A dry sand specimen failed in a triaxial text when the major and minor principal Stresses were relatively 900 KN/m2 and 250 KN/m2 Shear stress when same specimen is tested indirect shear test under a normal stress 100 KN/m2 will be________ KN/m2.
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Solution
Ans.e
τ = C + σntanΦ
Sand Specimen c = 0
σ = 100 KN/m2
τ = 100 × 0.685
τ = 68.516 KN/m2
Efficiency of Square pile group of 16 piles in which spacing of piles is 3 times the diameter of piles is nearly.
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Solution
According, to converse lebarre equation
\(\eta_{g} = 1 - \left ( \frac{m\left ( n - 1 \right ) + n\left ( m - 1 \right )}{mn} \right )\frac{\theta}{90}\)
where = θ = tan-1(d⁄s) - tan-1(d⁄3d)
= 18.43°
As its square pile group So m = n = 4
\(\eta_{g} = 1 - \frac{18.43}{90}\left ( \frac{4 \times 3 + 3 \times 4}{4 \times 4} \right )\)
= 69.28%
A clay sample has a void ratio of 0.53 in dry State. Its shrinkage limit when G = 2.7 is ___________%.
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Solution
Ans.19.6
Shrinkage limit = 0.53⁄G
= 0.53⁄2.7 × 100 = 19.629%
On a compression member, the plate element may buckle locally before the entire member fails. To avoid this, which of the following recommendations are made?
1. Thickness of members is taken in terms of width of compression element.
2. Length of compression element is increased.
3. Length of member is increased.
4. Outstand is decreased.
Select the correct answers using the codes given below.
The moment rotation curve shown in figure is that of a
A Straight pre-tensioned concrete beam 15m long with cross section of 400 × 400 mm is pre stressed with 900 mm2 of steel wires which are anchored to the bulk heads with a stress of 1050 N/mm2. If modular ratio is 6 then loss of stress due to pre-stressing will be ________ N/mm2.
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Solution
Loss of pre-stress = mp0⁄A
\(= 6 \times \frac{900 \times 1050}{400 \times 400}\)
\(= \frac{6 \times 945000}{400 \times 400}\)
= 35.4 N/mm2
ILD for reaction at E will be
Area of ILD for the shear force at the Hinge C will be________ m2.
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Solution
As, it is hinge at C, So slope of ILD can be different on either side of C
Area = 1⁄2 × 10 × 1
= 5m2
State which of the following Statements is true
A thin cylinder of thickness ‘t’ width ‘b’ and internal radius ‘r’ is subjected to a pressure ‘p’ on the entire internal surface what is the change in the radius of the cylinder? (m is the poisson’s ratio and E is the modular of elasticity)
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Solution
Due to pressure
\(\frac{\Delta d}{d} = E_{n} = \frac{\sigma_{n} - \mu \sigma e}{E}\)
\(\frac{\Delta r}{r} = \frac{pd}{2tE} = \frac{\mu pd}{4tE}\)
\(= \frac{pd}{2tE}\left ( 1 - \frac{\mu}{2} \right )\)
\(= \frac{p\left ( 2r \right )\left ( 2 - \mu \right )}{4tE}\)
\(\Delta r = \frac{pr^{2}\left ( 2 - \mu \right )}{2Et}\)