## User: manasahegde234

manasahegde234 •

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#### Posts by manasahegde234

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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
manasahegde234 •

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... $\sum F_y=0$
$s-W*dx-(s+ds)=0$
$s-Wdx-s-ds=0$
$-Wdx=ds$
$w=\frac{-ds}{dx}$
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
manasahegde234 •

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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
manasahegde234 •

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... 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
manasahegde234 •

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... $\sum F_y=0$
$V_B+V_E=20*4+50+25*2$
$V_B+V_E+180$
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
manasahegde234 •

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... 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
manasahegde234 •

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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
manasahegde234 •

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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
manasahegde234 •

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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
manasahegde234 •

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... ![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
manasahegde234 •

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