Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Maximum Shear Stress
The principal stresses are σ p1 = σ θ = 172.44 MPa and σ p2 = σ r = –45 MPa. The maximum
in-plane shear stress is thus
τ
max
σ
=
− σ
p1 p2
2
172.44 MPa − ( −45 MPa)
=
2
= 108.72 MPa Ans.
2
a pi
2 2
δr
=
[(1 − ν) r + (1 + ν) b ]
2 2
( b − a ) rE
Since the tube is open ended, the longitudinal stress is zero. Therefore, σ p3 = σ long = 0,
and it follows that τ abs max = τ max .
increase in inside and outside Diameters
The radial deformation for the case of internal pressure only is given by Equation (14.29).
The deformation of the inside radius is calculated with r = a = 104.5 mm:
2
(104.5 mm) (45 MPa)
2 2
=
[(1 − 0.3)(104.5 mm) + (1 + 0.3)(136.5 mm) ]
2 2
[(136.5 mm) − (104.5 mm) ](104.5 mm)(200,000 MPa)
= 0.0750 mm
The inside diameter increases by 2δ r = 0.150 mm.
Use r = b =136.5 mm to compute the deformation of the outside radius:
Ans.
2
a pi
2 2
δr
=
[(1 − ν) r + (1 + ν) b ]
2 2
( b − a ) rE
2
(104.5 mm) (45 MPa)
2 2
=
[(1 − 0.3)(136.5 mm) + (1 + 0.3)(136.5 mm) ]
2 2
[(136.5 mm) − (104.5 mm) ](136.5 mm)(200,000 MPa)
= 0.0870 mm
The outside diameter increases by 2δ r = 0.174 mm.
Ans.
Minimum Required Wall thickness
The largest normal stress is the circumferential stress at the inner surface of the tube. An
allowable normal stress of 140 MPa has been specified for the tube material, and the outside
diameter of the tube is to remain as 273 mm (i.e., b = 136.5 mm). Begin with Equation
(14.24). Substitute r = a and simplify to obtain
σ =
θ
2
a pi
b − a
2 2
⎛
1 b
+
⎝
⎜
r
2 2
⎞ b a
b a p
2 2 i
⎠
⎟ = +
−
2
2
Therefore,
2 2 2 2
p ( b + a ) = σ ( b − a )
i
θ
First, solve for a 2 :
a
2 2
σ θ
= b
⎛
⎝
⎜
σ
θ
−
+
pi
⎞
p ⎠
⎟
i
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