Determine the vertical force P to be applied at the joint B of a link mechanism as shown to cause motion to impend in 20kg block at C. The coefficient of friction between the block and the horizontal plane is 0.3, vertical force P is______ N
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Solution
A large stream has a reoxygenation constant of 0.3 per day.At a velocity of 0.85 m/sec and at a point at which an organic pollutant is discharged, it is saturated with oxygen at 10 m/g lit(D0= 0). Below the out fall, the ultimate demand for oxygen is found to be 20 mg/lit and the deoxygenation constant is 0.1 per day. Then the D.O at 48.3 downstream will be _____ mg/lit
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Solution
The cycle time data of 6m3 typical transit mixer is given
below loading time to transit mixer = 5 min
Travel time of loaded transit mixer to site = 30 min
Waiting time (average) = 30 min
Discharge time = 14 min
Travel time for return trip = 23 min.
If central batching plant having average output of 60 m3/hr is to run continous then number of transit mixers requirred will be _____
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Solution
The mean velocity of flow for the two dimensional flow velocity profile as shown will be ___ times Vmax.
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Solution
For a submerged embankment 25m high whose up stream face has an inclination of 45°. The soil has following properties; C = 40 kN/m2, φ= 20°, gs γsat= 18 kN/m3. There is a sudden draw down. Given for φ = 10° and β = 45°, Sn= 0.108 and for φ = 4.5°, b = 45°, Sn= 0.136. Factor of safety with respect to cohesion is_______.
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Solution
A 6m thick clay layer lies between two layers of sand each 1m thick, the top layer being at ground level. The water table being 1m below the ground level but the lower layer of sand is under artesian pressure, the piezometric surface being 5m above ground level. Saturated unit wt. of clay being 20kN/m3 and that of sand being 19 kN/m3. Above the water table, unit wt. of sand is 16.5 kN/m3. Effective stress at the bottom of clay layer will be ________(γw= 9.81 kN/m3)
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Solution
In case of low shear force, the design bending strength determined by \(Z_{e}\frac{f_{y}}{\gamma _{mo}}\) is limited below 1.2 Ze \(Z_{e}\frac{f_{y}}{\gamma _{mo}}\) for simple beams to ensure.
1.Lateral- torsional buckling is avoided
2.That onset of plasticity under unfactored loads is prevented
3.That yield does not occur at working loads of the above
According to Indian Railway Board, in respect of steel girders of single track span for metre / broad guage, the impact factor for span of 6m is
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Solution
Impact factor= 0.15 +\(\frac{8}{6+L}\)
For L= 6m
I.F =0.82
Hence, option (b) is correct.
A column is 4m high,is fixed at base and free at top end. It is subjected to following loading.
D.L = 40 kN
L.L = 100 kN
W.L = 4 KN/m
The design load for limit state of collapse is _____ kN
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Solution
As per IS code
Design load will be maximum of the following
(i)1.5 (D.L + L.L)
(ii)1.5 (D.L +W.L)
(iii)1.2 (D.L + L.L + W.L)
So, by 1.5 (D.L + L.L) = 1.5 × 140 = 210 kN
1.5 (D.L +W.L) = 1.5 (40 + 16) = 84 kN
1.2 (D.L + L.L + W.L) = 1.2 (40 + 100 + 16) = 187.2 kN
Hence, design load is 210 kN
A rectangular column section of 400 mm × 600 mm reinforced with 6× 28 mm φ bars. The ultimate concentric load capacity of the column will be _____ kN
(use M 25, fe 415 grade steel)
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Solution
Pu = 0.446 fck= AC+ 0.75 by Ast
= 0.446 fckAg + (0.75 fy– 0.446 fck) Ast
= (0.446 × 25 × 400) + [(0.75 × 415 – 0.446 × 25) 3694]
= 3784.57 kN