2 years ago

C++ question Thank you

Engineering

1 answer:

Salsk061 [2.6K]2 years ago

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You're welcome lol! !!!!!!!!!!!!!!

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The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is

romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

For cylinder AB:

Let Length of AB = 12 in

$c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\$

$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\$

$\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998} =1.3266*10^{-6}T_{BC}$

$2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}$

$T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in$

b) Maximum shear stress in BC

$\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi$

Maximum shear stress in AB

$\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi$

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3 years ago

What are some "vital signs" that we consider to tell us about the economy?

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Explanation:

if there is a deficit or a surplus

if the budget is good

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2 years ago

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6.1-2. Diffusion of CO, in a Binary Gas Mixture. The gas CO2 is diffusing at stcady state through a tube 0.20 m long having a di

zzz [600]

Answer:

Heat flux of CO₂ in cgs

= 170.86 x 10⁻⁹ mol / cm²s

SI units

170.86 x 10⁻⁸ kmol/m²s

Explanation:

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Your local hospital is considering the following solution options to address the issues of congestion and equipment failures at

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2 years ago

A specimen of commercially pure copper has a strength of 240 MPa. Estimate its average grain diameter using the Hall-Petch equat

romanna [79]

Answer:

3.115× $10^{-3}$ meter

Explanation:

hall-petch constant for copper is given by

$S_0$ =25 MPa

k=0.12 for copper

now according to hall-petch equation

$S_Y$ = $S_0$ + $\frac{K}{\sqrt{D}}$

240=25+ $\frac{0.12}{\sqrt{D}}$

D=3.115× $10^{-3}$ meter

so the grain diameter using the hall-petch equation=3.115× $10^{-3}$ meter

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2 years ago