Question 1.4.19: Choose a set of dimensions for a rectangular link that is to...

Choose a set of dimensions for a rectangular link that is to carry a maximum compressive load of 5000 lbf. The material selected has a minimum yield strength of 75 kpsi and a modulus of elasticity E = 30 Mpsi. Use a design factor of 4 and an end condition constant C = 1 for buckling in the weakest direction, and design for (a) a length of 15 in, and (b) a length of 8 in with a minimum thickness of  \frac {1}{2} in.

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(a) Using Eq. (4–44) we find the limiting slenderness ratio to be

  \frac{P_{cr}} {A} = \frac{C\pi ^2E} {\left({l}/{k} \right) ^2  }         (4–44)

 

\left(\frac{l}{k} \right) _1=\left(\frac{2\pi ^2CE}{S_y} \right) ^{{1}/{2}}=\left[\frac{2\pi ^2\left(1\right)\left(30\right)\left(10^6\right) }{75\left(10^3\right) } \right] ^{{1}/{2}}=88.9

 

By using  P_{cr}= n_dP = 4\left(5000\right) = 20 000 lbf, Eqs. (4–52)and (4–54) are solved, using various values of h, to form Table 4–3. The table shows that a cross section of \frac{5}{8} by \frac{3}{4} in, which is marginally suitable, gives the least area.

(b) An approach similar to that in part (a) is used with l = 8 in. All trial computations are found to be in the J. B. Johnson region of \frac{l}{k} values. A minimum area occurs when the section is a near square. Thus a cross section of \frac{1}{2}  \ by \ \frac{3}{4} in  is found to be suitable and safe.

b=\frac{12P_{crl^2}}{\pi ^2CEh^3}          h\leq p       (4–52)

b=\frac{P_{cr}}{hS_y\left(1-\frac{3l^2S_y}{\pi ^2CEh^2} \right) }      h\leq b        h\leq p      (4–54)

 

Table 4–3 Table Generated to Solve  Ex. 4–19, part (a)
h b A l/k Type Eq. No.
0.375 3.46 1.298 139 Euler (4–52)
0.5 1.46 0.73 104 Euler (4–52)
0.625 0.76 0.475 83 Johnson (4–54)
0.625 1.03 0.579 92 Euler (4–52)

 

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