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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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... 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 $\frac{P_s}{E_{steel}}=\frac{P_b}{E_{brass}}$ $P_s=\fra ... written 1 day ago by manasahegde23420 2 answers 53 views 2 answers ...$\sum F_y=0s-W*dx-(s+ds)=0s-Wdx-s-ds=0-Wdx=dsw=\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}=0M+\delta*dx-(w*dx)*\frac{dx}{2}-(M+dM ...
written 1 day ago by manasahegde23420
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... 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 $\frac{P_s}{E_s}=\frac{P_c}{E_c}$ $P_s=\frac{E_s}{ ... written 1 day ago by manasahegde23420 2 answers 44 views 2 answers ... Area of steel rod$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 $... written 1 day ago by manasahegde23420 2 answers 33 views 2 answers ...$\sum F_y=0V_B+V_E=20*4+50+25*2V_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}=0V_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}$At x=0, ... written 1 day ago by manasahegde23420 2 answers 40 views 2 answers ... Given data:$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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... 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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... 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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... 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, $\sigma_1=\frac{F_1}{A_1}$ $120=\fr ... written 1 day ago by manasahegde23420 1 answer 68 views 1 answers ... ![enter image description here][1] The stresses in the direction of X, Y and Z axes, Along X-axes,$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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