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User: manasahegde234

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Posts by manasahegde234

<prev • 177 results • page 1 of 18 • next >
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Answer: A: Problem on bars of Composite section.
... Area of steel rod=$\frac{\pi}{4}*150^2=17671.46 mm^2$ Area of brass tube=$\frac{\pi}{4}$ $=\frac{\pi}{4}*(170-150)^2$ $=314.15mm^2$ Now, let the stresses in steel and brass be $P_s$ and $P_b$ respectively. Strain in steel=strain in brass $\frac{P_s}{E_{steel}}=\frac{P_b}{E_{brass}}$ $P_s=\fra ...
written 1 day ago by manasahegde23420
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Answer: A: What is point of contra-flexure. Give the relation between shear force and bendi
... $\sum F_y=0$ $s-W*dx-(s+ds)=0$ $s-Wdx-s-ds=0$ $-Wdx=ds$ $w=\frac{-ds}{dx}$ The rate of change of the shearing force wrt c is equal to the load. OR The slope of the shear diagram at a given point equals to the load at that point. Now, $\sum M_{2-2}=0$ $M+\delta*dx-(w*dx)*\frac{dx}{2}-(M+dM ...
written 1 day ago by manasahegde23420
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Answer: A: Problem on Composite Section
... Case 1: When the column carries a load of 270KN Area of steel $As=\frac{\pi}{4}*16^2*8=1608.48mm^2$ Area of concrete=$A_c=250*250-1608.48=60891.52 mm^2$ Let the stress in steel and concrete be $P_s$ and $P_c$ Strain in steel=strain in concrete $\frac{P_s}{E_s}=\frac{P_c}{E_c}$ $P_s=\frac{E_s}{ ...
written 1 day ago by manasahegde23420
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Answer: A: Problem on Temperature stresses
... Area of steel rod $A_s=\frac{\pi}{4}*15^2=56.25 \pi mm^2$ Area of coppper rod $A-c=\frac{\pi}{4}*(50-40)^2=225\pi mm^2$ Free expansion: $\delta_{steel}=\alpha_s TL $ $\delta_{copper}=\alpha_c TL $ Let the actual expansion of each component be \delta $\alpha_c TL >\delta> \alpha_s TL $ ...
written 1 day ago by manasahegde23420
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Answer: A: Draw the shear force diagram and bending moment diagram for the beam loaded as s
... $\sum F_y=0$ $V_B+V_E=20*4+50+25*2$ $V_B+V_E+180$ and taking moment about E, $\sum M_E=0$ $-20*4*(\frac{4}{2}+5+5+2)+V_b*(5+5+2)-50*(5+2)-(25*2)*\frac{2}{2}=0$ $V_B=126.67KN$ Now, $V_G=180-V_B=53.33KN$ B.M analysis: BM in the section AB, distance * from A $BM_x=-20*\frac{x^2}{2}$ At x=0, ...
written 1 day ago by manasahegde23420
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Answer: A: Problems on Temperature Stresses
... Given data: $E_{steel}=2*10^5 N/mm^2$ (not given to be remembered) $T_1=16 ^{\circ} C T_2=66 ^{\circ}$ $L=15m=15*1000mm$ $\alpha_{steel}=12*10^{-6}$ co-efficient of linear expansion Now, change in temperature= $T=T_2-T_1=66-16=50^{\circ}$ Free expansion of rod is$ \delta=\alpha TL$ $=12*10^ ...
written 1 day ago by manasahegde23420
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Answer: A: Problem on Poisson's ratio, Young's modulus, Bulk modulus, and modulus of rigidi
... Given data: Tensile load(F)=55KN Diameter(d)=31mm Gauge Length (L)=300mm Extension $(\Delta L)=0.115mm$ $\Delta d $Change in diameter=0.00567 mm i) Poisson's ratio $\mu =\frac{lateral strain}{longitudinal strain}$ $\mu= \frac{\frac{0.00367}{31}}{\frac{0.115}{300}}$ $\mu=0.308$ 2) Young's mo ...
written 1 day ago by manasahegde23420
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Answer: A: For circular rod, find elongation, change in the diameter and change in volume o
... Given data: diameter d=30mm Length L=3mm=3000mm Force F=100KN Modulus of elasticity $(E)=70GN/m^2=70*10^3 N/mm^2$ Poisson's ratio $\mu=\frac{1}{3}$ Now, $\sigma stress=\frac{force or axial pull}{area of cross section}=$ $\frac{100*10^3}{\frac{\pi}{30^2}} N/mm^2$ Stress=$141.47 N/mm^2$ Now, m ...
written 1 day ago by manasahegde23420
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Answer: A: Determine the maximum weight 'w' that can be supported by two wires as shown in
... Stresses in wire 1 and 2 is $120 N/mm^2$. And area as $A1=200 mm^2$ and $A_2=250mm^2.$ Now considering the stresses as c and $\sigma_2$ in wire 1 and 2. Let forces be $F_1$ and $F_2$ in wire 1 and wire 2 respectively. Now, $stress=\frac{force}{area}$. Therefore, $\sigma_1=\frac{F_1}{A_1}$ $120=\fr ...
written 1 day ago by manasahegde23420
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Answer: A: A bar of steel is (60 mm x 60 mm) in section and 18mm along. It is subjected to
... ![enter image description here][1] The stresses in the direction of X, Y and Z axes, Along X-axes, $P_x=\frac{F_x}{A}=\frac{320*10^3}{60*60}(N/mm*mm)=88.89 N/mm^2$ Along Y axis, $P_y=\frac{F_y}{A}=\frac{760*10^3}{180*60}(N/mm*mm)=70.37 N/mm^2$ Along Z axis, $P_z=\frac{F_z}{A}=\frac{600*10^3}{180 ...
written 5 days ago by manasahegde23420

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