Answer:
a) 42.422 KN
b) 44.356 KN
Explanation:
Given data :
Diameter = 20 mm
yield strength = 350 MN/m^2
Torque ( T ) = 100 N.m
Bending moment = 150 N.m
<u>Determine the value of the applied axial tensile force when yielding of rod occurs </u>
first we will calculate the shear stress and normal stress
shear stress ( г ) = Tr / J = [( 100 * 10^3) * 10 ] / * ( 20)^4
= 63.662 MPa
Normal stress( Гb + Гa ) = MY/ I + P/A
= [( 150 * 10^3) * 10 ] / * ( 20)^4 + 4P /
= 190.9859 + 4P / MPa
<u>a) Using MSS theory </u>
value of axial force = 42.422 KN
solution attached below
<u>b) Using MDE theory </u>
value of axial force = 44.356 KN
solution attached below
Answer:
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Explanation:
Answer:
Max shear = 8.15 x 10^7 N/m2
Explanation:
In order to find the maximum stress for a solid shaft having radius r, we will be applying the Torsion formula which can be written as;
Allowable Shear Stress = Torque x Radius / pi/2 x radius^4
Putting the values we have;
T = 2000 N/m
Radius = Diameter/2 = 0.05 / 2 = 0.025 m
Putting values in formula;
Max shear = 2000 x 0.025 / 3.14/2 x (0.025)^4
Max shear = 8.15 x 10^7 N/m2
Answer:
net power = 108.89kw
Explanation:
check the attachment for explicit explanation.
Answer:
a) 0.975 m
b) 13.25 m
Explanation:
given data:
pressure gage reading 130 kPa
a) mercury with density 13,600 kg/m^3
let distance between two fluid level is h,
then
we know that
13600*9.8*h=130*10^3
so h = 0.975 m
b.) water with density 1000 kg/m^3
let distance between two fluid level is h,
then
we know that
1000*9.8*h= 130*10^3
so h= 13.25 m