Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and permit a 1.58 rotation of one flange with respect to the other before the flanges begin to rotate as a single unit. knowing that g 5 77.2 gpa, determine the maximum shearing stress in each shaft when a torque of t of magnitude 500 n?m is applied to the flange indicated. 3.57 the torque t is applied to flange

b.

The maximum shear stress in shaft AB, is 15 MPa

The maximum shear stress in shaft CD,   is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by  

J =

Therefore, we have

Where:

D = Diameter of shaft AB = 30 mm = 0.03 m

r = Radius of shaft AB = 15 mm = 0.015 m

∴ = 7.95 × 10⁻⁸ m⁴

and

Where:

D = Diameter of shaft CD = 36 mm = 0.036 m

r = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

= 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= rad = 2.76 × 10⁻² rad.

That is = 2.76 × 10⁻² rad.

However    

Where:

and

and = Torque on shaft AB and CD respectively

 = Required

= 500 N·m

and = Length of shafts AB an CD respectively

 = 600 mm = 0.6 m

= 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

 =

2.76 × 10⁻² rad =

=

Therefore;

T =  79.54 N.m

Where T = + T =

Therefore T = 500 - 79.54 = 420.46 N·m

τ =

= =   = 15 MPa

= =  = 45.9 MPa

no

sodium is solid at rt

c) up is the direction the magnetic force is acting on the charge.