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Question 7.75: Determine the maximum shear stress acting at section a–a in ...

Determine the maximum shear stress acting at section a–a in the beam.

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yˉ=(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 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=112(4)(0.753)+4(0.75)(3.075−0.375)2I=\frac{1}{12}(4)\left(0.75^{3}\right)+4(0.75)(3.075-0.375)^{2}

 

+112(0.5)(63)+0.5(6)(3.75−3.075)2\quad+\frac{1}{12}(0.5)\left(6^{3}\right)+0.5(6)(3.75-3.075)^{2}

 

+112(2)(0.753)+2(0.75)(7.125−3.075)2\quad+\frac{1}{12}(2)\left(0.75^{3}\right)+2(0.75)(7.125-3.075)^{2}

 

=57.05 in4=57.05  \mathrm{in}^{4}

 

Qmax⁡=Σyˉ′A′Q_{\max }=\Sigma \bar{y}^{\prime} A^{\prime}

 

=2.7(4)(0.75)+2.325(0.5)(1.1625)=2.7(4)(0.75)+2.325(0.5)(1.1625)

 

=9.4514 in3=9.4514  \mathrm{in}^{3}

 

τmax⁡=VQmax⁡It=2800(9.4514)57.05(0.5)=928 psi\tau_{\max }=\frac{V Q_{\max }}{I t}=\frac{2800(9.4514)}{57.05(0.5)}=928  \mathrm{psi}
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