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

SolutioN
(a) Determine the change in length of sides AB, BC, and BD.
Normal stresses: The three normal stresses on the block (which are actually principal
stresses) are equal:
σx = σ y = σ z =− p =−35 MPa
Normal strains: From Equations (13.16), we find that e x for a hydrostatic stress state
can be expressed as
1 1
ex = [ σx − νσ ( y + σz
)] = [ −p −ν
( −p − p )]
E
E
p
=− (1 − 2 ν)
E
Furthermore, we obtain this same expression for e y and e z . Consequently, we find that
the normal strains are equal in each of the three orthogonal directions for a hydrostatic
stress state. Thus, for a pressure p = 35 MPa, the strains in the aluminum alloy block are
35 MPa
−6
ex = ey = e z =− [1 − 2(0.33)] = − 163.0 × 10 mm/mm
73,000 MPa
Deformations: In the x, y, and z directions, the deformations are, respectively, as follows:
δ
δ
δ
AB
BC
BD
−6
= (160 mm)( − 163.0 × 10 mm/mm) = −0.0261 mm
−6
= (120 mm)( − 163.0 × 10 mm/mm) = −0.01956 mm
−6
= (200 mm)( − 163.0 × 10 mm/mm) = −0.0326 mm
Ans.
(b) Determine the Change in Volume of the Block: From the normal strains just calculated,
we can calculate the dilatation e (i.e., the volumetric strain) from Equation (13.21):
−
e = e + e + e = 3( − 163.0 × 10 ) =− 489.0 × 10
x y z
6 −6
The initial volume of the block was
6 3
V = (160 mm)(120 mm)(200 mm) = 3.84 × 10 mm
The change in volume is calculated from the product of the dilatation and the initial
volume:
− 6 6 3 3
∆ V = eV = ( − 489.0 × 10 )(3.84 × 10 mm ) =−1,878 mm Ans.
Notice that the volume of the block has decreased under the action of hydrostatic pressure.
ExAmpLE 13.6
A marble [E = 55 GPa; ν = 0.22] cube has dimensions a = 75 mm. Compressive strains
e x = −650 × 10 −6 and e y = e z = −370 × 10 −6 have been measured for the cube. Determine
a
y
a
(a) The normal stresses σ x , σ y , and σ z acting on the x, y, and z faces of the cube.
(b) The maximum shear stress in the material.
Plan the Solution
Equations (13.19) will be used to calculate the stresses σ x , σ y , and σ z from the given
strains e x , e y , and e z . Since there are no shear strains, we know that there are no shear
stresses, and consequently, the stresses σ x , σ y , and σ z must be principal stresses. From the
three principal stresses, the maximum shear stress in the cube can be calculated.
z
a
x
565