For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as
A=0.0048rads
<h3>What is the slope of end a of the cantilevered beam?</h3>Generally, the equation for the is mathematically given as
Therefore
A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}
A=0.00288+0.00192=0.0048rads
A=0.0048rads
In conclusion, the slope is
A=0.0048rads
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Answer:
principal stresses :б1 = 32.62mPa б2 = 31.38mPa
Max Shear stress : 16.31 mPa
Orientation of max principle plane = 44.43°
Orientation of minimum principal plane = 134.43°
Explanation:
Given data:
Torque = 50 N-m
weight = 80 kgs
half of weight is subjected to each leg
radius of bone = 10 mm = 0.010 m
<u>a) Determine the principal stresses and shear stress</u>
first calculate the max shear stress ( this will occur in the outermost element
= 16T / π*d^3 where : T = 50 , d = 0.020 m
hence max shear stress = 32 mPa
next determine compressive stress
= ( 40*g) / π/4*d^2 . where : d = 0.020 m , g = 9.81
hence compressive stress = 1.24 mPa
draw and calculate the radius of Mohr's circle
radius of Mohr's circle = 32.0060
Hence principal stresses = 32.0060 ± 0.62
б1 = 32.62mPa
б2 = 31.38mPa
attached below is the remaining part of the solution
