Internal Loading
Mechanics of Materials
Chapter 1
Engineering Mechanics:
Mechanics of Materials:
The branch of engineering mechanics that studies the internal effects of
_______________ and _______________ on a solid body that is subjected
to external loading.
Newton’s 1st Law of Motion:
Units:
Significant Figures:
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
Chapter 1
Statics Review: Trusses
Determine the force in members GE, GC, and BC of the truss shown. Indicate whether the
members are in tension or compression.
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
Chapter 1
Internal Forces (in 2D):
Positive sign convention:
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
Chapter 1
Then:
Steps for calculating internal forces (in 2D):
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
Chapter 1
Solve for the internal forces at C.
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
Chapter 1
Solve for the internal loading at C.
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
Chapter 1
Solve for the internal loading a section b-b through the centroid C on the beam.
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
Chapter 1
2D internal forces: RECAP AND IMPORTANT POINTS
HW # 1
Due________________________
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
ENGR 2070 – Mechanics of Materials
Chapter 1
9
Internal Loading
Mechanics of Materials
Chapter 1
Alternative methods of drawing
internal forces in 3D
Internal Forces (in 3D):
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
Chapter 1
Solve for the internal loadings at C.
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
Chapter 1
Solve for the internal loads at plane AB.
100
ENGR 2070 – Mechanics of Materials
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Internal Loading
Mechanics of Materials
Chapter 1
3D internal forces: RECAP AND IMPORTANT POINTS
HW # 2
Due________________________
ENGR 2070 – Mechanics of Materials
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Average Normal Stress
Sections 1.3-1.4
Normal Stress (σ) –
ENGR 2070 – Mechanics of Materials
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Average Normal Stress
Sections 1.3-1.4
Example:
Find the normal stress in each section of the beam.
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Average Normal Stress
Sections 1.3-1.4
The 80 kg lamp is supported by two rods AB and BC as shown in the figure. If AB has as
diameter of 10 mm and BC has a diameter of 8 mm, determine the average normal stress in
each rod.
ENGR 2070 – Mechanics of Materials
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Average Shear Stress
Section 1.5
Shear Stress (τ) –
ENGR 2070 – Mechanics of Materials
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Average Shear Stress
Section 1.5
Example:
A punch is used to make a slot in a 10mm thick plate. Find the minimum force required to
punch the slot if the plate shears at 250MPa.
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Average Shear Stress
Section 1.5
The board is subjected to a tensile force of 200 lb. Determine the average normal and average
shear stress in the wood fibers, which are oriented along plane a-a at 20o with the axis of the
baord.
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Average Shear Stress
Section 1.5
Types of Shear Connections:
a) Single Shear:
Lap Joint:
Bolted Joint:
ENGR 2070 – Mechanics of Materials
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Average Shear Stress
Section 1.5
b) Double Shear:
Double Lap Joint:
Bolted Joint:
ENGR 2070 – Mechanics of Materials
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Average Shear Stress
Section 1.5
Pins:
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Average Shear Stress
Section 1.5
Example:
The joint is fastened together using two bolts. Determine the required diameter of the bolts if
the failure shear stress is 140 MPa.
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Average Shear Stress
Section 1.5
General state of stress (infinitesemal stress element):
𝜏𝑧𝑥 = 𝑉𝐴𝑥
Complementary property of shear:
Through equilibrium, at any given point on a body, shear stresses along two perpendicular
planes will be equal
Must have equal magnitude and be pointing toward or away from each other
Pure shear condition:
ENGR 2070 – Mechanics of Materials
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Average Shear Stress
Section 1.5
Stress: RECAP AND IMPORTANT POINTS
HW # 3
Due________________________
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Average Shear Stress
Section 1.5
ENGR 2070 – Mechanics of Materials
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Allowable Stress Design
Section 1.6
Allowable Stress Design:
Why use ASD?
1.
2.
3.
4.
5.
6.
7.
ENGR 2070 – Mechanics of Materials
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Allowable Stress Design
Section 1.6
Design of Simple Connections:
1.
2.
Normal Force:
Shear Force:
Common Applications:
1. Sizing a tension or compression member
Example:
σFail = 36 ksi
P = 100 kips
F.S. = 2
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Allowable Stress Design
Section 1.6
2. Required area to resist bearing
Bearing Stress:
ENGR 2070 – Mechanics of Materials
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Allowable Stress Design
Section 1.6
Determine the required thickness of member BC and the diameter of the pins at A and B if σ fail =
58 ksi for member BC and τallowable = 10 ksi for the pins. The factor of safety = 2.
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Allowable Stress Design
Section 1.6
ENGR 2070 – Mechanics of Materials
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Allowable Stress Design
Section 1.6
The steel structure shown has a pin C with a diameter of 6mm and pins B and D have a
diameter of 10mm. The shear failure stress is fail = 150 MPa at all connections and a normal
failure stress of fail = 400 MPa in link BD.
(Note that link BD is not reinforced around the pin holes.) If a factor of safety
of 3 is to be used, determine the largest load P that can be applied at A.
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Allowable Stress Design
Section 1.6
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Allowable Stress Design
Section 1.6
Allowable Stress Design: RECAP AND IMPORTANT POINTS
HW # 4
Due________________________
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Strain
ENGR 2070 - Mechanics
Chapter 2
Intro:
Strain:
1) Load induced strain
2) Temperature induced strain
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Strain
ENGR 2070 - Mechanics
Chapter 2
Normal Strain:
P
P
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Strain
ENGR 2070 - Mechanics
Chapter 2
Example:
When force P is applied to the rigid arm ABC, point B displaces vertically downward through a
distance of 0.2 mm. Determine the normal strain in wire CD.
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Strain
ENGR 2070 - Mechanics
Chapter 2
Shear Strain:
Small strain assumption:
R
2 mm
500 mm
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Strain
ENGR 2070 - Mechanics
Chapter 2
Strain at corners A, B, C, D.
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Strain
ENGR 2070 - Mechanics
Chapter 2
Example:
The piece of rubber is originally rectangular. Determine the average shear strain γ xy at A.
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Strain
ENGR 2070 - Mechanics
Chapter 2
The rectangular plate is deformed into the shape shown by the dashed lines. Determine the
average normal strain along diagonal BD, and the average shear strain at corner B relative to
the x, y axes.
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Strain
ENGR 2070 - Mechanics
Chapter 2
Strain: RECAP AND IMPORTANT POINTS
HW # 5
Due________________________
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Stress-Strain Relationship
Section 3.1-3.3
Stress-Strain Diagram for Steel
Stress-strain diagram: Provides data about the material’s mechanical behavior without regard
for the element’s geometry and size.
How material behaves (very important in engineering)
How to we find the stress-strain relationship?
A test machine is used to stretch (or compress) the specimen at a controlled load rate until the
specimen reaches breaking point.
An extensometer is used to measure displacement or strain gages are used to measure strain
directly.
No two diagrams are exactly the same:
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Stress-Strain Relationship
Section 3.1-3.3
ENGR2070 – Mechanics of Materials
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Stress-Strain Relationship
Section 3.1-3.3
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Stress-Strain Relationship
Section 3.1-3.3
.2% offset method:
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Stress-Strain Relationship
Section 3.1-3.3
Ductile vs. Brittle Materials
Ductile:
Brittle:
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Hooke’s Law
Section 3.4
Hooke’s Law:
(only in elastic region!!!)
Note:
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Hooke’s Law
Section 3.4
Loading and Unloading:
Reloading and Unloading:
ENGR2070 – Mechanics of Materials
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Hooke’s Law
Section 3.4
Example:
Determine the approximate modulus of elasticity and the yield strength of the alloy using the
0.2% offset method. The diameter is 0.5 in and the gauge length is 2 in.
If the specimen is stressed to 30 ksi and unloaded, determine the percent elongation.
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Hooke’s Law
Section 3.4
A portion of the stress-strain curve for a stainless steel alloy is shown below. A 350-mm-long
bar is loaded in tension until it elongates 2.0 mm and then the load is removed.
(a) What is the permanent set in the bar?
(b) What is the length of the unloaded bar?
(c) If the bar is reloaded, what will be the proportional limit?
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Hooke’s Law
Section 3.4
The wire has a diameter of 5 mm and is made from A-36 steel. If a 80-kg man is sitting on seat
C, determine the elongation of wire DE.
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Hooke’s Law
Section 3.4
The material for the 50 mm long specimen has the stress-strain diagram shown. If P = 100 kN,
determine the elongation of the specimen.
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Strain Energy
Section 3.5
Strain Energy:
ENGR2070 – Mechanics of Materials
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Strain Energy
Section 3.5
Modulus of Resilience:
Modulus of Toughness:
Comparison of Materials:
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Strain Energy
Section 3.5
Example:
The stress-strain diagram for human skin is given. Find the Modulus of Elasticity, Modulus of
Toughness, and Modulus of Resilience
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Strain Energy
Section 3.5
The stress-strain diagram for a steel alloy having an original diameter of 0.5 in. and a gage
length of 2 in. is given in the figure. Determine approximately the
a) modulus of elasticity
b) load that causes yielding
c) ultimate load it can support
d) modulus of resilience
e) modulus of toughness
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Strain Energy
Section 3.5
Stress-Strain Relationship: RECAP AND IMPORTANT POINTS
HW # 6
Due_______________________
ENGR2070 – Mechanics of Materials
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Poisson’s Ratio
Section 3.6
Poisson’s Ratio:
Typical Material Values:
Material
Poisson’s Ratio
Aluminum
A36 Steel
Stainless Steel
Concrete
Wood
*Values located on back cover
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Poisson’s Ratio
Section 3.6
Example:
The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is
applied to it, determine the change in its length, the change in diameter, and change in volume.
Ep = 2.70 GPa, νp = 0.4.
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Shear Stress-Strain Diagrams
Section 3.7
Shear Stress-Strain Diagram:
Specimen subjected to torsion tests in which a shear stress-strain curve can be developed
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Shear Stress-Strain Diagrams
Section 3.7
The lap joint is connected together using a 1.25 in. diameter bolt. If the bolt is made from a
material having a shear stress–strain diagram that is approximated as shown, determine the
shear strain developed in the shear plane of the bolt when P = 75 kip. Also, determine the
permanent shear strain in the shear plane of the bolt when the applied force P = 150 kip is
removed.
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Creep and Fatigue
Section 3.8
Creep: When a material has to support a load for an extended period of time, the strain may
increase in some materials. A time-dependent deformation
Fatigue: When a material is subjected to repeated cycles of stress and strain, it causes the
material to break down and fracture at a stress less than the material’s yield stress.
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Creep and Fatigue
Section 3.8
Poisson’s Ratio: RECAP AND IMPORTANT POINTS
HW # 7
Due_______________________
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Elastic Deformation
Section 4.1 & 4.2
Saint Venant’s Principle:
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Elastic Deformation
Section 4.1 & 4.2
NOTES:
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Elastic Deformation
Section 4.1 & 4.2
The 20 mm diameter A-36 steel rod is subjected to the axial forces shown. Determine the displacement
of end C with respect to the fixed support at A.
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Elastic Deformation
Section 4.1 & 4.2
The 30 mm diameter A992 steel rod is subjected to the loading shown. Determine the displacement of
end C.
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Elastic Deformation
Section 4.1 & 4.2
The 20 mm diameter 2014-T6 aluminum rod is subjected to the uniform distributed axial load.
Determine the displacement of end A.
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Elastic Deformation
Section 4.1 & 4.2
Axially Loaded Deformation: RECAP AND IMPORTANT POINTS
HW # 8
Due_______________________
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Statically Indeterminate Axially Loaded Member
Section 4.3 & 4.4
Statically determinate:
Statically indeterminate:
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Statically Indeterminate Axially Loaded Member
Section 4.3 & 4.4
Compatibility equation:
ENGR2070 – Mechanics of Materials
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Statically Indeterminate Axially Loaded Member
ENGR2070 – Mechanics of Materials
Section 4.3 & 4.4
73
Statically Indeterminate Axially Loaded Member
Section 4.3 & 4.4
Example:
The assembly consists of two posts AD and CF made of A-36 steel and having a cross-sectional
area of 1000 mm2, and a 2014-T6 aluminum post BE having a cross sectional area of 1500 mm2.
If a central load of 400 kN is applied to the rigid cap, determine the normal stress in each post.
There is a small gap of 0.1 mm between the post BE and the rigid member ABC.
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Statically Indeterminate Axially Loaded Member
Section 4.3 & 4.4
Example:
The concrete column is reinforced using 4 rods (rebar), each with a diameter of 18mm. Determine the
stress in the concrete and the steel if the column is subjected to an axial load of 800 kN. Est = 200GPa, Ec
= 25 GPa
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Statically Indeterminate Axially Loaded Member
Section 4.3 & 4.4
Example:
A rigid bar AB is hinged to a support at A and supported by two vertical wires attached at points
C and D. Both wires have the same cross-sectional area (A=0.0272 in2) and are made of the
same material (E = 30x106 psi).
A) Determine the tensile stresses at C and D in the wires due to the load P = 340 lb acting at end
B of the bar.
B) Find the downward displacement B at end of the bar.
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Force Method of Analysis
Section 4.5
Superposition:
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Force Method of Analysis
Section 4.5
Example:
The A-36 steel rod shown in the figure has a diameter of 10 mm. It is fixed to the wall at A, and
before it is loaded there is a gap between the wall and the rod of 0.2 mm. Determine the
reactions at A and B’. Neglect the size of the collar at C. Take Est = 200 GPa.
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Force Method of Analysis
Section 4.5
Statically Indeterminate – Axially loaded members:
RECAP AND IMPORTANT POINTS
HW # 9
Due_______________________
ENGR2070 – Mechanics of Materials
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Force Method of Analysis
Section 4.5
ENGR2070 – Mechanics of Materials
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Thermal Stress
Section 4.5
Thermal Stress:
Typical values:
ENGR2070 – Mechanics of Materials
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Thermal Stress
Section 4.5
If specimen is free to move:
If specimen is fixed:
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Thermal Stress
Section 4.5
Example:
The C83400-red-brass rod AB and 2014-T6-aluminum rod BC are joined at the collar B and fixed
connected at their ends. If there is no load in the members when T1 = 50oF, determine the
average normal stress in each member when T2 =120oF Also, how far will the collar be
displaced? The cross-sectional area of each member is 1.75 in2.
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Thermal Stress
Section 4.5
Example:
The 50-mm-diameter cylinder is made from Am 1004-T61 magnesium and is placed in the
clamp when the temperature is T1 = 20° C.
A) If the 304-stainless-steel carriage bolts of the clamp each have a diameter of 10 mm,
and they hold the cylinder snug with negligible force against the rigid jaws, determine
the force in the cylinder when the temperature rises to T2 = 130° C.
B) The cylinder is placed in the clamp at T1 = 15° C. Determine the temperature at which
the average normal stress in either the magnesium or the steel first becomes 12 MPa.
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Thermal Stress
Section 4.5
Thermal Stress: RECAP AND IMPORTANT POINTS
HW # 10
Due_______________________
ENGR2070 – Mechanics of Materials
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Thermal Stress
Section 4.5
ENGR2070 – Mechanics of Materials
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Torsion
Section 5.1 & 5.2
Torsion:
ENGR2070 – Mechanics of Materials
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Torsion
Section 5.1 & 5.2
Derivation of Shear Stresses:
Polar Moment of Inertia:
Circular Cross Section:
Tube:
ENGR2070 – Mechanics of Materials
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Torsion
Section 5.1 & 5.2
Example:
The rod has a diameter of 0.5 in. and weight of 5 lb ft. Determine the maximum torsional stress in the
rod at a section located at A due to the rod’s weight. What is the torsional stress at point B?
ENGR2070 – Mechanics of Materials
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Torsion
Section 5.1 & 5.2
The solid shaft has a diameter of 40 mm. Determine the absolute maximum shear stress in the shaft
and sketch the shear-stress distribution along a radial line of the shaft where the shear stress is
maximum.
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Torsion
Section 5.1 & 5.2
The 60 mm diameter solid shaft is subjected to the distributed and concentrated torsional loadings
shown. Determine the shear stress at points A and B.
ENGR2070 – Mechanics of Materials
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Torsion
Section 5.1 & 5.2
Torsion: RECAP AND IMPORTANT POINTS
HW # 11
Due_______________________
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Angle of Twist
Section 5.4
Angle of Twist:
Right hand rule applies:
Example:
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Angle of Twist
Section 5.4
A series of gears are mounted on the 40-mm diameter steel shaft. Determine the angle of twist of gear
B relative to gear A. Take G = 75 GPa
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Angle of Twist
Section 5.4
The shaft is made from a solid steel section AB and a tubular portion made of steel and having a
brass core. If it is fixed to a rigid support at A, and a torque of T = 50 lb*ft is applied to it at C,
determine the angle of twist that occurs at C and compute the maximum shear stress and
maximum shear strain in the brass and steel. Take Gst = 11.5(103) ksi, Gbr = 5.6(103) ksi.
ENGR2070 – Mechanics of Materials
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Angle of Twist
Section 5.4
The A-36 steel shaft has a diameter of 50 mm and is subjected to the distributed and
concentrated loadings shown. Determine the absolute maximum shear stress in the shaft.
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Angle of Twist
Section 5.4
Angle of Twist: RECAP AND IMPORTANT POINTS
HW # 12
Due_______________________
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Angle of Twist
Section 5.4
ENGR2070 – Mechanics of Materials
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Statically Indeterminate Torque-Loaded Members
Section 5.4
Statically Indeterminate Torque-Loaded members:
ENGR2070 – Mechanics of Materials
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Statically Indeterminate Torque-Loaded Members
Section 5.4
Example:
A rod is made from two segments: AB is steel and BC is brass. It is fixed at its ends and
subjected to a torque of T = 680 N m. If the steel portion has a diameter of 30 mm, determine
the required diameter of the brass portion so the reactions at the walls will be the same. Gst =
75 GPa, Gbr = 39 GPa. Also determine the absolute maximum shear stress of the shaft.
ENGR2070 – Mechanics of Materials
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Statically Indeterminate Torque-Loaded Members
Section 5.4
The shaft is made of A-36 steel and is fixed at end D, while end A is allowed to rotate 0.005 rad
when the torque is applied. Determine the torsional reactions at these supports.
ENGR2070 – Mechanics of Materials
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Statically Indeterminate Torque-Loaded Members
Section 5.4
Statically Indeterminate-Torque loaded member:
RECAP AND IMPORTANT POINTS
HW # 13
Due_______________________
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Shear and Moment Diagrams
Section 6.1-6.2
Shear and Moment Diagrams:
Applied external loads cause
along the beam length.
and
that may vary
Diagrams show:
Why use them?
Beam Classifications:
Sign Convention (Chapter 1):
ENGR2070 – Mechanics of Materials
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Shear and Moment Diagrams
ENGR2070 – Mechanics of Materials
Section 6.1-6.2
104
Shear and Moment Diagrams
ENGR2070 – Mechanics of Materials
Section 6.1-6.2
105
Shear and Moment Diagrams
ENGR2070 – Mechanics of Materials
Section 6.1-6.2
106
Shear and Moment Diagrams
Section 6.1-6.2
Draw shear and moment diagrams:
ENGR2070 – Mechanics of Materials
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Shear and Moment Diagrams
Section 6.1-6.2
𝛥𝑉 = ∫ 𝑤(𝑥) 𝑑𝑥
𝛥𝑀 = ∫ 𝑉(𝑥) 𝑑𝑥
𝑑𝑉
𝑤(𝑥) =
𝑑𝑥
𝑑𝑀
𝑉(𝑥) =
𝑑𝑥
ENGR2070 – Mechanics of Materials
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Shear and Moment Diagrams
Section 6.1-6.2
1) Start at ends:
ENGR2070 – Mechanics of Materials
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Shear and Moment Diagrams
Section 6.1-6.2
2) From left to right, plot the change in V (caused by external forces and area under loading)
Change in V:
External forces:
Uniform distributed load:
Linear distributed load:
ENGR2070 – Mechanics of Materials
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Shear and Moment Diagrams
Section 6.1-6.2
3) From left to right, plot the change in M (caused by external moments and area under V diagram)
Change in M:
External moments:
Uniform V:
Linear V:
Exponential V:
ENGR2070 – Mechanics of Materials
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Shear and Moment Diagrams
Section 6.1-6.2
**Remember V is the slope of M***
V crosses zero:
V drops:
-If all else fails, you can make a “cut” at certain locations and calculate the V and M from Fy = 0
and M = 0
or
-Think about how V or M would change before and after the point in question.
ENGR2070 – Mechanics of Materials
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Shear and Moment Diagrams
Section 6.1-6.2
Example:
ENGR2070 – Mechanics of Materials
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Shear and Moment Diagrams
Section 6.1-6.2
Example:
ENGR2070 – Mechanics of Materials
114
Shear and Moment Diagrams
Section 6.1-6.2
Example:
ENGR2070 – Mechanics of Materials
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Shear and Moment Diagrams
ENGR2070 – Mechanics of Materials
Section 6.1-6.2
116
Shear and Moment Diagrams
Section 6.1-6.2
Draw the shear and moment diagrams for the compound beam which is pin connected at B.
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Shear and Moment Diagrams
ENGR2070 – Mechanics of Materials
Section 6.1-6.2
118
Shear and Moment Diagrams
Section 6.1-6.2
Shear and Moment Diagrams: RECAP AND IMPORTANT POINTS
HW # 14
Due_______________________
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Shear and Moment Diagrams
ENGR2070 – Mechanics of Materials
Section 6.1-6.2
120
Bending Stress
Section 6.4
The Flexure Formula:
ENGR2070 – Mechanics of Materials
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Bending Stress
Section 6.4
The moment (M) causes normal stress ()
Review: Centroid
ENGR2070 – Mechanics of Materials
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Bending Stress
Section 6.4
Calculate the centroid and moment of inertia about the centroidal axis of the cross sectional area.
All dimensions are mm.
20
250
200
mm
15
15
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Bending Stress
Section 6.4
ENGR2070 – Mechanics of Materials
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Bending Stress
Section 6.4
Example:
Two designs for a beam are to be considered. Determine which one will support a moment of 150 kN-m
with the least amount of bending stress. What is that stress? Draw the stress distribution for the beam.
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Bending Stress
Section 6.4
The wing spar ABD of a light plane is made from 2014–T6 aluminum and has a cross-sectional area of
1.27 in2, a depth of 3 in., and a moment of inertia about its neutral axis of 2.68 in4. Determine the
absolute maximum bending stress in the spar if the anticipated loading is to be as shown. Assume A, B,
and C are pins. Connection is made along the central longitudinal axis of the spar.
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Bending Stress
Section 6.4
The beam is subjected to a moment of M = 1 kip-ft. Determine the maximum bending stress in the
beam. Also, determine the resultant force the bending stress produces on the top board A of the beam.
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Bending Stress
Section 6.4
ENGR2070 – Mechanics of Materials
128
Bending Stress
Section 6.4
Bending Stress: RECAP AND IMPORTANT POINTS
HW # 15
Due_______________________
ENGR2070 – Mechanics of Materials
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Bending Stress
Section 6.4
ENGR2070 – Mechanics of Materials
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Transverse Shear
Section 7.1-7.2
Shear Stress:
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Transverse Shear
Section 7.1-7.2
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Transverse Shear
Section 7.1-7.2
Wide Flange Beams:
Plot the intensity of the shear stress distributed over the cross section of the strut if it subjected
to a shear force of V = 600 kN
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Transverse Shear
Section 7.1-7.2
For a rectangular cross section:
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Transverse Shear
Section 7.1-7.2
Example:
A laminated wood beam on simple supports is built up by gluing together four 2 in. x 4 in.
boards (actual dimensions) to form a solid beam 4 in. x 8 in. in cross section, as shown in the
figure. The allowable shear stress in the glued joints is 65 psi, and the allowable bending stress
in the wood is 1800 psi. If the beam is 9 ft long, what is the allowable load P acting at the onethird point along the beam as shown? (Include the effects of the beam’s own weight, assuming
that the wood weighs 35 lb/ft3.)
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Shear Flow in Build-Up Members
Section 7.3
Shear Flow in Built-Up Members:
Built-up members:
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Shear Flow in Build-Up Members
Section 7.3
Example:
The beam is subjected to a shear of V = 2 kN. Determine the average shear stress developed in each nail
if the nails are spaced 75 mm apart on each side of the beam. Each nail has a diameter of 4 mm.
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Shear Flow in Build-Up Members
Section 7.3
Example:
The strut consists of 3 pieces of wood nailed together as shown and loaded as shown (w = 200
lb/ft).
a) Determine the location and magnitude of the maximum compressive and tensile
bending stress.
b) Calculate the bending stress at the interface between the boards at 2 ft to the left of
point A.
c) Determine the location and magnitude of the maximum transverse shear stress.
d) Calculate the transverse shear stress 1 inch from the bottom of the cross section and 2
ft to the left of point A.
e) Determine the required nail spacing if each nail can support 650 lb in single shear.
f) Determine the maximum distributed load the beam can support if each nail can support
500 lb in sing shear and the nails are spaced 6 inches apart.
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Shear Flow in Build-Up Members
Section 7.3
Transverse Shear: RECAP AND IMPORTANT POINTS
HW # 16
Due_______________________
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Shear Flow in Build-Up Members
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Section 7.3
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Thin-Walled Pressure Vessels
Section 8.1
Thin walled pressure vessels:
Cylindrical Vessels:
σ1 = hoop stress (Tangent to radius):
σ2 = longitudinal stress (Axial Direction):
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Thin-Walled Pressure Vessels
Section 8.1
Spherical Vessels:
Example 1. A cylindrical pressure vessel has an inner diameer of 4 ft and a thickness of .5 in. Determine
the maximum internal pressure it can sustain so that neither its circumferential nor it longitudinal stress
component exceeds 20 ksi.
Under the same conditions, what is the maximum internal pressure that a similar-sized spherical vessel
can sustain?
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Thin-Walled Pressure Vessels
Section 8.1
Example. Given the wooden tank (r = 18 inches) shown below, determine the normal stress in the hoop
restraints with p = 2 psi (gauge pressure). The hoops are 0.5” thick and 2” wide. In addition, determine
the tensile stress in each 0.25” diameter bolt.
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Stress due to Combined Loading
Section 8.2
Combined Loading:
In 2D:
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Stress due to Combined Loading
Section 8.2
Example:
The frame supports the distributed load shown. Determine the state of stress acting at point E.
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Stress due to Combined Loading
Section 8.2
In 3D:
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Stress due to Combined Loading
Section 8.2
Example:
Determine the state of stress at point A on the cross section of the pipe assembly at section a-a.
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Stress due to Combined Loading
ENGR2070 – Mechanics of Materials
Section 8.2
148
Stress due to Combined Loading
Section 8.2
Example:
The sign is subjected to the uniform wind loading. Determine the stress components at point C
and D on the 100-mm-diameter supporting post.
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Stress due to Combined Loading
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Section 8.2
150
Stress due to Combined Loading
Section 8.2
Combined Loading: RECAP AND IMPORTANT POINTS
HW # 17
Due_______________________
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Stress due to Combined Loading
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Section 8.2
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Plane-Stress Transformation
Section 9.1
Plane Stress:
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Plane-Stress Transformation
Section 9.2
Equilibrium on the segment:
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Plane-Stress Transformation
Section 9.2
Example:
Determine the equivalent state of stress on an element if it is oriented 30° clockwise from the element
shown. Use the stress-transformation equations.
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Plane-Stress Transformation
Section 9.2
The grains of wood in the board make an angle of 20o with the horizontal as shown. Determine the
normal and shear stress that act perpendicular and parallel to the wood grains if the board is subjected
to an axial load of 250 N.
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Plane-Stress Transformation
Section 9.2
Principal Stresses and Maximum In-Plane Shear Stress:
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Plane-Stress Transformation
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Section 9.2
158
Plane-Stress Transformation
ENGR2070-Mechanics of Materials
Section 9.2
159
Plane-Stress Transformation
Section 9.2
Example:
Determine the equivalent state of stress on an element at the same point which represents (a) the
principal stress, and (b) the maximum in-plane shear stress and the associated average normal stress.
Also, for each case, determine the corresponding orientation of the element with respect to the
element shown. Sketch the results on each element.
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Plane-Stress Transformation
Section 9.2
Determine the principal stress at point A on the cross section of the arm at section a–a. Specify the
orientation of this state of stress and indicate the results on an element at the point.
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Plane-Stress Transformation
Section 9.2
The 3-in. diameter shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B.
Determine the principal stresses and maximum in-plane shear stress at a point on the outer surface of
the shaft at section a–a.
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Plane-Stress Transformation
Section 9.2
Stress Transformation: RECAP AND IMPORTANT POINTS
HW # 18
Due_______________________
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Plane-Stress Transformation
ENGR2070-Mechanics of Materials
Section 9.2
164
Plane-Stress Transformation
Section 9.3
Mohr’s Circle:
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Plane-Stress Transformation
ENGR2070-Mechanics of Materials
Section 9.3
166
Plane-Stress Transformation
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Section 9.3
167
Plane-Stress Transformation
ENGR2070-Mechanics of Materials
Section 9.3
168
Plane-Stress Transformation
ENGR2070-Mechanics of Materials
Section 9.3
169
Plane-Stress Transformation
ENGR2070-Mechanics of Materials
Section 9.3
170
Plane-Stress Transformation
ENGR2070-Mechanics of Materials
Section 9.3
171
Plane-Stress Transformation
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Section 9.3
172
Plane-Stress Transformation
Section 9.3
Mohr’s Circle: RECAP AND IMPORTANT POINTS
HW # 19
Due_______________________
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Plane-Stress Transformation
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Section 9.3
174
Deflection of Beams
Section 12.1
Deflection of Beams:
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Deflection of Beams
Section 12.1
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Column Buckling
Section 13.2
Column Buckling
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Column Buckling
Section 13.2
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Column Buckling
Section 13.2
Example:
The W14 x 30 is used as a structural A992 steel column that can be assumed pinned at both of its ends.
Determine the largest axial force P that can be applied without causing it to buckle.
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Column Buckling
Section 13.2
Determine the maximum force F that can be supported by the assembly without causing member AC to
buckle. The member is made of A992 steel and has a diameter of 2 in. Take F.S. = 2 against buckling.
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Column Buckling
Section 13.2
Effective Length Concept:
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Column Buckling
Section 13.2
Example:
The 10-ft wooden column has the dimensions shown. Determine the critical load if the bottom is fixed and the top
is pinned. Ew = 1.6(103) ksi, σY = 5 ksi.
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Column Buckling
Section 13.2
The A992 steel column can be considered pinned at its top and fixed at the bottom and braced against
weak axis bending at mid-height. Determine the maximum allowable force P that the column can
support without buckling. Apply a FS of 2 against buckling. Take A = 7.4(10-3) m2, Ix = 87.3(10-6) m4, and
Iy = 18.8(10-6) m4.
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Column Buckling
Section 13.2
Column Buckling: RECAP AND IMPORTANT POINTS
HW # 20
Due_______________________
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