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

SolutioN
(a) Normal stresses: From Equations (13.19), calculate the normal stresses acting on
the cube:
σ
σ
E
=
[(1 − νe ) + νe ( + e )]
(1 + ν)(1−
2 ν) 55,000 MPa
− 6
=
[(1 − 0.22)( − 650) + (0.22)( −370 − 370)](10 )
(1 + 0.22)[1−
2(0.22)]
=−53.9 MPa
x x y z
E
=
[(1 − νe ) + νe ( + e )]
(1 + ν)(1−
2 ν) 55,000 MPa
− 6
=
[(1 − 0.22)( − 370) + (0.22)( −650 − 370)](10 )
(1 + 0.22)[1−
2(0.22)]
=−41.3 MPa
y y x z
E
σ z =
[(1 − νe ) z + νe ( x + e y)]
(1 + ν)(1−
2 ν) 55,000 MPa
− 6
=
[(1 − 0.22)( − 370) + (0.22)( −650 − 370)](10 )
(1 + 0.22)[1−
2(0.22)]
=−41.3 MPa
Ans.
Ans.
Ans.
(b) Maximum shear stress in the material: There are no shear stresses acting on the x,
y, or z faces of the marble cube; consequently, σ x , σ y , and σ z must be principal stresses:
σ
σ
σ
p1
p2
p3
= σ = −53.9 MPa
x
= σ = −41.3 MPa
y
= σ = −41.3 MPa
z
From Equation (12.18), we know that the absolute maximum shear stress is equal to
one-half of the difference between the maximum and minimum principal stresses:
σmax
− σmin
τ absmax =
2
Thus, the absolute maximum shear stress in the marble cube is
τ absmax
53.9 MPa ( 41.3 MPa)
= − − − = 6.30 MPa
Ans.
2
Special case of plane Stress
When stresses act only in the x–y plane (Figure 13.10), σ z = 0 and τ yz = τ zx = 0. Consequently,
Equation (13.16) reduces to the following:
mecmovies 13.7 presents
an animated derivation of
the generalized Hooke’s Law
equations for biaxial stress.
566
1
ex = ( σx −νσ
y )
E
1
ey = ( σ y −νσ
x )
E
v
ez =− ( σx + σ y)
E
(13.24)