Answer:
<em>a) 42 mm</em>
<em>b) 144.4 MPa</em>
<em></em>
Explanation:
Load P = 200 kN = 200 x 10^3 N
Torque T = 1.5 kN-m = 1.5 x 10^3 N-m
maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa
diameter of shaft d = ?
From T = τ * 
* 
substituting values, we have
1.5 x 10^3 = 100 x 10^6 x 
x
= 7.638 x 10^-5
d = ![\sqrt[3]{7.638 * 10^-5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B7.638%20%2A%2010%5E-5%7D)
= 0.042 m = <em>42 mm</em>
b) Normal stress = P/A
where A is the area
A = 
= 
= 1.385 x 10^-3
Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = <em>144.4 MPa</em>
they're essentially the same thing so i'd say yes
Answer:
a)R= sqrt( wt³/12wt)
b)R=sqrt(tw³/12wt)
c)R= sqrt ( wt³/12xcos45xwt)
Explanation:
Thickness = t
Width = w
Length od diagonal =sqrt (t² +w²)
Area of raectangle = A= tW
Radius of gyration= r= sqrt( I/A)
a)
Moment of inertia in the direction of thickness I = w t³/12
R= sqrt( wt³/12wt)
b)
Moment of inertia in the direction of width I = t w³/12
R=sqrt(tw³/12wt)
c)
Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)
R= sqrt ( wt³/12xcos45xwt)
Answer:
The text file attached has the detailed solution of all the parts individually.
Answer:
SIR IT IS D HOPE THIS HELPS (☞ﾟヮﾟ)☞☜(ﾟヮﾟ☜)
Explanation:
