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MECHANICS OF MATERIALS REVIEW
Notation: σ - normal stress (psi or Pa) τ- shear stress (psi or Pa) ε - normal strain (in/in or m/m) γ - shearing strain (in/in or m/m) I - area moment of inertia (in^4 or m^4 ) J - polar area moment of inertia (in^4 or m^4 ) N - revolutions per minute E - modulus of elasticity (psi or Pa) G - modulus of rigidity (psi or Pa)
ν - Poisson’s ratio α - coefficient of thermal expansion (/°F or /°C) M - bending moment in beams T - torque in shafts ∆T - temperature change (ΕF or ΕC) hp - horsepower (1 hp = 550 ft-lb/sec) F.S. - factor of safety =
failure load allowed load εt = α∆T - thermal strain
Section 1: Introduction
a. Stress: force per unit area acting on a plane Normal stress (σ): force acts perpendicular to the plane. Shear stress (τ): force acts parallel to the plane
b. Strain: deformation per unit length of dimension Normal strain (ε): stretches or compresses material Shear strain (γ): changes the angle between lines within the material
c. Average Shear Stress in Fasteners
τ =
V
A Single Shear V = Shear force on pin V = P A = Cross sectional area of pin Double Shear
V =
P
d. Bearing Stress in Fasteners
σ =
P
dt P = force d = diameter of fastener t = thickness of part
Section 2: Axial Loading
a. Axial relationships If the line of action of the load, P, passes through the centroid of the resisting cross-section:
If the material is also linear, then:
Where E is the modulus of elasticity for the material.
The relationship between axial loading and deformation becomes:
b. Statically Determinate Members Static Equilibrium ΣF = 0 = -F 1 + F 2 – F 3 + F 4
Internal Forces PAB = F 1 (Tension) PBC = F 1 – F 2 (Tension) PCD = F 1 – F 2 + F 3 (Tension)
Deformation
δAD = δAB + δBC + δCD =
PABLAB
AABEAB +
PBCLBC
ABCEBC +
PCDLCD
ACDECD
Since the P’s were assumed in tension, negative values will indicate compression and contraction for the deformation rather than elongation.
Thermal Deformation
Thermal deformation may be added to any mechanical deformation caused by internal forces acting on the material to obtain a total deformation.
A
P
axial stress = σ=
L
axial strain
E
UniaxialHookes Law
A E
PL
axial deformation : δ=
δ ABthermal = α ABLAB ∆ T
Section 3: Torsion of Circular Sections
a. Shear stress If the shaft has a circular cross section and the material remains in the linear-elastic region, the shear stress in the shaft varies as a linear function of the distance (∆) from the center of the shaft and is given by:
The maximum shear stress in the shaft is on the outer surface independent of whether the shaft is solid or hollow and is given by:
The polar area moment of inertia is:
The calculated stresses act on the element as shown.
The deformation is measured by the angle of twist (θ) of one end relative to the other and is giver by:
where G is the modulus of rigidity for the material and L is the length of shaft.
The shaft also has maximum and minimum normal stresses acting on a element rotated 45° from the element for which the shear stress was calculated. The maximum tensile and compressive stresses are related to the shear stress by:
J
T
shearstress ρ : τ =
J
T r max shear stress : τmax = o
( 4 4 )
4
sec :
sec :
o i
o
hollow tion J r r
r solid tion J
J G
TL
angle oftwist : θ=
J
T ro
σ ten .= −σ comp .= τmax=
Section 4: Beams
a. Flexural or Bending Stress If the loads on a beam act in its plane of symmetry and the beam is linear elastic, the bending stresses acting normal to the cross section vary linearly with the distance from the neutral axis (N.A.) and are given by:
In the absence of axial loads:
In the sketch the cross-section is shown rectangular. However, the cross section, in general, can be circular, triangular, etc. The properties of many structural sections such as T-, I-, H-sections can be found in handbooks. If the section is not standard, you must be prepared to determine the centrodial location as well as the value of INA. The maximum bending stress occurs at the location in the beam where (My) is maximum. The section modulus provides a single parameter for design purposes.
where c is the maximum distance of material from the neutral axis.
b. Shearing Stresses in Beams
The transverse and longitudinal horizontal shearing stress in a beam is given by:
where Q is the first moment of the shaded area about the neutral axis if the shearing stress is being evaluated along
the inside edge of the shaded area. For a rectangular section Q = (c – a)(t)
(c + a)
The maximum shearing stress will occur where
Q
t is maximum. Q is always
maximum at the neutral surface. However,
Q
t may or may not be maximum at the neutral surface. Check all possibilities.
I NA
My
bending stresses : σ=
neutral axis = centrodial axis
S
M
max imum bendingstress σmax= max
c
I
S = sec tion mod ulus =
I t
VQ
NA
e. Beam Deflections
Two Integration Method The deflection of straight beams is determined from the equation:
Here y(x) is the lateral displacement of the beam from its original position as a function of position along the beam, the primes denote derivatives with respect to x, and M(x) is the bending moment as a function of position along the beam. Integration of this equation once yields the equation of the slope as a function of position along the beam:
A second integration yields the deflection or elastic curve equation:
The two integration constants C 1 and C 2 are evaluated using the boundary conditions imposed on the slope and deflection by the supports.
The majority of beam loading requires that the bending moment be defined using more than one analytic function. Each function is valid over its own portion of the beam length and results in its own set of slope and deflection equations that are valid in that portion of the beam. Each set of equations has its own pair of integration constants. The additional boundary conditions come from requiring that the slope and deflection given by the equations on both sides of a boundary between changes in bending moment functions give the same value when evaluated at the boundary.
Superposition Method The solutions for these equations for many different types of supports and loads are given in many of the common engineering handbooks. The principal of superposition allows the solutions of different loads to be added together to give the solution for the combined loads. The limitations of this method depend on how extensive the available beam tables are. It must be kept in mind that the table entry must be able to exactly match the portion of the load being represented using only a scaling factor and/or mirror imaging. Loads in the tables may have either positive or negative values.
EIy ′′ ( x ) = M ( x )
EIy ′ ( x ) =∫ M ( x ) dx + C 1
EIy ( x ) = ∫ ∫[ M ( x ) dx ] dx + C 1 x + C 2
y ( atboundary ) y ( at boundary )
y atboundary y atboundary L R
L R
YOK
Section 7: Plane Transformations
a. Stresses
Transformation Equations It is assumed that all the stresses in one direction are zero. The coordinate axes are orientated to place the z-axis in that direction. This situation is common in engineering applications. A free surface is the classic example.
The stresses representing the state of stress at a point are different when measured with respect to two different coordinate systems that are rotated with respect to each other. If the first system is labeled xy then the x'y' is rotated counter-clockwise by an angle θ.
The primed stresses may be determined from the unprimed by the equations:
σy' = σx'(θ + 90°)
Principal Stress and Maximum Shearing Stress There will always be a maximum and minimum stress value, referred to as the principal stresses, occurring at some orientation. There will also be a maximum shearing stress that occurs on two different planes.
The values of the principal stresses are given by:
The plus sign is used for the larger σ 1 and the minus sign for the smaller σ 2.
The value of the maximum shearing stress is given by:
The orientation of the σ 1 plane relative to the σx plane is given by:
( ) ( ) ( )
( ) ( θ) τ ( θ)
Sin Cos
Cos Sin
xy
x y xy
xy
x y x y x
′ ′
′
( )^2
2 1 , 2 2 2 xy
σ σ x σ y σ x σ y + τ
( )^2
2 max 2 xy
τ σ x σ y + τ
x y
xy P Tan
b. Strains
Transformation Equations The analysis is based on a plane strain state in which all strains in the z- direction are zero. The analysis can also be used for a plane stress state with one minor modification. A material can not have both plane stress and plane strain states at the same time.
The relationship between the strains at a point measured relative to a set of axes x-y and a set x'-y' which have the same origin but are rotated counter- clockwise from the original axes by an angle θ are given by
for the normal strains and by
for the shearing strain. Note the similarity of form between these equations and the stress transformation equations.
Principal Strains and Maximum Shearing Strain As with the stresses there are maximum and minimum (principal) values of the normal strains for particular orientations at the point and maximum shearing strains. The principal strains are given by
and the maximum shearing strain is given by
The orientation of the larger principal strain to the positive x-direction is given by
The direction of the smaller principal strain is perpendicular to the first. The directions involved with the maximum shearing strain are the two directions at 45° to both of the principal directions.
Mohr's Circle for Strain
( ) ( θ)
γ θ +
ε −ε
ε +ε ε (^) ′ = 2 2
x x y x yCos^ xySin
( ) ( θ)
γ θ +
ε − ε = −
γ (^) ′ ′ 2 2
xy x ySin xyCos
2 2 1 , 2 2 2 2
γ +
ε −ε ±
ε +ε ε = x y x y xy
2 2 max 2 2 2
γ +
ε −ε =±
γ (^) x y xy
ε −ε
γ
θ = −
2
x y
xy
P Tan
A Mohr's Circle mapping between the strains acting with respect to a set of x-y axes at a point and a point in the strain plane can be made. The same rules apply as for the stress circle with ε replacing σ and
γ 2 replacing^ τ. This makes the radius of the circle equal to half the in-plane maximum shearing strain.
Sign convention for the shear strain is based on which way that axis has to twist to have the right angle close for a positive shear strain and open for a negative shear strain.
The circle is centered at εavg =
εx + εy 2 and^
γ 2 = 0, with a radius R =^
γmax
As with the stresses, the principal strains are located where the circle crosses the horizontal axis. Maximum shearing strains are located at the top and bottom of the circle.
Section 8: Material Properties
a. Poisson's Ratio
When a material is stretched in one direction it contracts in the lateral directions. The resulting longitudinal and lateral strains occur in a fixed ratio known as Poisson's ratio. The value of Poisson's ratio for a given material may be
determined from a simple tension test as ν = -
εlateral εlongitudinal
. The minus sign
b. Generalized Hooke's Law
Three dimensional stress state.
These relationships are valid within the linear region of the materials stress-strain response.
G is the modulus of rigidity (shearing modulus of elasticity)
G, E, and ν are related by the formula: G =
E
2(1 + ν)
Plane stress state: σz = τxz = τyz = 0
[ ( )] ( )( )
[( ) ( )]
[ ( )] ( )( )
[( ) ( )]
[ (^ )] ( )( )
[( ) (^ )]
yz yz
yz yz
xz xz
xz xz
xy xy
xy xy
z z x y z z x y
y y x z y y x z
x x y z x x y z
G
G
G
G
G
G
E
E
E
E
E
E
( ) ( )
( )
( ) ( )
( )
( )
( )
xy xy
xy xy
x y
z x y z
y y x y y x
x x y x x y
G
G
E
E
E
E
E
2
2
REVIEW PROBLEMS
- An aluminum bar having a constant cross sectional area of 0.25 in^2 carries the axial loads applied at the positions shown. Find the deformation of the bar.
a. ____ 0.0192 in. b. ____ 0.2880 in. c. ____ 0.3264 in. d. ____ 0.3840 in. e. ____ None of these.
- A steel rod with a cross sectional area of 0.5 in^2 is stretched between two rigid walls. The temperature coefficient of strain is 6.5× 10 -6^ in./in./°Ε and E is 30× 106 psi. If the tensile load is 2000 lb. at 80°F, find the tensile load at 0°F.
a. ____ 5800 lb. b. ____ 7800 lb. c. ____ 8800 lb. d. ____ 9800 lb. e. ____ 19,600 lb.
- The composite bar shown is firmly attached to unyielding supports at the ends and is subjected to the axial load P shown. If the aluminum is stressed to 10,000 psi, find the stress in the steel.
a. ____ 1000 psi b. ____ 2000 psi c. ____ 5000 psi d. ____ 10,000 psi e. ____ 20,000 psi
- Determine the maximum bending moment in the beam.
a. ____ 3600 ft-lb. b. ____ 5400 ft-lb. c. ____ 7200 ft-lb. d. ____ 8100 ft-lb. e. ____ 4050 ft-lb.
- Find the maximum transverse shearing force in the beam shown.
a. ____ 450 lb. b. ____ 1800 lb. c. ____ 2250 lb. d. ____ 3600 lb. e. ____ 4050 lb.
- By means of strain gages, the flexural stresses are found to be –12,000 psi at A and +4000 psi at B. Assuming the elastic limit of the material has not been exceeded; find the flexural stress at the bottom of the beam.
a. ____ 6000 psi. b. ____ 8000 psi. c. ____ 9000 psi. d. ____ 10,000 psi. e. ____ 12,000 psi.
- For the cast iron beam shown, the maximum permissible compressive stress is 12,000 psi and the maximum permissible tensile stress is 5000 psi. Find the maximum safe load P that can be applied to the beam as shown.
a. ____ 220 lb. b. ____ 333 lb. c. ____ 1250 lb. d. ____ 3000 lb. e. ____ 7500 lb.
- A 12-inch, 35-lb I-beam 30 ft. long is supported at 5 ft. from each end and carries a uniform distributed load of 1600 lbs per ft. (which includes its own weight). Determine the maximum flexural stress in the beam.
- Find the maximum vertical shearing force which may be applied to a box beam having the cross section shown without exceeding a horizontal shearing stress of 500 psi.
a. ____ 3065 lb. b. ____ 4000 lb. c. ____ 6000 lb. d. ____ 6130 lb. e. ____ 6300 lb.
- Find the reaction at the right end of the beam shown.
a. ____ wL/ b. ____ wL/ c. ____ 3wL/
- Two beams, simply supported at their ends, jointly support a load P = 3500 lb. applied to the upper 6-ft. beam at its midpoint. The beams are identical except for length and cross at their midpoints. find the load carried by the lower 9-ft. beam.
a. ____ 700 lb. b. ____ 800 lb. c. ____ 1000 lb. d. ____ 1750 lb. e. ____ 2700 lb.
- For stress conditions on the element shown, find the principal stresses and the plane on which the maximum principal stress acts.
- A circular shaft of brittle material subjected to torsion fractures along a 45° angle. Failure is due to what kind of stress?
a. ____ Shearing stress. b. ____ Compressive stress. c. ____ Tensile stress. d. ____ Combined stress. e. ____ None of these.
- Which has the higher shear stress for a given elastic torque?
a. a one-inch diameter rod, or b. a two-inch diameter rod.
- Identical rods of aluminum and steel are each subjected to the same elastic torque. Which rod will have the higher shear stress?
a. steel b. aluminum c. both have the same stress
- If G represents the modulus of rigidity (or shear modulus of elasticity), E is the modulus of elasticity, and ν is Poisson’s ratio, which of the following statements is true for any homogeneous material?
a. ____ G is independent of E. b. ____ G is 0.4E. c. ____ G is 0.5E d. ____ G depends upon both E and ν. e. ____ None of these.
ANSWERS EIT REVIEW
MECHANICS OF MATERIALS
- c
- d
- e
- d
- τB = 5,333.3 psi τA = 2,666 psi
- τS = 81,528 psi τA = 54,352 psi θ = 1.086 rad
- e
- c
- b
- d
- σmax = 19,100 psi
- d
- c
- b
- 19.8 in.
- a
- 8,000 psi
- σmax = 605 psi σnmin = -6606 psi
- c
- a
- c
- d
