Question 8.60: A WT305 × 41 standard steel shape is subjected to a tension ...

A WT305 × 41 standard steel shape is subjected to a tension force P that is applied 250 mm above the bottom surface of the tee shape, as shown in Figure P8.60. If the tension normal stress of the upper surface of the WT-shape must be limited to 150 MPa, determine the allowable force P that may be applied to the member.

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Section properties (from Appendix B) 

Depth 300 mm

Centroid \bar{y}=88.9 mm (from flange to centroid)

\begin{aligned}&A=5,230  mm ^{2} \\&I_{z}=48.7 \times 10^{6}  mm ^{4}\end{aligned}

Stresses 

\begin{aligned}&\sigma_{\text {axial }}=\frac{F}{A}=\frac{P}{5,230  mm ^{2}}=P\left(1.9120 \times 10^{-4}  mm ^{-2}\right) \\&\sigma_{\text {bending }}=\frac{M_{z} c}{I_{z}}=\frac{P(250  mm -88.9  mm )(300  mm -88.9  mm )}{48.7 \times 10^{6}  mm ^{4}} \\&=\frac{P(161.1  mm )(211.1  mm )}{48.7 \times 10^{6}  mm ^{4}} \\&=P\left(6.9832 \times 10^{-4}  mm ^{-2}\right)\end{aligned}

Normal stress on the upper surface of the WT-shape
The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses. Since these stresses are expressed in terms of the unknown force P, the tension normal stress is given by:

\begin{aligned}\sigma_{\text {upper surface }} &=P\left(1.9120 \times 10^{-4}  mm ^{-2}\right)+P\left(6.9832 \times 10^{-4}  mm ^{-2}\right) \\&=\left(8.8953 \times 10^{-4}  mm ^{-2}\right) P\end{aligned}

The normal stress on the upper surface of the WT-shape must be limited to 150 MPa; therefore,

\begin{aligned}\left(8.8953 \times 10^{-4}  mm ^{-2}\right) P & \leq 150  MPa \\& \therefore P \leq \frac{150  N / mm ^{2}}{8.8953 \times 10^{-4}  mm ^{-2}}=168,629  N =168.6  kN\end{aligned}

 

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