The side of the building in Fig. 2–33a is subjected to a wind loading that creates a uniform normal pressure of 15 kPa on the windward side and a suction pressure of 5 kPa on the leeward side. Determine the horizontal and vertical components of reaction at the pin connections A, B, and C of the sup

Question

The side of the building in Fig. [latex] 2-33 (\mathrm{a})[/latex] is subjected to a wind loading that creates a uniform normal pressure of [latex]15 \mathrm{kPa}[/latex] on the windward side and a suction pressure of [latex]5 \mathrm{kPa}[/latex] on the leeward side. Determine the horizontal and vertical components of reaction at the pin connections [latex]A, B[/latex], and [latex]C[/latex] of the supporting gable arch.

Accepted Answer

Since the loading is evenly distributed, the central gable arch supports a loading acting on the walls and roof of the dark-shaded tributary area. This represents a uniform distributed load of [latex]\left(15 \mathrm{kN} / \mathrm{m}^{2} ight)[/latex] [latex](4 \mathrm{~m})=60 \mathrm{kN} / \mathrm{m}[/latex] on the windward side and [latex]\left(5 \mathrm{kN} / \mathrm{m}^{2} ight)(4 \mathrm{~m})=[/latex] [latex]20 \mathrm{kN} / \mathrm{m}[/latex] on the suction side, Fig. 2-33b.

Free-Body Diagrams. Simplifying the distributed loadings, the freebody diagrams of the entire frame and each of its parts are shown in Fig. 2-33c.

Equations of Equilibrium. Simultaneous solution of equations is avoided by applying the equilibrium equations in the following sequence using previously computed results.*

Entire Frame:

[latex] \begin{array}{c} \curvearrowleft +\Sigma M_{A}=0 ;-(180+60)(1.5)-(254.6+84.9) \cos 45^{\circ}(4.5) \ -\left(254.6 \sin 45^{\circ} ight)(1.5)+\left(84.9 \sin 45^{\circ} ight)(4.5)+C_{y}(6)=0 \ C_{y}=240.0 \mathrm{kN} \ +\uparrow \Sigma F_{y}=0 ; \quad-A_{y}-254.6 \sin 45^{\circ}+84.9 \sin 45^{\circ}+240.0=0 \ A_{y}=120.0 \mathrm{kN} \end{array} [/latex]

Member [latex]A B[/latex] :

[latex]\begin{array}{cc}\curvearrowleft+\Sigma M_{B}=0 ; & -A_{x}(6)+120.0(3)+180(4.5)+254.6(2.12)=0 \ & A_{x}=285.0 \mathrm{kN} \ \stackrel{+}{ ightarrow} \Sigma F_{x}=0 ; & -285.0+180+254.6 \cos 45^{\circ}-B_{x}=0 \ & B_{x}=75.0 \mathrm{kN} \ +\uparrow \Sigma F_{y}=0 ; & -120.0-254.6 \sin 45^{\circ}+B_{y}=0 \ & B_{y}=300.0 \mathrm{kN}\end{array}[/latex]

Member [latex]C B[/latex] :

[latex] \begin{array}{c} \stackrel{+}{ ightarrow} \Sigma F_{x}=0 ; \quad-C_{x}+60+84.9 \cos 45^{\circ}+75.0=0 \ C_{x}=195.0 \mathrm{kN} \end{array} [/latex]

*The problem can also be solved by applying the six equations of equilibrium only to the two members. If this is done, it is best to first sum moments about point [latex]A[/latex] on member [latex]A B[/latex], then point [latex]C[/latex] on member [latex]C B[/latex]. By doing this one obtains two equations to be solved simultaneously for [latex]B_{x}[/latex] and [latex]B_{y}[/latex].

Question 2.13: The side of the building in Fig. 2–33a is subjected to a win...

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