Q21GATE 2013MCQ

The pin-jointed 2-D truss is loaded with a horizontal force of 15 kN at joint S and another 15 kN vertical force at joint U, as shown. Find the force in member RS (in kN) and report your answer taking tension as positive and compression as negative. __________

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📖 Explanation

\begin{aligned} \Sigma M_{T}&=0 \\ \Rightarrow R_{v} \times 8-15 \times 4+15 \times 4 & =0 \\ \therefore \quad R_{V} & =0\\ \text{Take moment about V, } M_{V}&=0 \\ \Rightarrow \quad R_{H} \times 4&=0 \\ \therefore \quad R_{H}&=0 \end{aligned}

We know that if two members meet at a joint which are not collinear and also there is no external forces acting on that joint then both members will carry zero forces. $\therefore F_{O V}=F_{Q R}=0$ Now, consider joint R , $\Sigma F_{y}=0$

\begin{aligned} \Rightarrow F_{R U} \sin 45^{\circ}&=0 \\ \Rightarrow F_{R U}&=0\\ \text{Now},\quad \Sigma F_{x}&=0 \\ \Rightarrow \quad F_{R S}+F_{R U} \cos 45^{\circ}&=0 \\ \therefore F_{R S}&=0 \end{aligned}

So, force is member is "RS" will be zero.

Q22GATE 2010MCQ

For the truss shown in the figure, the force in the member QR is

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📖 Explanation

Using method of joints and considering joint S, we get,

\begin{aligned} \Sigma F_{x}&=0 \\ \Rightarrow \quad F_{T S}&=0\\ \text{and }\quad \Sigma F_{y}&=0 \\ \Rightarrow F_{R S}-P&=0 \\ \Rightarrow F_{R S}&=P(\text { tensile }) \end{aligned}

Considering joint R, we get

\begin{aligned} \Sigma F_{v}&=0 \\ \Rightarrow \quad-F_{I R} \sin 45^{\circ}-P&=0 \\ \Rightarrow \quad F_{I R}&=-P \sqrt{2} \text{(compressive)}\\ \Sigma F_{x}&=0 \\ \Rightarrow \quad-F_{T R} \cos 45^{\circ}-F_{Q R}&=0 \\ \Rightarrow \quad P \sqrt{2} \times \frac{1}{\sqrt{2}}-F_{\alpha R}&=0\\ \Rightarrow \quad F_{Q R}&=P\text{(tension)} \end{aligned}

Q23GATE 2008MCQ

The members EJ and IJ of a steel truss shown in the figure below are subjected to a temperature rise to 30 $^{\circ}$ C. The coefficient of thermal expansion of steel is 0.000012 per $^{\circ}$ C per unit length. The displacement (mm) of joint E relative to joint H along the direction HE of the truss, is

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📖 Explanation

As only EI and IJ are subjected to temperature rise only, so we will calculated forces in both these members due to unit load Relative displacement of joint E and H

\begin{array}{l} =\frac{1}{\sqrt{2}} \times 3000 \times 0.000012 \times 30 \\ =0.764 \mathrm{mm} \end{array}

Q24GATE 2007MCQ

The right triangular truss is made of members having equal cross sectional area of 1550 $mm^{2}$ and Young's modulus of $2\times 10^{5}$ MPa. The horizontal deflection of the joint Q is

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📖 Explanation

\begin{aligned} H_{R}-135 &=0 \\ H_{R} &=135 \mathrm{kN} \\ \tan \theta &=\frac{6}{4.5} \\ \sin \theta=& \frac{6}{7.5} ; \cos \theta=\frac{4.5}{7.5} \\ \text{Taking} \quad \Sigma M_{R}&=0 \\ \Rightarrow-V_{P} \times 4.5+135 \times 6 &=0 \\ \Rightarrow \quad & V_{P}=180 \mathrm{kN} \\ \therefore \quad & V_{R}=-180 \mathrm{kN} \end{aligned}

\begin{array}{ll} \Sigma F_{y}=0 & \Sigma F_{x}=0 \\ F_{R Q}-180=0 & F_{P O} \cos \theta-135=0 \\ F_{R Q}=180 \mathrm{kN} & F_{P Q}=\frac{135 \times 7.5}{4.5} \\ \text { and } \Sigma F_{x}=0 & F_{P Q}=225 \mathrm{kN} \\ F_{R P}-135=0 & \\ F_{R P}=135 \mathrm{kN} \end{array}

Tension (+); compression (-) and $K=\frac{P}{135}$ for each member,

\begin{aligned} \delta_{0} &=\Sigma \frac{P K L}{A E} \\ &=\frac{2818.125+1436.4+607.5}{A E} \\ &=\frac{4862.025 \times 10^{-6} \times 10^{3}}{1550 \times 10^{-6} \times 2 \times 10^{5} \times 10^{-3}} \\ &=15.68 \mathrm{mm} \end{aligned}

Q25GATE 2005MCQ

A truss is shown in the figure. Members are to equal cross section A and same modulus of elasticity E. A vertical force P is applied at point C. Deflection of the point C is

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📖 Explanation

It is evidence from the diagram that all the interior members will carry zero force. Assuming tensile forces as positive and compressive forces as negative Deflection at C, $\delta_{c}=\Sigma \frac{P K L}{A E}$ Where K is the force in member when a unit load is applied at C.

\begin{aligned} &=\frac{P L}{\sqrt{2} A E}+\frac{P L}{\sqrt{2} A E}+\frac{P L}{2 A E}\\ &=\frac{\sqrt{2} P L+\sqrt{2} P L+P L}{2 A E} \\ &=\left(\frac{2 \sqrt{2}+1}{2}\right) \frac{P L}{A E} \end{aligned}

Q26GATE 2004MCQ

For the plane truss shown in the figure, the number of zero force members for the given loading is

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Q27GATE 2003MCQ

In a redundant joint model, three bar members are pin connected at Q as shown in the figure. Under some load placed at Q, the elongation of the members MQ and OQ are found to be 48 mm and 35 mm respectively. Then the horizontal displacement 'u' and the vertical displacement 'v' of the node Q, in mm, will be respectively,

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📖 Explanation

$\text { Let } \angle M Q N=\theta_{1} \text { and } \angle O Q N=\theta_{2}$

\begin{aligned} \therefore \quad \tan \theta_{1}&=\frac{M N}{N Q}=\frac{400}{500}=0.8 \\ \Rightarrow \quad \sin \theta_{1}&=0.625 \\ \text{ and }\cos \theta_{1}&=0.781 \\ \text{and }\tan \theta_{2}&=\frac{N O}{N Q}=\frac{500}{500}=1 \\ \Rightarrow \quad \sin \theta_{2}&=0.707 \\ \text{ and }\cos \theta_{2}&=0.707 \\ \end{aligned}

Elongation of member MQ in terms of u and v will be,

\begin{aligned} u \sin \theta_{1}+v \cos \theta_{1}&=48 \\ \text{or, }0.625 u+0.781 v&=48 \quad &\ldots(i) \end{aligned}

Elongation of member OQ in terms of u and v will be,

\begin{aligned} \text{or, }-u \sin \theta_{2}+v \cos \theta_{2}&=35 \\ -0.707 u+0.707 v&=35 \quad \ldots(ii) \end{aligned}

Solving (i) and (ii) we get, $u=6.64 \mathrm{mm}\text{ and }v=56.14 \mathrm{mm}$

Q28GATE 2003MCQ

A truss, as shown in the figure, is carrying 180 kN load at node $L_2$ . The force in the diagonal member $M_2U_4$ will be

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📖 Explanation

Let reactions at $L_{0}$ and $L_{6}$ be $R_{L_{0}}$ and $R_{L_{6}}$ respectively. $\therefore R_{L_{0}}+R_{L_{6}}=180 \qquad\ldots(i)$ Taking moments about $L_{0}$ , we get, $R_{L_{6}} \times 24-180 \times 8=0$ $\Rightarrow \quad R_{L_{6}}=60 \mathrm{kN}$ $\therefore$ From (i), we get, $R_{L_{0}}=120 \mathrm{kN}$ Assuming a section passing through members $U_{3} U_{4}, M_{2} L_{3}$ and $L_{3} L_{4}$ and considering left portion Taking moments about $L_{3}=0$ $\Rightarrow 120 \times 12-180 \times 4+F_{U_{3}} u_{4} \times 3=0$ $\Rightarrow F_{U_{3} U_{4}}=-240 \mathrm{kN}=240 \mathrm{kN}$ (compression) Now assuming a section passing through members $U_{3} U_{4}, M_{2} U_{4}, M_{2} L_{4}$ and $L_{3} L_{4}$ and considering right portion. $\cos \theta=\frac{3}{5} ; \sin \theta=\frac{4}{5}$ Taking moments about $L_{4}=0$

\begin{array}{l} F_{U_{3} U_{4}} \times 3-\left(F_{M_{2} U_{4}} \sin \theta\right) \times 3-60 \times 8=0 \\ \Rightarrow \quad 240 \times 3-480=F_{M_{2} U_{4}} \times 3 \times \frac{4}{5} \\ \Rightarrow \quad F_{M_{2} U_{4}}=100 \mathrm{kN} (tension) \end{array}

Q29GATE 2001MCQ

Identify the FALSE statement from the following, pertaining to the effects due to a temperature rise $\Delta T$ in the bar BD alone in the plane truss shown below:

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