Question 8.S-P.1: A 160-kN force is applied as shown at the end of a W200 × 52...
A 160-kN force is applied as shown at the end of a W200 × 52 rolled-steel beam. Neglecting the effect of fillets and of stress concentrations, determine whether the normal stresses in the beam satisfy a design specification that they be equal to or less than 150 MPa at section A-A^{\prime}.
Shear and Bending Moment. At section A-A^{\prime}, we have
M_{A}=(160 kN )(0.375 m )=60 kN \cdot mV_{A}=160 kN
Normal Stresses on Transverse Plane. Referring to the table of Properties of Rolled-Steel Shapes in Appendix C, we obtain the data shown and then determine the stresses s_{a} and s_{b}.
At point a:
s _{a}=\frac{M_{A}}{S}=\frac{60 kN \cdot m }{512 \times 10^{-6} m ^{3}}=117.2 MPaAt point b:
s _{b}= s _{a} \frac{y_{b}}{c}=(117.2 MPa ) \frac{90.4 mm }{103 mm }=102.9 MPaWe note that all normal stresses on the transverse plane are less than 150 MPa.
Shearing Stresses on Transverse Plane
At point a:
Q = 0 t _{a}=0
At point b:
Q=(204 \times 12.6)(96.7)=248.6 \times 10^{3} mm ^{3}=248.6 \times 10^{-6} m ^{3}t _{b}=\frac{V_{A} Q}{I t}=\frac{(160 kN )\left(248.6 \times 10^{-6} m ^{3}\right)}{\left(52.7 \times 10^{-6} m ^{4}\right)(0.0079 m )}=95.5 MPa
Principal Stress at Point b. The state of stress at point b consists of the
normal stress s _{b}=102.4 MPa and the shearing stress t _{b}=95.5 MPa. We draw Mohr’s circle and find
=\frac{102.9}{2}+\sqrt{\left(\frac{102.9}{2}\right)^{2}+(95.5)^{2}}
s _{\max }=159.9 MPa
The specification, s _{\max } \leq 150 MPa, is not satisfied
Comment. For this beam and loading, the principal stress at point b is 36% larger than the normal stress at point a. For L ≥ 874 mm, the maximum normal stress would occur at point a.
