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lord [1]
2 years ago
8

The W16x50, steel beam below has a span of 26' and is subjected to a 2.3 k/ft uniform distributed loading. If a 8 kip load is al

so applied at the mid point, what is the maximum bending stress in the beam
Engineering
1 answer:
MariettaO [177]2 years ago
5 0

Answer: hello the steel beam related to your question is missing attached below is the missing diagram

answer : max stress = -14.32 ksi

Explanation:

<em>For a W16 * 50 beam </em>

I = 659 in

d = 16.26 in

For steel : E = 29 * 10^3 ksi

Span ( length ) of steel beam = 26'

uniformly distributed load on the steel beam = 2.3 k/ft

load at middle = 8 kip ,  Distance from middle to either Ra or Rb = 13'

<em>First step : calculate the value of  Rb and Ra </em>

∑Ma =  0

      0  = ( 8 * 13 )+ (26 * 2.3 * 13 ) - Rb * 26

∴ 25 Rb = 881.4

Hence: Rb = 35.256 kip

∑fy = 0

         0 = Ra + Rb - 2.3 * 26 - 8

Ra = 67.8 -  Rb = 67.8 - 35.256

     = 32.54 kip

second step :

Mmax = Mmidpoint

∴  Mmax = Ra * 13 - 2.3 ( 13 ) * (26/4 )

              = 32.54 *  3 - 2.3 ( 13 ) * (26/4 )

              = 97.62 - 194.35 = -96.73 kip fit

<u>Finally determine the maximum bending stress </u>

бmax = M * ( d/2 ) / I

         = - 96.73 * ( 16.26 / 2 )*12  /  659

         = -96.73 * 97.56 / 659 = -14.32 ksi

<u />

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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

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Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

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\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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Exceeding critical mach may result in the onset of compressibility effects such as:______.
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Explanation:

Sound barrier is a sudden increase in drag and other effects when an aircraft travels faster than the speed of sound. Other undesirable effects are experienced in the transonic stage, such as relative air movement creating disruptive shock waves and turbulence. One of the adverse effect of this sound barrier in early plane designs was that at this speed, the weight of the engine required to power the aircraft would be too large for the aircraft to carry. Modern planes have designs that now combat most of these undesirable effects of the sound barrier.

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Which basic principle influences how all HVACR systems work?
bezimeni [28]

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B) An increase in pressure can lower the boiling point of a liquid and change the temperature at which it turns to a gas.

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In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o
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The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

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P = 10 atm

  = 10\times 101325 \ Pa

  = 1013250 \ Pa

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

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V = 1000 r

   = 1 \ m^3

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ PV=nRT

o,

⇒ n=\frac{PV}{RT}

By substituting the values, we get

       =\frac{1013250\times 1}{8.3145\times 298}

       =408.94 \ moles

As we know,

⇒ Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}

or,

⇒        MW=\frac{m}{n}

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