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EAT203Lecture 1: Stress-strain relationship
EAT203, Engineering Mechanics
Lecture Topic Tutorial
1 Stress-strain relationship2 Shear force (SF) and bending moment (BM) for non-uniform loading 13 SF and BM for statically indeterminate problem4 Shear stress in bending 25 Stress transformation6 Experimental stress / strain 37 Yield criteria (theories of failure)8 Stress concentration + Fatigue 49 Vibration (Forced with damping) 510 Kinematics - particle (velocity diagram)11 Kinematics - ridig body (velocity diagram)12 Kinematics - acceleration diagram 613 Gear system14 Balancing of rotating system 7
[1] Benham, P.P., Crawford, R.J. & Armstrong, C.G., Mechnics of Engineering Materials, 2nd ed., Prentice Hall, UK (1996)[2] Hibbeler, R. C., Mechanics of Materials, 7th ed., Prentice Hall, Singapore (2008)[3] Hibbeler, R. C., Engineering Mechanics Dynamics, 8th ed., Prentice-Hall International
Stress
• The effect of external applied forces on a solid body or the member of a framework can be measured in terms of the internal reacting forces.
• The intensity of internal force at a point is called stress.
• Stress is defined as the internal force per unit cross section area at right angle to the direction of the force.
Average Normal Stress Distribution
When a bar is subjected to a constant deformation
APAP
dAdFA
σ = average normal stressP = resultant normal forceA = cross sectional area of bar
Shear Stress, • If an applied load consists of two equal and opposite
forces which are not in the same line, then the material being loaded will tend to shear as shown.
• Shear stress is the force (acting tangent to the area) per unit area
AF
StrainDirect strain
• Consider a uniform bar subjected to an axial tensile load, F. If the resulting extension of the bar is and its unloaded (original) length is L, then the direct tensile strain is
L
Shear strain• A shear stress produces a shear strain as is
shown below. The shear strain is defined as• For small deflection,
• The shear strain is dimensionless and is measured in radians
yx tan
Example 1The plate is deformed into the dashed shape. If, in this deformed shape, horizontal lines on the plate remain horizontal and do not change their length, determine (a) the average normal strain along the side AB, and (b) the average shear strain in the plate relative to the x and y axes.
Line AB, coincident with the y axis, becomes line AB’ after deformation, thus the length of this line is
The average normal strain for AB is
The negative sign indicates the strain causes a contraction of AB.
Solution: Part (a)
mm 018.24832250' 22 AB
(Ans) 1093.7250
250018.248'
3
ABABAB
avgAB
As noted, the once 90°angle BAC between the sides of the plate, referenced from the x, y axes, changes to θ’ due to the displacement of B to B’.
Since then is the angle shown in the figure.
Solution: Part (b)
'2 xy xy
(Ans) rad 0121.02250
3tan 1
xy
G
Hooke’s Law defines the linear relationship between the normal stress and strain within the elastic region, i.e., Young’s modulus.
Eσ = normal stressE = Young’s modulusε = normal strain
While Young’s modulus describe the material’s response to direct strain, shear modulus, G describe the material’s response to shear strain
vGEand 12
• states that in the elastic range, the lateral strain is proportional to the direct strain (due to stress), where
• Negative sign, as direct strain (negative) causes lateral expansion (positive strain), and vice versa.
• Typical values for Poisson’s ratio are 1/3 or 1/4.
direct
lateralv
Poisson’s Ratio, v
Example 2
A steel bar (E = 210GPa, v = 0.32) has the dimensions shown. If an axial force of is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. The material behaves elastically.
Solution:
The normal stress in the bar is
69
6
107610210100.16
st
zz E
Pa 100.16
05.01.01080 6
3
AP
z
The axial elongation of the bar is therefore
(Ans) m1145.11076 6z
zz L
The contraction strains in x and y directions are
-66 103.24107632.0 zstyx v
The changes in the dimensions of the cross section are (Ans) m215.105.0103.24
(Ans) m43.21.0103.246
6
yyy
xxx
L
L
zyx
zyxx
Ev
E
Ev
Ev
E
zxy
y Ev
E
3D stress-strain
xyz
z Ev
E
yzyxy vE
vvvE
1211
xzyxx vE
vvvE
1211
zzyxz vE
vvvE
1211
Plane stress
Ev
Eyx
x
Ev
Exy
y
yxz Ev
Coefficient of Linear Thermal Expansion
• Most materials increase in volume as their temperature increases
• Change in length of a structural member of length L is proportional to temperature increase T:
= .T.L
• Constant of proportionality (K-1) is coefficient of linear thermal expansion {C.T.E.}
• Increasing temperature causes expansion and thus a positive strain and vice versa.
• Then thermal strain, = /L = .T
L
A B
Unconstrained Bar
Axial Stress in Constrained Bar• If there is any restriction on the material, then a
thermal stress will result. For an initially unstressed prismatic bar AB, an axial stress (F) developed as it undergoes an increase in temperature.
L
L
F
A
A
B
B
• In this case, principle of superposition is to be applied
• Within elastic limit, assess effects of temperature change and load separately
• Add results, noting sign, to determine total effect, i.e. total strain = sum of strains due to both external loads and thermal strains
Total strain, = strain due to external load, F + thermal strain
0 = -(F/EA) + .T
F = .T.E.A
Example 3