The built-up beam is formed by welding together the thin plates of thickness 5 mm. 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 +\sum{\left(M_{ R} ight)_{A}}=\sum{ M_{A}} ; \quad-P e=-\left(F_{w} ight)_{1}(0.3)[/latex]

[latex]e=\frac{0.3\left(F_{w} ight)_{1}}{P}[/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}(0.005)\left(0.4^{3} ight)+\frac{1}{12}(0.005)\left(0.2^{3} ight)=30\left(10^{-6} ight) \mathrm{m}^{4}[/latex]

Referring to Fig. b, [latex]\bar{y}^{\prime}=(0.1-s)+\frac{s}{2}=(0.1-0.5 s) \mathrm{m} [/latex]. Thus, Q as a function of s is

[latex]Q=\bar{y}^{\prime} A^{\prime}=(0.1-0.5 s)(0.005 s)=\left[0.5\left(10^{-3} ight) s-2.5\left(10^{-3} ight) s^{2} ight] \mathrm{m}^{3}[/latex]

Shear Flow:

[latex]q=\frac{V Q}{I}=\frac{P\left[0.5\left(10^{-3} ight) s-2.5\left(10^{-3} ight) s^{2} ight]}{30\left(10^{-6} ight)}=P\left(16.6667 s-83.3333 s^{2} ight)[/latex]

Resultant Shear Force: The shear force resisted by the shorter web is

[latex]\left(F_{w} ight)_{1}=2 \int_{0}^{0.1 \mathrm{m}} q d s=2 \int_{0}^{0.1 \mathrm{m}} P\left(16.6667 s-83.3333 s^{2} ight) d s=0.1111 P[/latex]

Substituting this result into Eq. (1),

[latex]e=0.03333 \mathrm{m}=33.3 \mathrm{m}[/latex]

Question 7.60: The built-up beam is formed by welding together the thin pla...

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