1. Chapter 2 Mechanics of Materials
Tensile stress (+)
Compressive stress (-)
Normal stress =
At a point:
Example: Estimate the normal stress on a shin bone (脛骨)

2. Shear stress (切應力) =  = F tangential to the area / A
At a point,

3. Shear strain (?) = deformation under shear stress =
Normal strain (正應變)  = fractional change of length= l x
Shear strain (?) = deformation under shear stress = x F
fixed

4. Hooke's law: In elastic region,   , or /  = E
Stress-strain curve   o
Yield pt.
Work hardening
break
Elastic
deformation
Hooke's law: In elastic region,   , or /  = E
E is a constant, named as Young’s modulus or modulus of elasticity
Similarly, in elastic region, / = G, where G is a constant, named as shear modulus or modulus of rigidity.

5. Exercise set 2 (Problem 3) Find the total extension of the bar.
15mm W
5mm
1.2m
0.6m o dx
Width of a cross-sectional element at x:
Stress in this element :
Strain of this element:
The extension of this element :
The total extension of the whole bar is :
= 2.13 x 10-4 m

6. Bulk modulus 

7. Poisson's ratio : For a homogeneous isotropic materialx d
normal strain :
lateral strain :
Poisson's ratio :
value of  :

8. Double index notation for stress and strain
1st index: surface, 2nd index: force
For normal stress components : x  xx, y  yy , z  zz,
x  xx y z
zx
zy
yz
xz
xy
yx

9. Joint effect of three normal stress componentsy z

10. Symmetry of shear stress components
Take moment about the z axis, total torque = 0,
(xy yz) x = (yx x z) y, hence, xy = yx .
Similarly, yz = zy and xz = zx z y x
xy
yx x y z

11. Original shear strain is “simple” strain =x dy
rot 2
There is no real deformation during pure rotation,
but “simple” strain  0.
Define pure rotation angle rot and
pure shear strain, such that the angular displacements of the two surfaces are:
1= rot+ def and 2= rot- def . Hence,
rot = (1+ 2)/2 and def = (1- 2)/2
def x
2 = -
Example: 1 = 0 and 2 = - ,
so def = (0+)/2 = /2 and rot= (0-)/2 = -/2
Pure shear strain is /2

12. For hydrostatic pressure
Example: Show that
Proof:
For hydrostatic pressure l
xx = yy = zz = , hence
3 = xx+yy+zz = (1-2v)(xx+yy+zz)/E
xx =yy =zz = -p (compressive stress)

13. Example : Show that nn = /2
2019/7/17
Example : Show that nn = /2
Point C moves further along x- and y-direction by distances of AD(/2) and AD(/2) respectively.
nn = [(AD  /2)2 + (AD  /2)2]1/2 / [(AD)2 + (AD)2]1/2 = /2
True shear strain: yx = /2
Therefore, the normal component of strain is equal to the shear component of strain:
nn = yx and nn = /2 A C’ C D D’

14. Consider equilibrium along n-direction:
Example : Show that nn = nn/(2G)
Consider equilibrium along n-direction:
 yx (lW) sin 45o x2 = 2 (l cos 45o) W nn l n
yx
xy
2 l cos 45o
Therefore yx = nn
From definition :  = xy /G = nn /G = 2 nn

15. Example : Show xx = xx/E -  yy/E- v zz/E
Set xx =  nn = - yy,  zz = 0, xx = nn
nn = (1+)  nn /E =  nn /2G (previous example)
nn
-nn

16. xx = (xx- v yy- v zz) /E 0 = [xx- 0 – 0.3(- 3107)]/60109
12kN x
Ex. 12 kN forces are applied to the top & bottom of a cube (20 mm edges), E = 60 GPa,  = 0.3. Find (i) the force exerted by the walls, (ii) yy
xx = 0, yy = 0 and
zz= -12103 N/(2010-3 m)2 = 3107 Pa
xx = (xx- v yy- v zz) /E
0 = [xx- 0 – 0.3(- 3107)]/60109
 xx = -9106 Pa (compressive)
Force = Axx = (2010-3 m)2(-9106 Pa) = -3.6 103 N
(ii) yy = (yy- v zz- v xx) /E
= [0 – 0.3(- 3107) – 0.3(- 9106)]/60109 = 1.9510-4

17. Elastic Strain Energy The energy stored in a small volume:
 Energy density in the material :
e=extension x

18. Similarly for shear strain :