A hydraulic jack is shown in the figure, the force exerted on the handle will be ___________ N
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Solution
As per Pascal’s law equal pressure will be exerted in all directions in space as the fliud is in rest.
P1=P2
\(\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}\)
\(\frac{200KN}{\frac{\lambda }{4}(75)^{2}}=\frac{F_{2}}{\frac{\lambda }{4}(15)^{2}}\)
F2= 8 KN
Then 50 × F= 5 × F2
F=5⁄50×8
F =0.8 KN
=800 N
An aeroplane of weight 20 KN has a wing area of 30m2 and it flies at the speed of 50m/sec. If the lift coefficient Varies linearly from 0.35 at 0° to 0.80 at 6°. Then the angle of at tack which the wings should make with the horizontal will be____________ (Take d= 1.2 kg/m3 for air)
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Solution
weight =CLA\(\frac{\rho V^{2}}{2}\)
CL=\(\frac{20\times 1000}{\frac{1}{2}\times 1.2\times 50^{2\times 30}}=0.44\)
By interpolation
Angle of attack =\(\frac{6-0}{0.80-0.35}\times 0.44-0.35=1.2^{\circ}\)
A large tank is shown in figure has a vertical pipe 100cm long and 2c min diameter. The tank contains an oil of density 900 kg/m3 and Viscosity 1.5 poise. The discharge through the tube when the height of oil level of the tank is 2m above the pipe inlet will be _________ lit/min
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Solution
3m=hf+\(\frac{V_{b}^{2}}{2g}\)(By energyequ.)
3=\(\frac{128\mu Q\iota }{\pi \gamma D^{4}}+\left ( \frac{Q}{\frac{\pi }{4}D^{2}} \right )^{2}\frac{1}{2g}\)
3=\(\frac{128\times 0.15Q\times \iota }{\pi (900\times 9.81)(0.02)^{4}}+\frac{1}{2\times 9.81}\times \frac{Q^{2}}{(0.785)^{2}(0.02)^{2}}\)
3 = 4326.332Q + 516,941.86 Q2
Q = 6.438 × 10-4 m3/sec
Q = 38.628 lit/min
Which of the following statements are correct with referance to gradually varied flow.
1.Gradually Varied flow is a steady uniform flow.
2.If the shape of the channel remains constant through out its reach then it is said to be prismatic.
3.As the depth of flow decreases,head loss also decreases.
4.In case of Horizontal and Adverse slope, uniform flow is not possible.
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Solution
1.Gradually Varied Flow- steady but Non Uniform flow.
2.Prismatic channel - shape as well as slope is constant.
3.As depth of flow decreases, head loss increases.
hLαVα1⁄Y
4. As\(\frac{dE}{dx}\)=SO-Sf
for Horizontal channel So = 0 Adverse channel So < 0 so[latex]\frac{dE}{dx}[/latex]< 0 Hence cannot sustain uniform flow only 4 is correct.
An unconfined compression test was done on a saturated clay specimen of diameter 40mm and height 80mm. The failure load was 400N and the axial deformation then was 7mm.Find the unconfined strength of soil ___________ KN/m2.
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Solution
Diameter of sample dO= 40mm
Height of sample = 80mm
Failure load = 400N
Axial deformation= 7mm
Area of the sample initially =π⁄4(40)2
Ao= 1256.64 mm2
Area after axial deformation =\(\frac{A_{O}}{1-\varepsilon _{f}}=\frac{1256.64}{1-\left ( \frac{7}{80} \right )}\)
=1377.14 mm2
unconfined compressive strength =\(\frac{400}{1377.14}\)
=290.45KN/m2
A square pile group of 16 piles through a filled up soil of 3m depth. The pile diameter is 250mm and pile spacing is 0.75m.If the unit cohesion of the material is 18 KN/m2 and unit weight is 15 KN/m3. Compute the negative friction on the group in KN __________.
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Solution
Block failure:- Since it is a square group, 4 rows of 4
piles each will be used
individual piles, cohesion =18 KN/m2
Qng= n P D C
= 16 ×(π × 0.25) × 3 × 18
= 678.58 KNB
lock, B = 3s + d
= (3 × 0.75) + 0.25
= 2.5m
Qng= P D C + γ D A
= ( 4 × 3 × 2.5 × 18) + (15 × 3 × 2.52)
= 821.25 KN
∴ Negative skin friction on the group will be larger of the two i.e. 821.25 KN
An unconfined compression test was performed on saturated soil with initially capillary tension of 20 KN/m2. If major principal stress applied on the soil sample is 80 KN/m2 then pore water pressure at failure will be _________ KN/m2.Take S kempton’s pore pressure coefficient A = 0.6
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Solution
we have Δu=B(σ3+A(σ1+σ3))
saturated so B = 1
unconfined test so σ3= 0
Δu =A σ1
Δu =80 × 0.6
= 48 KN/m2
Δu = uf– (–uc )
= uf+ uc= 48
uf= 48 + 20
= 68 KN/m2
A stepped steel shaft of circular cross-section is built in at C and loaded at the free end A as shown. The ratio of diameters d2: d1 in order that plastic hinge condition will develop simultaneously at B and C will be __________.
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Solution
Bending moment at B = \(\frac{P_{u}l}{4}\)
Bending moment at C = Pul
\(\frac{M_{B}}{M_{C}}=\frac{1}{4}\)
For hinge of develop simultaneously
Mp1=\(\frac{p_{u}l}{4}\) Mp2=pul
\(\frac{d_{1}^{3}}{6}=\frac{p_{u}l}{4}\: \frac{d_{2}^{3}}{6}=p_{u}l\)
\(\frac{d_{1}^{3}}{d_{2}^{3}}=\frac{1}{4}\Rightarrow \frac{d_{2}}{d_{1}}=(4)^{1/3}=1.587\)
A beam of rectangular section 3cm wide and 8cm deep, is simply supported over a span of 1m, the steel has a yields tress 2500 kg/cm2. The beam carries a central point load of 4 ton. The depth of the elastic core at mid-span will be________cm.
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Solution
Moment at the centre =\(\frac{p\iota }{4}=\frac{4000\times 100}{4}\)= 1,00,000 kgcm
Zp of the section =1⁄4bh2=1⁄4×3(8)2=48cm3
Mp = sy Zp = 2500 × 48 = 1,20,000 kgcm
1,00,000=1,20,000\(\left ( 1-\frac{\alpha ^{2}}{3} \right )\)
Depth of core = α d = 5.656 cm
A pre-stressing concrete beam (100 × 300)mm in section is pre stressing with a straight cable at an eccentricity of 50mm.The effective pre stressing is 75KN. The span of beam is 5m and total load on the beam including the self weight is 2.4KN/m. The eccentricity of the pressure line at quaters pan will be.
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Solution
Reaction at each support =\(\frac{2.4\times 5}{2}\)=6KN
Bending moment at quater span
= 6 × 1.25 –\(\frac{2.4\times 1.25^{2}}{2}\)= 5.625 KN/m
e=M⁄P=\(\frac{5.625\times 10^{6}}{75\times 10^{3}}\)=75 mm
e = –50 + 75= +25mm