SHEARING DEFORMATION:
S
- Shearing forces cause deformation. An element subjected to shear does
YN
O
N
R
AM
O
not change in length but undergoes a change in shape
E
- The change in angle at the corner of an original rectangular element is
𝛿
𝐿
G
𝛾=
R
R
called the shear strain and is expressed as;
N
- The ratio of the shear stress 𝜏 and the shear strain 𝛾 is called the
E
modulus of elasticity for shear or modulus of rigidity and is denoted
as, 𝐺;
S
𝜏
𝛾
O
𝐺=
R
AM
- The relationship between the shearing deformation and the applied
shearing force is;
=
𝜏𝐿
𝐺
N
𝛿𝑠 =
𝑉𝐿
𝐴𝑠 𝐺
YN
O
where:
𝑉 = 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒
R
E
𝐴𝑠 = 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑎𝑟𝑒𝑎
R
Poisson's Ratio:
G
When a bar is subjected to tensile loading, there is an increase in
N
length of the bar in the direction of the applied load, but there is
E
also, a change in lateral dimension perpendicular to the load. The ratio
Of the sideways deformation (or strain) to the longitudinal deformation
(or strain) is called Poisson’s Ratio and is denoted by, 𝑣 . For most
E
N
G
R
R
E
YN
O
N
R
AM
O
S
steel, it lies in the range of 0.25 to 0.30, and 0.20 for concrete.
𝜀𝑥
=
𝜀𝑧
−
𝜀𝑥
S
𝑣=−
𝜀𝑦
O
where:
R
AM
𝜀𝑥 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑥 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝜀𝑧 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑧 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
N
𝜀𝑦 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑦 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
YN
O
* The negative sign indicates a decrease in transverse dimension, and
𝜀𝑥 is positive
R
E
BIAXIAL DEFORMATION:
R
If an element is subjected simultaneously by tensile stresses,
𝜎𝑥 and
N
G
𝜎𝑦 , in the x and y directions, the strain in the x-direction is
E
the strain y-direction is
𝜎𝑦
𝐸
𝜎𝑥
and
𝐸
. Simultaneously, the stress in the
x-direction will produce a lateral contraction on the y-direction
of the amount
−𝑣𝜀𝑦 or −
𝑣𝜎𝑦
𝐸
.
The resulting strain in the x-direction
𝐸
−
𝐸
𝜎
𝑣 𝑥
𝐸
, 𝜎𝑦 =
O
, 𝜎𝑥 =
𝜀𝑥 +𝑣𝜀𝑦 𝐸
1−𝑣 2
𝜀𝑦 +𝑣𝜀𝑥 𝐸
1−𝑣 2
N
𝜀𝑦 =
𝑣
𝜎𝑦
R
AM
𝜀𝑥 =
𝜎𝑥
−
𝐸
𝜎𝑦
S
will be;
YN
O
TRIAXIAL DEFORMATION:
If an element is subjected simultaneously by three mutually
𝜀𝑧 =
R
𝜎𝑦 − 𝑣 𝜎𝑥 + 𝜎𝑧
R
𝜎𝑥 − 𝑣 𝜎𝑦 + 𝜎𝑧
N
𝜀𝑦 =
1
𝐸
1
𝐸
1
𝐸
E
𝜀𝑥 =
𝜀𝑦 , 𝜀𝑧 respectively,
G
strains 𝜀𝑥 ,
E
perpendicular normal stresses 𝜎𝑥 ,
𝜎𝑧 − 𝑣 𝜎𝑥 + 𝜎𝑦
𝜎𝑦 , 𝜎𝑧 , which are accompanied by
Note:
S
* Tensile stress and elongation are taken as positive. Compressive
R
AM
O
stress and contraction are taken as negative.
Relationships between E, G and 𝒗;
N
𝐺=
𝐸
2 1+𝑣
YN
O
Bulk Modulus of Elasticity or Modulus of Volume Expansion, 𝒌;
- Is a measure of a resistance of a material to change in volume without
R
=
𝜎
∆𝑉ൗ
𝑉
R
𝑘=
𝐸
3 1−2𝑣
E
a change in shape or form;
N
𝑉 = 𝑣𝑜𝑙𝑢𝑚𝑒
G
where:
E
∆𝑉 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒
∆𝑉Τ
𝑉
= 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝜎
𝑘
=
3 1−2𝑣
𝐸
O
S
∆𝑉
𝑉
R
AM
Illustrative Problem:
A rectangular steel block is 3′′ long in the x-direction, 2′′ long in the
y-direction, and 4′′ long in the z-direction. The block is subjected to a
N
triaxial loading of three uniformly distributed forces as follows;
YN
O
48 𝑘𝑖𝑝𝑠 tension in x-direction, 60 𝑘𝑖𝑝𝑠 compression in the y-direction, and
54 𝑘𝑖𝑝𝑠 tension in the z-direction. If 𝑣 = 0.30 and 𝐸 = 29𝑥106 psi, determine
E
The single uniformly distributed load in the x-direction that would
R
produce the same deformation in the y-direction as the original loading
R
𝐺𝑖𝑣𝑒𝑛:
𝐹𝑥 = 48 𝑘𝑖𝑝𝑠 𝑇
𝑦 = 2′′
𝐹𝑦 = 60 𝑘𝑖𝑝𝑠 𝐶
N
E
z = 4′′
G
𝑥 = 3′′
𝐹𝑧 = 54 𝑘𝑖𝑝𝑠 𝑇
R
G
N
E
E
R
YN
O
N
R
AM
O
S
R
G
N
E
E
R
YN
O
N
R
AM
O
S
R
G
N
E
E
R
YN
O
N
R
AM
O
S
R
G
N
E
E
R
YN
O
N
R
AM
O
S
Illustrative Problem:
S
A 150 mm long bronze tube closed at its ends is 80 mm in diameter and
R
AM
O
has a wall thickness of 3 mm. It fits without clearance in an 80 mm hole
in a rigid block. The tube is then subjected to an internal pressure of
4.0 𝑀𝑃𝑎. Assuming 𝑣 =
1
3
and 𝐸 = 83 𝐺𝑃𝑎, determine the tangential stress in
N
the tube.
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑:
YN
O
𝐺𝑖𝑣𝑒𝑛:
𝜌 = 4.0 𝑀𝑃𝑎
𝜎𝑥
E
𝐷 = 80 𝑚𝑚
𝐿 = 150 𝑚𝑚
R
G
𝐸 = 83 𝐺𝑃𝑎
N
1
3
E
𝑣=
R
𝑡 = 3 𝑚𝑚
R
G
N
E
E
R
YN
O
N
R
AM
O
S
R
G
N
E
E
R
YN
O
N
R
AM
O
S
Illustrative Problem:
S
The 50 mm rubber rod is placed in a hole with a rigid lubricated walls.
R
AM
O
There is no clearance between the rod and the sides of the hole.
Determine the change in the length of the rod when the 8 𝑘𝑁 load is
N
applied. Use 𝐸 = 40 𝑀𝑃𝑎 and 𝑣 = 0.45 for rubber.
YN
O
𝐺𝑖𝑣𝑒𝑛:
𝑃 = 8 𝑘𝑁
E
𝐷 = 50 𝑚𝑚
R
𝐿 = 300 𝑚𝑚
G
N
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑: ∆𝐿
E
𝑣 = 0.45
R
𝐸 = 40 𝑀𝑃𝑎
Illustrative Problem:
O
5
8
− 𝑖𝑛 diameter rod and it
R
AM
experimental plastic. The test specimen is a
S
A standard tension test is used to determine the properties of an
is subjected to 800 𝑙𝑏 tensile force. If an elongation of 𝛿𝐿 =0.45 𝑖𝑛 and a
decrease in diameter of 𝛿𝐷 = 0.025 𝑖𝑛 are observed in a 5-in gage length,
N
determine;
YN
O
a) the modulus of elasticity of the material
b) the Poisson’s ratio of the material
E
N
G
R
R
E
c) the modulus of rigidity of the material
R
G
N
E
E
R
YN
O
N
R
AM
O
S
Problem Set No. 01:
O
S
A 2-m length of an aluminum pipe 240-mm outer diameter and 10-mm wall
If 𝐸 = 73 𝐺𝑃𝑎 and 𝑣 = 0.33, determine;
b) the change in its outer diameter
E
N
G
R
R
E
YN
O
c) the change in its wall thickness
N
a) the change in the length of the pipe
R
AM
thickness is used as a short column and carries a centric load of 640 𝑘𝑁.
Problem Set No. 02:
O
S
A 20-mm square has been scribed on the side of a large steel pressure
R
AM
vessel. After pressurization, the biaxial stress condition of the
is as shown. Using 𝐸=200 𝐺𝑃𝑎 and 𝑣=0.30, determine the percent change
E
N
G
R
R
E
YN
O
N
in the slope of diagonal DB due to the pressurization of the vessel.
Problem Set No. 03:
O
S
The rectangular block of material of length 𝐿 and cross-sectional area
R
AM
𝐴 fits snugly between two rigid, lubricated walls. Derive the
the expression for the change in length of the block due to the axial
E
N
G
R
R
E
YN
O
N
load 𝑃.
Problem Set No. 04:
S
Two 1.75-in thick rubber pads are bonded to three steel plates to form
O
the shear mount as shown in the figure. Determine the displacement of
R
AM
the middle plate when the 1200-lb load is applied. Consider the
E
N
G
R
R
E
YN
O
N
the deformation of the rubber only. Use 𝐸 = 500 𝑝𝑠𝑖 and 𝑣 = 0.48 for rubber.