Question

Determine the reactions at the supports for the structure shown

Answer 1

Let \(E\) be the crown point.

Considering Half of the member

\(\begin{array}{ll}\text { Length, } & x=15 \mathrm{~m} \\ \text { Rise, } & y=8 \mathrm{~m} \text { and }\end{array}\)

Length of Incline member, \(\mathrm{AE}\) or \(\mathrm{BE}=\sqrt{(15)^{2}+(8)^{2}}=17 \mathrm{~m}\)

Therefore, \(\sin \theta=\frac{8}{17} \quad\) and \(\quad \cos \theta=\frac{15}{17}\)

\(\begin{aligned} \text { Vertical component of loads } &=10 \times 17 \times \cos \theta \\ &=10 \times 17 \times \frac{15}{17}=150 \mathrm{kN} \\ \text { Horizontal component of loads } &=10 \times 17 \times \sin \theta \\ &=10 \times 17 \times \frac{8}{17}=80 \mathrm{kN} \end{aligned}\)

Since the \(B\) is place on the rollers, with the roller base horizontal, the reaction at B should be vertical. Hence there will be no horizontal reaction at \(B\).

Let vertical reaction at \(\mathrm{B}\) be \(\mathrm{V}_{\mathrm{B}}\)

Taking moments about \(\mathrm{A}\)

$$ \begin{aligned} &\mathrm{V}_{\mathrm{B}} \times 30+150 \times 7.5=150 \times 22.5 \quad \Rightarrow \mathrm{V}_{\mathrm{B}}=\frac{3375-1125}{30}=75 \mathrm{kN} \uparrow \\ &\sum \mathrm{F}_{\mathrm{Y}}=0 \\ &\mathrm{~V}_{\mathrm{A}}=(150-150)-75=-75 \mathrm{kN} \downarrow \end{aligned} $$

Horizontal reaction at \(\mathrm{A}\)

$$ \begin{aligned} &\sum F_{X}=0 \\ &H_{A}=80+80=160 \mathrm{kN} \rightarrow \end{aligned} $$

(Since both loads are acting towards left)