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Alexeev081 [22]
2 years ago
8

The rigid bar CDE is attached to a pin support at E and rests on the 30 mm diameter brass cylinder BD. A 22 mm diameter steel ro

d AC passes through a hole in the bar and is secured by a nut which is snugly fitted when the temperature of the entire assembly is 20oC. The temperature of the brass cylinder is then raised to 50oC while the steel rod remains at 20oC.
Assuming that no stresses were present before the temperature change,
determine the stress in the cylinder.
Engineering
1 answer:
Leno4ka [110]2 years ago
3 0

Answer:

stress = 38.84 MPa

Explanation:

S_{D} = \alpha _{brass} * (delta T) *(L_{BD} )\\= (18.8 * 10^(-6) )*(30)*(0.3)\\= 0.0001692 m\\\\E_{BD} = stress / strain\\stress = E_{BD} * S_{D}\\stress = (200 *10^9) * (0.0001692)\\stress = 33.84 MPa

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Link BD consists of a single bar 36 mm wide and 18 mm thick. Knowing that each pin has a 12-mm diameter, determine the maximum v
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Answer:

hello the diagram attached to your question is missing attached below is the missing diagram

answer :

a) 48.11 MPa

b) - 55.55 MPa

Explanation:

First we consider the equilibrium moments about point A

∑ Ma = 0

( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0

therefore ;<em> Fbd = 36 ( cos ∅tan30° - sin∅ ) kN  ----- ( 1 )</em>

A ) when ∅ = 0

Fbd = 20.7846 kN

link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation

A = ( b - d ) t

b = 12 mm

d = 36 mm

t = 18

therefore loading area ( A ) = 432 mm^2

determine the maximum value of average normal stress in link BD  using the relation below

бbd = \frac{Fbd}{A}  = 20.7846 kN / 432 mm^2  =  48.11 MPa

b) when ∅ = 90°

Fbd = -36 kN

the negativity indicate that the loading direction is in contrast to the assumed direction of loading

There is compression in link BD

next we have to calculate the loading area using this equation ;

A = b * t

b = 36mm

t = 18mm

hence loading area = 36 * 18 = 648 mm^2

determine the maximum value of average normal stress in link BD  using the relation below

бbd = \frac{Fbd}{A} = -36 kN / 648mm^2 = -55.55 MPa

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Chapter 19: Diesel Engine Operation and Diagnosis -Chapter Quiz
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Explanation:

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The 10-kg block slides down 2 m on the rough surface with kinetic friction coefficient μk = 0.2. What is the work done by the fr
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Answer:

153.2 J

Explanation:

Let's first list our given parameters;

mass (m) of the block = 10 kg

which slides down ( i.e displacement) = 2 m

kinetic coefficient of friction (μk) = 0.2

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W = 153.2 J

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