Frame Truss and Beam MCQ Quiz in मराठी - Objective Question with Answer for Frame Truss and Beam - मोफत PDF डाउनलोड करा

Explanation:

Taking moment about point P

F ×  L - T _RS × L = 0

T RS = F

Force in member RS  = F (compressive)

Now FBD of point R

Taking summation of force in Y direction 

-T_PR sin45° + T_RS = 0

-TPR sin45° + F = 0

T_PR = F √ 2

Forces in the members PR  F √ 2 (tensile)

Explanation:

When rod swings to right, Linear acceleration & angular acceleration (∝) comes into picture

Let R = reaction at hinge

Linear acceleration

\(a = \propto r = \frac{L}{2} \times \propto \Rightarrow \propto = \frac{{2a}}{L}\)

and \({\rm{\Sigma }}{M_G} = {I_G} \times \propto\)

\(R\left( {\frac{L}{2}} \right) + P\left( {\frac{L}{6}} \right) = \frac{{M{L^2}}}{{12}}\left( {\frac{{2a}}{L}} \right)\\ \Rightarrow a = \frac{{3R}}{M} + \frac{P}{M}\)           ______________(1)

\(P-R = Ma = M\left( {\frac{{3R}}{M} + \frac{P}{M}} \right)\)

⇒ P – R = 3R + P ⇒ R = 0

Concept:

The force in the member AB will be determined by the method of joints, i.e. by analyzing each joint.

For this we will be fixing some rules, they are

• for every joint we will consider the unknown forces are going away from the joint
• and the forces are considered positive.
• So after determination, if one force comes positive then, it will mean that it is tensile and if it comes negative then it would mean that it is compressive.

Calculations:

Given:

\(Sin θ =\frac{3}{5}\)

\(Cos θ =\frac{4}{5}\)

θ = 36.86° 

Consider Point A:

Vertical Force:

P + F_ABsinθ = 0 

FABsinθ = -P

\(F_{AB}=\frac{-P}{Sin θ}=\frac{-P}{3/5}=\frac{-5P}{3}~N\)

Horizontal Force:

F_AC + F_ABCosθ = 0

FAC = -FABCosθ 

\(F_{AC}=\frac{-(-5P)}{3}\times \frac45=\frac{4P}{3}~N\)

Concept:

Since  θ = 45° , so

PQ = QR = 4 m

And \(\mathop \sum \nolimits {M_Q} = 0\)

100 × Cos 60° × 4 = RR × 4

R_R = 50 kN

Now from equilibrium, F_PR cos 45° = 100 cos 60°

\(\Rightarrow {F_{PR}} = 70.71~ kN\)

Explanation:

Conditions of Zero force member in truss : 

(1)   If a joint has only two non – collinear members and there is no external load or support reaction at that joint, then those two members are zero force members.

(2)   If three members from a truss joint for which two of the members are collinear and there is no external load or reaction at that joint, then the third non-collinear member is a zero force member.    

Here in our problem Condition 2 is satisfied so at a joint B member BC will become a zero force member. After this at joint C member CD becomes a zero force member and this process continues at joint D and joint E so members DE and EF also become zero force members.

Concept:

1) When a rotating body is involved then there comes a force which is known as Centrifugal force (F_c)

It is calculated by using formula

F_c = mr(ω)²

2) Now in the Question, it is mentioned that end Q is just lifted off the ground. So as it is just lifted off the ground there will be no reaction force from that point.

∴ FBD of mobile

M = mass of mobile

m = eccentric mass

r = eccentricity

Hence, Taking a moment about point P = O

∴ Mg × 0.06 - F_c × (0.09) = 0

∴ 90 × 10^-3 × 9.81 × 0.06 = mr(ω)² × (0.09)

90 × 10^-3 × 9.81 × 0.06 = 2 × 10^-3 × 2.19 × 10^-3 (ω)² × 0.09

∴ ω² = 134380.8964

∴ ω = 366.58

\(\therefore \frac{{2\pi N}}{{60}} = 366.58\)

∴ N = 3500 rpm

Explanation:

(1) All forces directed along member in truss only. In frame forces may or may not be directed along member.

(2) For rigid frame even after removing support, structure of frame doesn’t collapsed.

(3) Support reaction can be find out by considering whole frame structure.

(4) Ideal truss structure has pin joint only.

Explanation:

The strength of two beams of the same material can be compared by the section modulus values.

The beam is stronger when section modulus is more, the strength of the beam depends on section modulus.

The strength of the beam also depends on the material, size, and shape of the cross-section. 

Section modulus of a beam can be expressed as

 \({\sigma _b} = \frac{M}{Z}\)

Therefore, 

\(Z = \frac{M}{{\sigma _b}}\)

where Z is the section modulus and found out as \(Z = \frac{I}{y}\)

\(Z=\frac{I}{y}=\frac{\frac{\pi d^4}{64}}{\frac{d}{2}}={\frac{\pi d^3}{32}}\)

Strength of the beam ∝ Z ∝ d3

According to method of section:

\(\sum {M_E} = 0 \Rightarrow {F_{AB}} \times 2 = W\left( {1.5} \right) + 3W\left( {4.5} \right) = 15W\)

FAB = 7.5 W

7.5 W = 15 kN

W = 2 kN