Question 4.SP.8: Knowing that for the cast iron link shown the allowable stre...
Knowing that for the cast iron link shown the allowable stresses are 30 MPa in tension and 120 MPa in compression, determine the largest force P which can be applied to the link. (Note: The T-shaped cross section of the link has previously been considered in Sample Prob. 4.2.)
Properties of Cross Section. From Sample Prob. 4.2, we have
A= 3000 mm^{2} = 3 \times 10^{-3} m^{2} \quad\quad \bar{Y} = 38 mm = 0.038 m \\ I = 868 \times 10^{-9} m^{4}We now write: d = (0.038 m) – (0.010 m) = 0.028 m
Force and Couple at C. We replace P by an equivalent force-couple system at the centroid C.
P = P M = P(d) = P(0.028 m) = 0.028P
The force P acting at the centroid causes a uniform stress distribution (Fig. 1). The bending couple M causes a linear stress distribution (Fig. 2).
\sigma_{0} = \frac{P}{A} = \frac{P}{3 \times 10^{-3}} = 333P \quad\quad (Compression) \\ \sigma_{1} = \frac{Mc_{A}}{I} = \frac{(0.028P)(0.022)}{868 \times 10^{-9}} = 710P \quad\quad (Tension) \\ \sigma_{2} = \frac{Mc_{B}}{I} = \frac{(0.028P)(0.038)}{868 \times 10^{-9}} = 1226P \quad\quad (Compression)Superposition. The total stress distribution (Fig. 3) is found by superposing the stress distributions caused by the centric force P and by the couple M. Since tension is positive, and compression negative, we have
\sigma_{A} = – \frac{P}{A} + \frac{Mc_{A}}{I} = -333P + 710P = +377P \quad\quad (Tension) \\ \sigma_{B} = – \frac{P}{A} – \frac{Mc_{B}}{I} = – 333P – 1226P = – 1559P \quad\quad (Compression)Largest Allowable Force. The magnitude of P for which the tensile stress at point A is equal to the allowable tensile stress of 30 MPa is found by writing
\sigma_{A} = 377P = 30 MPa \quad\quad\quad P = 79.6 kNWe also determine the magnitude of P for which the stress at B is equal to the allowable compressive stress of 120 MPa.
\sigma_{B} = – 1559P = – 120 MPa \quad\quad\quad P = 77.0 kNThe magnitude of the largest force P that can be applied without exceeding either of the allowable stresses is the smaller of the two values we have found.
P = 77.0 kN
