Determine the maximum shear stress acting at section a–a in the beam.
Question 7.75: Determine the maximum shear stress acting at section a–a in ...
\bar{y}=\frac{(0.375)(4)(0.75)+(3.75)(6)(0.5)+(7.125)(2)(0.75)}{4(0.75)+6(0.5)+2(0.75)}=3.075 \mathrm{in}
I=\frac{1}{12}(4)\left(0.75^{3}\right)+4(0.75)(3.075-0.375)^{2}
\quad+\frac{1}{12}(0.5)\left(6^{3}\right)+0.5(6)(3.75-3.075)^{2}
\quad+\frac{1}{12}(2)\left(0.75^{3}\right)+2(0.75)(7.125-3.075)^{2}
=57.05 \mathrm{in}^{4}
Q_{\max }=\Sigma \bar{y}^{\prime} A^{\prime}
=2.7(4)(0.75)+2.325(0.5)(1.1625)
=9.4514 \mathrm{in}^{3}
\tau_{\max }=\frac{V Q_{\max }}{I t}=\frac{2800(9.4514)}{57.05(0.5)}=928 \mathrm{psi}
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