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Find ratio for ultimate flexural strength at the centre – span section given that Aps = 3000mm2, d = 1150mm, fcu = 50n/mm2, bw = 150mm, fpu = 1500n/mm2, b = 600mm, ht = 250mm, design ultimate moment mud = 3400knm?

Correct Answer: 0.23
Aps = 3000mm2, d = 1150mm, fcu = 50n/mm2, bw = 150mm, fpu = 1500n/mm2, b = 600mm, ht = 250mm, design ultimate moment mud = 3400knm, according to BS: 8110-1985, Aps = (Apw+Apf) = Apf = 0.45×50(600-150)(250/1500) = 0.45xfcu(b-bw)(hf/fpu) = 1680mm2, Apw = (1300-1680) = 1320mm2, ratio(fpuApw/fcubwd) = (1500×1320/50x150x1150) = 0.23.

Check for ultimate flexural strength if given Aps is 154mm2, fpu is 1600n/mm2, b is 160mm, fcu is 50n/mm2and diameter is 265mm?

Calculate ultimate moment and shear of effective span is 30m, live load is 9kn/m, dead load excluding self weight is 2kn/m, load factors for dead load is 1.4 for live load is 1.6 cube strength of concrete fcu is 50n/mm2 cube strength at transfer is fci is 35n/mm2, tensile strength of concrete Ec is 34kn/mm2 loss ratio ɳ is 0.85 and 8mm diameter high tensile strength fpu is 1500n/mm2 are available for use and the modulus of elasticity of high tensile wires is 200kn/mm2?

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Sinking of an intermediate support of a continuous beam
i) reduces the negative moment at support
ii) increases the negative moment at support
iii) reduces the positive moment at center of span
iv) increases the positive moment at center of span
The correct answer is

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In the figure given below A and D are fixed points, link AB rotates at a uniform angular velocity of 900 rev/min ccw, find the instantaneous velocity of point E in m/s. Rab = 100mm, Bc = 450, distance of E from BC = 100mm, Cf = 75mm, CD = 275mm, AD = 250mm, perpendicular DF = 17mm, perpendicular BE = 250mm.

In the figure given below A and D are fixed points, link AB rotates at a uniform angular velocity of 900 rev/min ccw, find instantaneous velocity of point F in m/s. Rab = 100mm, Bc = 450, distance of E from BC = 100mm, Cf= 75mm, CD = 275mm, AD =250mm, perpendicular DF = 17mm, perpendicular BE = 250mm.

In the figure given below A and D are fixed points, link AB rotates at a uniform angular velocity of 900 rev/min ccw, find angular velocity of BC in rad/s. Rab = 100mm, Bc = 450, distance of E from BC = 100mm, Cf= 75mm, CD = 275mm, AD =250mm, perpendicular DF = 17mm, perpendicular BE = 250mm.