Question 1 |

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)

The magnitude (absolute value) of force (in kN) in member EF is _______. (round off to the nearest integer)

12 | |

30 | |

87 | |

92 |

Question 1 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

Question 2 |

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 _________

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 _________

370 | |

650 | |

760 | |

500 |

Question 2 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}

Question 3 |

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 ________

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 ________

0.4 | |

0.6 | |

0.9 | |

0.2 |

Question 3 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}

Question 4 |

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 _________.

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 _________.

30 | |

-30 | |

40 | |

-40 |

Question 4 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}

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}

Question 5 |

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

Neglecting self-weight of the members, the number of zero-force members in the truss under the action of the load P, is

6 | |

7 | |

8 | |

9 |

Question 5 Explanation:

As \Delta _{AB}=0, hence F _{AB}=0

Total number of zero force member = 8

Question 6 |

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

(30, -30, 30) kN | |

(20, 0, 10) kN | |

(10, 30, -10) kN | |

(0, 60, -30) kN |

Question 6 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}

Question 7 |

A plane truss is shown in the figure

Which one of the options contains ONLY zero force members in the truss?

Which one of the options contains ONLY zero force members in the truss?

FG, FI, HI, RS | |

FG, FH, HI, RS | |

FI, HI, PR, RS | |

FI, FG, RS, PR |

Question 7 Explanation:

So zero force members are FI, FG, RS, PR

Question 8 |

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:

For the loads shown on the truss, the statement that correctly represents the nature of forces in the members of the truss is:

There are 3 members in tension, and 2 members in compression | |

There are 2 members in tension, 2 members in compression, and 1 zero-force member | |

There are 2 members in tension, 1 member in compression, and 2 zero-force members | |

There are 2 members in tension, and 3 zero-force members |

Question 8 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).

Question 9 |

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 ______

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 ______

2.1 | |

3.6 | |

1.2 | |

2.7 |

Question 9 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}

\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}

Question 10 |

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?

Which one of the following sets gives the correct values of the force, stress and change in length of the horizontal member QR?

Compressive force = 25 kN; Stress = 250kN/{m^{2}}
; Shortening = 0.0118 m | |

Compressive force = 14.14 kN; Stress = 141.4kN/{m^{2}}
; Extension = 0.0118 m | |

Compressive force = 100 kN; Stress = 1000kN/{m^{2}}
; Shortening = 0.0417 m | |

Compressive force = 100 kN; Stress = 1000kN/{m^{2}}
; Extension = 0.0417 m |

Question 10 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}

There are 10 questions to complete.

have to work on indeterminate trusses