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**Q.1 Consider the pin-jointed plane truss shown in figure (not drawn to scale). 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 [2019 : 1 Mark, Set-I]** (B)

(a) (10, 30,-10) kN

(b) (30,-30, 30) kN

(c) (20, 0,10) kN

(d) (0,60,-30) kN

Ans.

Using,

...(i)

Taking moment about P,

Again considering the equilibrium of RHS of section X-X

∴ R

For the loads shown on the truss, the statement that correctly represents the nature of forces in the members of the truss is: [2018 : 1 Mark, Set-II]

(a) There are 3 members in tension, and 2 members in compression

(b) There are 2 members in tension, 2 members in compression, and 1 zero-force member

(c) There are 2 members in tension, 1 member in compression, and 2 zero-force members

(d) There are 2 members in tension, and 3 zero- force Members

Ans.

Since member BD neither elongate nor contract.

Hence, F

So there are 2 tension members (AB and DC) and 3 zero force members (AD, BD, BC).

Given that E = 2 x 10

Force in each member due to unit load.

Which one of the following sets gives the correct values of the force, stress and change in length of the horizontal member QR? [2016 : 2 Marks, Set-II]

(a) Compressive force = 25 kN;

Stress = 250 kN/m

(b) Compressive force = 14.14 kN;

Stress = 141.4 kN/m

(c) Compressive force = 100 kN;

Stress = 1000 kN/m

(d) Compressive force = 100 kN;

Stress = 1000 kN/m

Ans.

Given data:

Consider joint 'S’,

⇒

As the truss is symmetrical,

∴ (Tensile)

Now consider joint 'Q,

∴ (Compressive)

Stress in member OR,

⇒

∴

As the member QR is subjecterd to compression, it will go under shortening.

∴ Shortening,

∴

∴ Δ = 0.0471 m

The members which do not carry any force are [2016 : 2 Marks, Set-I]

(a) FT, TG, HU, MP, PL

(b) ET, GS, UR, VR, QL

(c) FT, GS, HU, MP, QL

(d) MP, PL, HU, FT, UR

Ans.

If there members meet at a joint and out of them are collinear, then non collinear member will carry zero force provided that there is no external load at the joint.

Use this statement to check the members with zero force.

Which one of the following sets gives the correct values of V

(a) V

(b) V

(c) V

(d) V

Ans.

⇒ V

First calculating reactions,

⇒

R

Let us cut a section 1-1 as shown in figure and consider the lower part.

Given AE = 30 kN and L = 3m,

Calculation for force Pin XY member.

Strain energy,

We have to find out vertical displacement of joint R in mm,

Apply unit load at R as shown below

Due to symmetric loading,

Consider P:

∑V = 0,

(a) 30 kN Compressive

(b) 30 kN Tensile

(c) 50 kN Compressive

(d) 50 kN Tensile

Ans.

Consider joint Q,

Taking moment about A,

R

⇒ R

R

At joint D,

⇒

Taking free body diagram as shown in figure,

Take moment about V,M

⇒ R

∴ R

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.

∴ F

Now, consider joint P,

Joint V,

Now, member RS and RU area non-collinear members meeting at a joint with no external force acting on the joint.

⇒

(a) zero

(b) P/√2

(c) P

(d) √2 P

Ans.

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

Considering joint R, we get

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