Q1GATE 2025MSQ

In the pin-jointed truss shown in the figure, the members that carry zero force are identified. Which of the following options is/are zero-force members?

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📖 Explanation

To identify zero-force members, we use the following rules: Rule 1: If two non-collinear members meet at a joint unstressed by an external force or support, then both members are zero-force members. Rule 2: If three members form a truss joint and two of them are collinear, the third member is a zero-force member provided no external load or support reaction is applied at that joint. Apply Rule 1: At joint B, members BC and BD meet, and there is no external load or support at B. By Rule 1, $\text{BC is a zero-force member}$ . Apply Rule 1 again: At joint J, members JK and JL meet, with no external force or support. By Rule 1, $\text{JK is a zero-force member}$ . Final Answer: $\text{Zero-force members are: BC and JK}$

Q2GATE 2024NAT

The plane truss shown in the figure has 13 joints and 22 members. The truss is made of a homogeneous, prismatic, linearly elastic material. All members have identical axial rigidity. $\mathrm{A}$ to $\mathrm{M}$ indicate the joints of the truss. The truss has pin supports at joints $\mathrm{A}$ and $L$ and roller support at joint $K$ . The truss is subjected to a $10 \mathrm{kN}$ vertically downward force at joint $\mathrm{H}$ and a $10 \mathrm{kN}$ horizontal force in the rightward direction at joint $\mathrm{B}$ as shown The magnitude of the reaction (in $\mathrm{kN}$ ) at the pin support $\mathrm{L}$ is ____ (rounded off to 1 decimal place).

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📖 Explanation

Considering left portion of section 1-1

\begin{array}{rlrl} \Sigma M_{D} & =0 \\ \Rightarrow & V_{A} \times 2-H_{A} \times 2 & =0 \\ \Rightarrow & V_{A} & =H_{A} \end{array}

Considering right portion of section 2 - 2

\begin{aligned} \Sigma M_{I}=0 \Rightarrow & H_{L} \times 2=0 \\ \Rightarrow \quad & H_{L} \times 2=0 \end{aligned}

Considering left portion of section 2 - 2 $\Sigma M_{I}=0 \Rightarrow V_{A} \times 6-H_{A} \times -10 \times 1=0$ $\Rightarrow \quad 6 V_{A}-2 H_{A}=10$ $\Rightarrow \quad V_{A}=2.5 \mathrm{kN} \quad\left[\because H_{A}=V_{A}\right]$ Now, $\quad \Sigma F_{y}=0$ $\Rightarrow \quad V_{A}+V_{L}-10=0$ $\Rightarrow \quad V_{L}=7.5 \mathrm{kN}$

Q3GATE 2023NAT

An idealised bridge truss is shown in the figure. The force in Member $U_{2} L_{3}$ is _____ kN (round off to one decimal place).

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📖 Explanation

$\sum F_{y}=0, \quad 50-20-20-F_{U_{2} L_{3}} \sin 45^{\circ}=0$ $\therefore \quad F_{U_{2} L_{3}}=\frac{10}{\sin 45^{\circ}}=14.14 \mathrm{kN}$ $\therefore$ Rounding off to one decimal.

Q4GATE 2023MCQ

Consider the pin-jointed truss shown in the figure (not to scale). All members have the same axial rigidity, $\mathrm{AE}$ . Members $\mathrm{QR}, \mathrm{RS}$ , and $\mathrm{ST}$ have the same length L. Angles OBT, RCT, SDT are all $90^{\circ}$ . Angles BQT, CRT, DST are all $30^{\circ}$ . The joint $T$ carries a vertical load $P$ . The vertical deflection of joint $T$ is $k \frac{P L}{A E}$ . What is the value of $k$ ?

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📖 Explanation

Considering joint $T$ , $\sum F_{y}=0, \quad F_{Q T} \sin 60^{\circ}=P$ $\therefore \quad \mathrm{F}_{\mathrm{QT}}=\frac{2 \mathrm{P}}{\sqrt{3}}$ $\sum \mathrm{F}_{\mathrm{X}}=0, \mathrm{~F}_{\mathrm{BT}}=-\mathrm{F}_{\mathrm{QT}} \cos 60^{\circ}=-\frac{P}{\sqrt{3}}$ $\therefore \quad$ Strain energy $(U)=\left(\frac{P^{2} L}{2 A E}\right)_{Q T}+\left(\frac{P^{2} L}{2 A E}\right)_{B T}$ $=\frac{\left(\frac{2 \mathrm{P}}{\sqrt{3}}\right)^{2}(3 \mathrm{~L})}{2 \mathrm{AE}}+\frac{\left(-\frac{\mathrm{P}}{\sqrt{3}}\right)^{2}\left(\frac{3 \mathrm{~L}}{2}\right)}{2 \mathrm{AE}}$ $=\frac{2 P^{2} \times L}{A E}+\frac{P^{2} \times L}{4 A E}$ $\frac{8 P^{2} L+P^{2} L}{4 A E}=\frac{9 P^{2} L}{4 A E}$ $\therefore \quad$ Vertical deflection of joint $T$ $=\frac{\partial \mathrm{U}}{\partial \mathrm{P}}=\frac{18 \mathrm{PL}}{4 \mathrm{AE}}=4.5 \frac{\mathrm{PL}}{\mathrm{AE}}$ $\therefore \quad \mathrm{K}=4.5$

Q5GATE 2022NAT

The plane truss shown in the figure is subjected to an external force $P$ . It is given that $P = 70 kN, a = 2 m,$ and $b = 3 m$ . The magnitude (absolute value) of force (in kN) in member $EF$ is _______. (round off to the nearest integer)

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📖 Explanation

\begin{aligned} \Sigma M_J&=0\\ V_A \times 8-H_A \times 1-70 \times 4&=0\\ 8V_A-H_A&=280 \;\;\;(i) \end{aligned}

To find $H_A$ (Cut the truss by 1-1) Consider left hand side

\begin{aligned} \Sigma M_E&=0\\ V_A \times 4-H_A \times 4&=0\\ V_A&=H_A \;\;\;(ii)\\ \text{using (i) and (ii)}&\\ 8V_A-V_A&=280\\ 7V_A&=280\\ V_A&=40kN\\ H_A&=40kN\\ H_J&=40kN\\ V_J&=70-40=30kN\\ \end{aligned}

To find force in member EF (Cit the truss by 2-2) Consider right hand side Force in member EF $F_{EF}=V_J=30kN$

Q6GATE 2021NAT

Refer the truss as shown in the figure (not to scale). If load, $F=10 \sqrt{3} \mathrm{kN}$ , moment of inertia, $I=8.33 \times 10^{6} \mathrm{~mm}^{4}$ , area of cross-section, $A=10^{4} \mathrm{~mm}^{2}$ , and length, L=2 m for all the members of the truss, the compressive stress (in $\mathrm{kN} / \mathrm{m}^{2}$ ,in integer) carried by the member Q-R is _________

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📖 Explanation

$V_{P}=V_{S}=5 \sqrt{3} \mathrm{kN}$ Considering equilibrium of LHS of section (1 )-(1) Taking moment about 'T'

\begin{aligned} \Sigma M_{T}(C W)&=0\\ (5 \sqrt{3} \times a)+F_{Q R}\left(\frac{\sqrt{3} a}{2}\right)&=0\\ \Rightarrow \qquad \qquad \qquad \qquad \quad F_{O A}&=-10 \mathrm{kN} or 10 \mathrm{kN} (C) \\ \text{ Compressive} &\text{ stress \in member QR} \left(\sigma_{\mathrm{C}}\right)\\ \sigma_{\mathrm{C}}&=\frac{F_{Q A}}{2 A} \\ &=\frac{10 \mathrm{kN}}{2 \times \left(10^{4} \times 10^{-6}\right) \mathrm{m}^{2}} \\ &=500 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Q7GATE 2021NAT

A truss EFGH is shown in the figure, in which all the members have the same axial rigidity R. In the figure, P is the magnitude of external horizontal forces acting at joints F and G. If $R=500 \times 10^{3} \mathrm{kN}$ , $P=150 \mathrm{kN}$ and L=3 m, the magnitude of the horizontal displacement of joint G (in mm,round off to one decimal place) is ________

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📖 Explanation

Note: No need to calculate 'I< force in all members because 'P force is zero fore all members except FG By unit load method

\begin{aligned} \text { 1. } \Delta_{\mathrm{HG}}=& \sum_{i=1}^{n} \frac{P_{i} k_{i} L_{i}}{A_{i} E_{i}} \\ \Delta_{\mathrm{HG}}=& \text { Horizontal deflection at joint } \mathrm{G} \\ \therefore \qquad & \qquad \qquad\Delta_{\mathrm{HG}} =\underbrace{\frac{P \times 1 \times L}{A E}}_{\text {FG-member }} \times \underbrace{Q^{\prime}}_{\text {(For all other members) }} \\ & \qquad \qquad\Delta_{\mathrm{HG}} =\frac{P L}{A E}=\left(\frac{150 \times 3}{500 \times 10^{3}} \times 10^{3}\right) \mathrm{mm} \\ & \qquad \qquad \qquad=0.90 \mathrm{~mm} \end{aligned}

Q8GATE 2020MCQ

Consider the planar truss shown in the figure (not drawn to the scale) Neglecting self-weight of the members, the number of zero-force members in the truss under the action of the load P, is

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📖 Explanation

As $\Delta _{AB}=0$ , hence $F _{AB}=0$ Total number of zero force member = 8

Q9GATE 2020NAT

The plane truss has hinge supports at P and W and is subjected to the horizontal forces as shown in the figure. Representing the tensile force with '+' sign and the compressive force with '-' sign, the force in member XW (in kN, round off to the nearest integer), is _________.

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📖 Explanation

Force in PQ Considering the section above (1) - (1) Taking moment about 'R'

\begin{aligned} \Sigma M_{RL}&=0\\ (10 \times 4)+(10 \times 8)&+F_{PQ} \times 4=0\\ F_{PQ}&=-\frac{120}{4}\\ &=-30kN=30kN (Comp.) \end{aligned}

Q10GATE 2019MCQ

A plane truss is shown in the figure Which one of the options contains ONLY zero force members in the truss?

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Q11GATE 2019MCQ

Consider the pin-jointed plane truss shown in the figure. Let $R_P,R_Q$ and $R_R$ denote the vertical reactions (upward positive) applied by the supports at P, Q, and R, respectively, on the truss. The correct combination of ( $R_P,R_Q,R_R$ )is represented by

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📖 Explanation

Adopting method of sections and taking LHS of the section

\begin{aligned} &\Sigma F_y= 0\\ &R_P=30kN \\ &\text{For complete truss,} \\ &\Sigma M_R =0 \\ &9R_P-30 \times 6 -R_Q \times 3=0 \\ &R_Q=30kN(\downarrow ) \\ &\text{Taking RHS of section,} \\ &\Sigma F_y=0\Rightarrow R_R=-R_Q \\ &\text{Thus},\;\; R_Q= 30kN(\downarrow )\\ &R_R=30kN(\uparrow ) \end{aligned}

Q12GATE 2018NAT

Consider the deformable pin-jointed truss with loading, geometry and section properties as shown in the figure. Given that $E =2 \times 10^{11} N/m^{2}$ , $A = 10 mm^{2}$ , L = 1 m and P = 1 kN. The horizontal displacement of Joint C (in mm, up to one decimal place) is ______

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📖 Explanation

Force is each member due to applied loading

\begin{aligned} F_{A B}&=P(\text { Comp. }) \\ F_{B C}&=3 P(\text { Comp. }) \\ F_{A C}&=\sqrt{2} P(\text { Tension }) \end{aligned}

Force in each member due to unit load.

\begin{aligned} F_{A B}&=P(\text { Comp. }) \\ F_{B C}&=3 P(\text { Comp. }) \\ F_{A C}&=\sqrt{2} (\text { Tension }) \end{aligned}

\begin{array}{l} \therefore \text{ Total deflection }=\delta_{H_{c}}=\Sigma \frac{P k L}{A E} \\ =\frac{5.414 \times (1000 \mathrm{N}) \times (1000 \mathrm{mm})}{\left(10 \mathrm{mm}^{2}\right) \times \left(2 \times 10^{5} \mathrm{N} / \mathrm{mm}^{2}\right)}=2.7 \mathrm{mm} \end{array}

Q13GATE 2018MCQ

All the members of the planar truss (see figure), have the same properties in terms of area of cross-section (A) and modulus of elasticity (E). For the loads shown on the truss, the statement that correctly represents the nature of forces in the members of the truss is:

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📖 Explanation

Since member BDneither elongate nor contract. Hence, $\quad F_{B D}=0$ . So, there are 2 tension members (AB and DC) and 3 zero force members (AD, BD, BC).

Q14GATE 2016MCQ

A plane truss with applied loads is shown in the figure.The members which do not carry any force are The members which do not carry any force are

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Q15GATE 2016MCQ

Consider the plane truss with load P as shown in the figure. Let the horizontal and vertical reactions at the joint B be $H_{B}$ and $V_{B}$ , respectively and $V_{C}$ be the vertical reaction at the joint C. Which one of the following sets gives the correct values of $V_{B}, H_{B}$ and $V_{C}$ ?

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📖 Explanation

\begin{aligned} V_{B}+V_{C} &=P &\ldots(i)\\ \Sigma M_{B} &=0\\ V_{C} \times 2 L-P \times 2 L&=0 \\ \Rightarrow \quad V_{C}&=P \\ \text{From(i), }\quad V_{B}&=0 \\ \Sigma F_{x}&=0 \\ \Rightarrow \quad H_{B}&=0 \end{aligned}

Q16GATE 2016MCQ

Consider the structural system shown in the figure under the action of weight W. All the joints are hinged. The properties of the members in terms of length (L), area (A) and the modulus of elasticity (E) are also given in the figure. Let L, A and E be 1 m, $0.05m^{2}$ and $30\times 10^{6}N/m^{2}$ , respectively, and W be 100 kN. Which one of the following sets gives the correct values of the force, stress and change in length of the horizontal member QR?

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📖 Explanation

Given data:

\begin{array}{l} L=1 \mathrm{m}, A=0.05 \mathrm{m}^{2} \\ E=30 \times 10^{6} \mathrm{N} / \mathrm{m}^{2} \end{array}

Consider joint 'S.

\begin{aligned} & F_{\mathrm{SO}}=F_{\mathrm{SR}} \\ 2 F_{\mathrm{SQ}} \cos 45^{\circ}=& \mathrm{W} \\ \Rightarrow \quad F_{\mathrm{SQ}}=& \frac{W}{2} \times \sqrt{2}=\frac{W}{\sqrt{2}} \\ \therefore \quad & F_{\mathrm{SO}}=\frac{W}{\sqrt{2}} \end{aligned}

As the truss is symmetrical, $\therefore \quad F_{\mathrm{OP}}=F_{\mathrm{PR}}=\frac{W}{\sqrt{2}} \qquad \text{(Tensile)}$ Now consider joint 'Q',

\begin{aligned} F_{\mathrm{OP}}&=F_{\mathrm{OS}}=\frac{W}{\sqrt{2}} \\ \Sigma F_{x}&=0\\ \Rightarrow \frac{F_{Q P}}{\sqrt{2}}&+\frac{F_{\alpha S}}{\sqrt{2}}+F_{Q R}=0\\ \Rightarrow \quad F_{\mathrm{QR}}&=\mathrm{W} &\text{(Compressive)}\\ \therefore \quad F_{\mathrm{OR}}&=100 \mathrm{kN} &\text{(Compressive)} \end{aligned}

Stress in member Q R,

\begin{aligned} \sigma _{QR} &=\frac{F_{QR}}{2A} \\ \sigma _{QR} &=\frac{100}{2 \times 0.05} \\ \sigma _{QR}&=1000kN/m^2 \end{aligned}

As the member QR consist compressive tone, so, it will go under shortening. Sortening,

\begin{aligned} \Delta &=\frac{P(Length)}{2AE}=\frac{F_{QR}L_{QR}}{2AE}\\ L_{QR}&=\sqrt{L^{2+}L^2}=\sqrt{2}L\\ \therefore \;\; \Delta &=\frac{100 \times 10^3 \times \sqrt{2} \times 1}{2 \times 0.05 \times 30 \times 10^6}\\ &=\frac{100 \times 10^3 \times \sqrt{2}}{0.1 \times 30 \times 10^6}\\ &=\frac{\sqrt{2}}{30}\\ \therefore \;\; \Delta &=0.471 m \end{aligned}

Q17GATE 2015NAT

For the 2D truss with the applied loads shown below, the strain energy in the member XY is _______ kN-m. For member XY, assume AE = 30 kN, where A is cross-section area and E is the modulus of elasticity.

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📖 Explanation

First calculating reactions

\begin{aligned} \Sigma M_{A}&=0\\ \Rightarrow\quad R_{B} \times 3-10 \times 9-5 \times 3 &=0 \\ R_{B} &=35 \mathrm{kN} \\ H_{A} &=10 \mathrm{kN} \\ \Sigma M_{B} &=0 \\ \Rightarrow\quad R_{A} \times 3+10 \times 9 &=0 \\ R_{A} &=-30 \mathrm{kN} \end{aligned}

Let us cut a section 1-1 as shown in figure and consider the lower part. Given AE = 30 kN and L = 3 m. Calculation for force Pin XY member

\begin{aligned} \Sigma F_{\text {horizontal }} &=0 \\ \Rightarrow \quad F_{X Y}&=10 \mathrm{kN} \end{aligned}

Strain energy,

\begin{aligned} U&=\frac{P^{2} L}{2 A E}\\ U &=\frac{10^{2} \times 3}{2 \times 30} \\ &=5 \mathrm{kNm} \end{aligned}

Q18GATE 2014NAT

For the truss shown below, the member PQ is short by 3 mm. The magnitude of vertical displacement of joint R (in mm) is _______________

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📖 Explanation

PQ is short by 3 mm We have to find out vertical displacement of joint R in mm, $\Delta_{R}=\Sigma u(\lambda)$ Let in apply unit load at R as shown below

\begin{aligned} u_{P R} \sin \theta &=\frac{1}{2} \\ u_{P Q}+U_{P A} \cos \theta &=0 \\ u_{P O}=-U_{P R} \cos \theta &=-\frac{1}{2 \sin \theta} \cdot \cos \theta \\ u_{P O} &=-\frac{1}{2} \cot \theta \\ &=\frac{-1 \times 4 / 3}{2}=\frac{-2}{3} \\ \Delta_{R} &=u_{P O} \times \lambda_{P O} \\ &=\frac{-2}{3}(-3) \\ &=2 \mathrm{mm}\text{ upwards} \end{aligned}

Q19GATE 2014MCQ

Mathematical idealization of a crane has three bars with their vertices arranged as shown in the figure with a load of 80 kN hanging vertically. The coordinates of the vertices are given in parentheses. The force in the member QR, $F_{QR}$ will be

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📖 Explanation

\begin{aligned} V_{0}+V_{A} &=80 &\ldots(i)\\ \Sigma M_{R} &=0 \\ \Rightarrow\quad 80 \times 3 &=V_{0} \times 2 \\ \therefore\quad V_{O} &=120 \mathrm{KN} \\ \Rightarrow\quad V_{R} &=-40 \mathrm{kN} \quad \text { From eq }(1) \\ \tan \phi &=\frac{1}{4} \\ \Rightarrow\quad \sin \phi &=\frac{1}{\sqrt{17}}, \cos \phi=\frac{4}{\sqrt{17}} \\ \therefore\quad \tan \theta &=\frac{3}{4} \\ \Rightarrow\quad \cos \theta &=\frac{4}{5} \end{aligned}

Consider joint Q

\begin{aligned} \alpha &=104.03^{\circ}-90^{\circ}=14.03^{\circ} \\ \Sigma F_{y} &=0 \\ \Rightarrow\quad F_{P Q} \cos \alpha &=+V_{Q}=0 \\ \Rightarrow\quad F_{P Q} \cos 14.03^{\circ} &+120=0 \\ \Rightarrow\quad F_{P Q} &=-123.6897 \mathrm{kN} \\ \Sigma F_{x} &=0 \\ \Rightarrow\quad F_{P Q} \sin \alpha &=F_{O R} \\ \therefore\quad F_{O R} &=-29.986 \mathrm{kN} \\ & \simeq 30 \mathrm{kNm}(\text { comp. }) \end{aligned}

Q20GATE 2013MCQ

A uniform beam weighting 1800 N is supported at E and F by cable ABCD. Determine the tension (in N) in segment AB of this cable (correct to 1-decimal place). Assume the cable ABCD, BE and CF to be weightless. _____

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📖 Explanation

Taking moment about A,

\begin{aligned} R_{D} \times 4 &=1800 \times 1.5 \\ \Rightarrow R_{D}&=675 \mathrm{N} \\ R_{A}= 1800-R_{D}&=1125 \mathrm{N} \end{aligned}

At joint D, $T_{\infty }=\frac{R_{D}}{\sin 45}=675 \sqrt{2} \mathrm{N}$ Taking free body diagram as shown in figure,

\begin{aligned} \Sigma V &=0 \\ T_{A B} \sin \theta+675 \sqrt{2} \sin 45^{\circ} &=1800 \\ T_{A B} \sin \theta &=1125 &\ldots(i)\\ \Sigma H &=0 \\ T_{A B} \cos \theta &=675 \sqrt{2} \cos 45^{\circ} \\ \Rightarrow \quad T_{A B} \cos \theta &=675 &\ldots(ii)\\ \text{From} e q \cdot(i) \text { and }(i i) & \\ \tan \theta &=\frac{5}{3} \\ \sin \theta &=0.8575 \end{aligned}

Putting value of $\sin \theta$ in eq. (i)

\begin{aligned} T_{A B} \times 0.8575 &=1125 \\ T_{A B} &=1311.96 \mathrm{N} \end{aligned}

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