A cast iron link is 40 mm wide by 200 mm high by 500 mm long. The allowable stresses are 40 MPa in tension and 80 MPa in compression. Compute the largest compressive load P that can be applied to the ends of the link along a longitudinal axis that is located 150 mm above the bottom of the link.

Solution

 Original Loading

 

M=50P N ⋅mm

Equivalent Loading
 

 

 

$$

Top fiber (σ = 80 MPa compression)
σtop=−σa−σ f

σ top=−PA−6Mbd2${\sigma }_{top}=-{\displaystyle \frac{P}{A}}-{\displaystyle \frac{6M}{b{d}^2}}$

−80=−P40(200)−6(50P)40(200d2)$-80=-{\displaystyle \frac{P}{40(200)}}-{\displaystyle \frac{6(50P)}{40(200d^2)}}$

−80=−P3,200$-80=-{\displaystyle \frac{P}{3,200}}$

${}_{}$

P=256,000 N=256 kN$\mathrm{<span\; class="mjxassistivemathml"><i\; style="mso-bidi-font-style:\; normal;"><span\; style="border:\; none\; windowtext\; 1.0pt;\; color:\; \#002060;\; font-family:\; "Calibri","sans-serif";\; font-size:\; 14.0pt;\; mso-ascii-theme-font:\; minor-latin;\; mso-bidi-theme-font:\; minor-latin;\; mso-border-alt:\; none\; windowtext\; 0in;\; mso-hansi-theme-font:\; minor-latin;\; padding:\; 0in;">P<mo>=</mo><mn>256</mn><mo>,</mo><mn>000</mn><mtext> </mtext><mtext>N</mtext><mo>=</mo><mn>256</mn><mtext> </mtext><mtext>\; k\; N</mtext></span></i></span><i\; style="mso-bidi-font-style:\; normal;"><span\; style="color:\; \#002060;\; font-family:\; "Calibri","sans-serif";\; font-size:\; 14.0pt;\; mso-ascii-theme-font:\; minor-latin;\; mso-bidi-theme-font:\; minor-latin;\; mso-hansi-theme-font:\; minor-latin;"><br>\; <o:p></o:p></span></i>}$

 

Check the bottom fiber
σ bottom=−σa +σf${\sigma }_{bottom}=-{\sigma }_a+{\sigma }_f$

σ bottom=−PA+6Mbd2${\sigma }_{bottom}=-{\displaystyle \frac{P}{A}}+{\displaystyle \frac{6M}{b{d}^2}}$

σbottom=−256(1000)40(200)+6(50×256)(1000)40(2002)${\sigma }_{bottom}=-{\displaystyle \frac{256(1000)}{40(200)}}+{\displaystyle \frac{6(50\times 256)(1000)}{40({200}^2)}}$

σ bottom=16 MPa tension<40 MPa${\sigma }_{bottom}=16\text{MPa tension}<40\text{MPa}$       (okay)
 

$$

Thus,
P=256 k N$P=256\text{kN}$       answer