The built-up beam is fabricated from the three thin plates having a thickness t. Determine the location of the shear center O.

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Accepted Answer

Shear Center. Referring to Fig. a and summing moments about point A, we have

[latex]\curvearrowleft +\Sigma\left(M_{R} ight)_{A}=\Sigma M_{A} ; \quad P e=F_{f}(a)[/latex] [latex] ext{(1)}[/latex]

Section Properties: The moment of inertia of the cross section about the axis of symmetry is

[latex]I=\frac{1}{12}(t)(2 a)^{3}+2\left[ ext { at }\left(\frac{a}{2} ight)^{2} ight]=\frac{7}{6} a^{3} t[/latex]

Referring to Fig. c,[latex] \bar{y}^{\prime}=\frac{a}{2^{.}}[/latex] Thus, Q as a function of s is

[latex]Q=\bar{y}^{\prime} A^{\prime}=\frac{a}{2}(s t)=\frac{a t}{2} s[/latex]

Shear Flow:

[latex]q=\frac{V Q}{I}=\frac{P\left(\frac{a t}{2} s ight)}{\frac{7}{6} a^{3} t}=\frac{3 P}{7 a^{2}} s[/latex]

Resultant Shear Force: The shear force resisted by the flange is

[latex]F_{f}=\int_{0}^{a} q d s=\int_{0}^{a} \frac{3 P}{7 a^{2}} s d s=\left.\frac{3 P}{7 a^{2}}\left(\frac{s^{2}}{2} ight) ight|_{0} ^{a}=\frac{3}{14} P[/latex]

Substituting this result into Eq. (1),

[latex]e=\frac{3}{14} a[/latex]

Question 7.66: The built-up beam is fabricated from the three thin plates h...

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