity of the hook.

Subject: Machine Design -I

Topic: Curved beams and Thin cylinder

Difficulty: High

md1(66) • 414 views
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 modified 5 months ago by written 9 months ago by
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 written 6 months ago by
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Step 1: Calculation of permissible tensile stress:

$\sigma_{max}=\frac{S_yt}{fs}=\frac{380}{3.5}=108.57N/mm^2$

Step 2: Calculation of eccentricity(e):

For the cross section XX

$R_N=\frac{\frac{b_i+b_o}{2}h}{\frac{b_ir_o-b_oR_i}{h}*log_e\frac{R_o}{R_i}-(b_i-b_o)}$

$R_N=\frac{\frac{90+30}{2}*120}{\frac{90*170-30*50}{120}*log_e\frac{170}{50}-(90-30)}$

=89.1816mm

$R=R_i+\frac{h(b_i+2b_o)}{3(b_i+b_o)}$

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

$e=R-R_N=100-89.181=10.8184mm$

Step 3:Calculation of bending stress:

$h_i=R_N-R_i=89.1816-50=39.1816mm$

$A=0.5[h(b_i+b_o)]=0.5[(120)(90+30)]=7200mm^2$

$M_b=PR=100P N-mm$

$\sigma_{bi}=\frac{M_bh_i}{AeR_i}=\frac{100P*39.1816}{7200*10.8184*50}=\frac{7.2435P}{7200} N/mm^2$ (i)

Step 4: Calculation of direct tensile stress:

$\sigma_t=\frac{P}{A}=\frac{P}{7200} N/mm^2$ (ii)

Step 5: Calculation of load carrying capacity:

Superimposing the two stresses and equating the resultant to permissible stress,

$\sigma_{bi}+\sigma_t=\sigma_{max}$

$\frac{7.2435P}{7200}+\frac{P}{7200}=108.57$

P=94827.95N

ADD COMMENTlink
 written 5 months ago by
0

Step 1: Calculation of permissible tensile stress:

$\sigma_{max}=\frac{S_yt}{fs}=\frac{380}{3.5}=108.57N/mm^2$

Step 2: Calculation of eccentricity(e):

For the cross section XX

$R_N=\frac{\frac{b_i+b_o}{2}h}{\frac{b_ir_o-b_oR_i}{h}*log_e\frac{R_o}{R_i}-(b_i-b_o)}$

$R_N=\frac{\frac{90+30}{2}*120}{\frac{90*170-30*50}{120}*log_e\frac{170}{50}-(90-30)}$

=89.1816mm

$R=R_i+\frac{h(b_i+2b_o)}{3(b_i+b_o)}$

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

$e=R-R_N=100-89.181=10.8184mm$

Step 3:Calculation of bending stress:

$h_i=R_N-R_i=89.1816-50=39.1816mm$

$A=0.5[h(b_i+b_o)]=0.5[(120)(90+30)]=7200mm^2$

$M_b=PR=100P N-mm$

$\sigma_{bi}=\frac{M_bh_i}{AeR_i}=\frac{100P*39.1816}{7200*10.8184*50}=\frac{7.2435P}{7200} N/mm^2$ (i)

Step 4: Calculation of direct tensile stress:

$\sigma_t=\frac{P}{A}=\frac{P}{7200} N/mm^2$ (ii)

Step 5: Calculation of load carrying capacity:

Superimposing the two stresses and equating the resultant to permissible stress,

$\sigma_{bi}+\sigma_t=\sigma_{max}$

$\frac{7.2435P}{7200}+\frac{P}{7200}=108.57$

P=94827.95N

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 written 5 months ago by
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