A uniform beam weighing 1800 N is supported at E and F by

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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.

1. 1272.9198
2. 127
3. 1272.91
4. 1272

Correct Option: C

Taking Moment about A.

T₃ =	1800 × 1.25	= 1125 N
	2

T₄ =	1800 × 0.75	= 675 N
	2

We get, V_B × 4.5 = (1125 × 1.5 + 675 × 3.5)
⇒ V_B = 900 kN ...(i)
⇒ V_A = (1125 + 675 – 900) = 900 kN
M_{1, beam} = (V_A × 1.5) = 900 × 1.5 = 1350 Nm
M_{2, beam} = (V_A × 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]
⇒ H_y₁ = M_{1, beam} = 1350 ...(ii)
Taking bending moment about point (ii)
⇒ H_y₂ = 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)]