Mechanical and structural analysis miscellaneous
- A uniform beam weighing 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 cables ABCD, BE and CF to be weightless.
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Taking Moment about A.T3 = 1800 × 1.25 = 1125 N 2 T4 = 1800 × 0.75 = 675 N 2
We get, VB × 4.5 = (1125 × 1.5 + 675 × 3.5)
⇒ VB = 900 kN ...(i)
⇒ VA = (1125 + 675 – 900) = 900 kN
M1, beam = (VA × 1.5) = 900 × 1.5 = 1350 Nm
M2, beam = (VA × 3.5 – 1125 × 2)
= (900 × 3.5 – 2250) = 900 Nm
Taking bending moment about point (i) [as we know that bending moment at every point in the cable is zero]
⇒ Hy1 = M1, beam = 1350 ...(ii)
Taking bending moment about point (ii)
⇒ Hy2 = 900⇒ H = 900 = 900 N 1
⇒ 900 y1 = 1350∴ y1 = 
1350 
= 1.5 m ....(iii) 900 ∴ θ1 = tan-1 y1 = tan-1 1.5 = 45 1.5 1.5 θ2 = tan-1 y1 - y2 = tan-1 1.5 - 1 = 14.03 2 2
From Lamis Theorem⇒ T1 = 1125 sin(90 + θ2) sin{[180 - (θ1 + θ2)]} ⇒ T1 = 1125sin(90 + 14.03) = 1272.91 sin[180 - (45 + 14.03)] Correct Option: C
Taking Moment about A.T3 = 1800 × 1.25 = 1125 N 2 T4 = 1800 × 0.75 = 675 N 2
We get, VB × 4.5 = (1125 × 1.5 + 675 × 3.5)
⇒ VB = 900 kN ...(i)
⇒ VA = (1125 + 675 – 900) = 900 kN
M1, beam = (VA × 1.5) = 900 × 1.5 = 1350 Nm
M2, beam = (VA × 3.5 – 1125 × 2)
= (900 × 3.5 – 2250) = 900 Nm
Taking bending moment about point (i) [as we know that bending moment at every point in the cable is zero]
⇒ Hy1 = M1, beam = 1350 ...(ii)
Taking bending moment about point (ii)
⇒ Hy2 = 900⇒ H = 900 = 900 N 1
⇒ 900 y1 = 1350∴ y1 = 1350 = 1.5 m ....(iii) 900 ∴ θ1 = tan-1 y1 = tan-1 1.5 = 45 1.5 1.5 θ2 = tan-1 y1 - y2 = tan-1 1.5 - 1 = 14.03 2 2
From Lamis Theorem⇒ T1 = 1125 sin(90 + θ2) sin{[180 - (θ1 + θ2)]} ⇒ T1 = 1125sin(90 + 14.03) = 1272.91 sin[180 - (45 + 14.03)]
- A uniform beam (EI = constant) PQ in the form of a quartercircle of radius R is fixed at end P and free at the end Q, where a load W is applied as shown. The vertical downward displacement, δq, at the loaded point Q is given by: δq =
β WR3 EI
Find the value of b(correct to 4-decimal places).
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Correct Option: A
- All members in the rigid-jointed frame shown are prismatic and have the same flexural stiffness EI. Find the magnitude of the bending moment at Q (in kNm) due to the given loading.
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Correct Option: D
- The pin-jointed 2-D truss is loaded with a horizontal force of 15 kN at joint S and another 15kN 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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∑ MT = 0
⇒ Rv × 8 – 15 × 4 + 15 × 4 = 0
∴ Rv = 0
Take moment about V, MV = 0
⇒ RH × 4 = 0
∴ RH = 0
We know that if two members meet at a joint which are not col-linear and also there is no external forces acting on that joint then both members will carry zero forces.
∴ FQV = FQR = 0
Now, consider joint R
∑ Fy = 0
⇒ FRU sin 45º = 0
⇒ FRU = 0
Now, 0 ∑ Fx = ⇒ FRS + FRU cos 45º = 0
⇒ FRS + FRU cos 45º = 0
∴ FRS = 0
So, force is member is “RS” will be zero.Correct Option: A
∑ MT = 0
⇒ Rv × 8 – 15 × 4 + 15 × 4 = 0
∴ Rv = 0
Take moment about V, MV = 0
⇒ RH × 4 = 0
∴ RH = 0
We know that if two members meet at a joint which are not col-linear and also there is no external forces acting on that joint then both members will carry zero forces.
∴ FQV = FQR = 0
Now, consider joint R
∑ Fy = 0
⇒ FRU sin 45º = 0
⇒ FRU = 0
Now, 0 ∑ Fx = ⇒ FRS + FRU cos 45º = 0
⇒ FRS + FRU cos 45º = 0
∴ FRS = 0
So, force is member is “RS” will be zero.
Direction: A rigid beam is hinged at one end and supported on linear elastic springs (both having a stiffness of ‘k’) at points ‘1’ and ‘2’ and an inclined load acts at ‘2’, as shown.
- If the load P equals 100 kN, which of the following options represents forces R1 and R2 in the springs at points ‘1’ and ‘2’?
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At 1,
R1 = kδ1 = 2p = 2 × 100 = 40 kN 5 5
At 2,R2 = kδ2 = 4p = 80 kN 5 Correct Option: D
At 1,
R1 = kδ1 = 2p = 2 × 100 = 40 kN 5 5
At 2,R2 = kδ2 = 4p = 80 kN 5
