Title: CTC / MTC 222 Strength of Materials
1CTC / MTC 222Strength of Materials
2Final Exam
- Wednesday, December 12, 1015 -1215
- 30 of grade
- Graded on the basis of 30 points in increments of
½ point - Open book
- May use notes from first two tests plus two
additional sheets of notes - Equations, definitions, procedures, no worked
examples - Also may use any photocopied material handed out
in class - Work problems on separate sheets of engineering
paper - Hand in test paper, answer sheets and notes
stapled to back of answer sheets
3Course Objectives
- To provide students with the necessary tools and
knowledge to analyze forces, stresses, strains,
and deformations in mechanical and structural
components. - To help students understand how the properties of
materials relate the applied loads to the
corresponding strains and deformations.
4Chapter One Basic Concepts
- SI metric unit system and U.S. Customary unit
system - Unit conversions
- Basic definitions
- Mass and weight
- Stress, direct normal stress, direct shear stress
and bearing stress - Single shear and double shear
- Strain, normal strain and shearing strain
- Poissons ratio, modulus of elasticity in tension
and modulus of elasticity in shear
5Direct Stresses
- Direct Normal Stress , ?
- s Applied Force/Cross-sectional Area F/A
- Direct Shear Stress, ?
- Shear force is resisted uniformly by the area of
the part in shear - ? Applied Force/Shear Area F/As
- Single shear applied shear force is resisted by
a single cross-section of the member - Double shear applied shear force is resisted by
two cross-sections of the member
6Direct Stresses
- Bearing Stress, sb
- sb Applied Load/Bearing Area F/Ab
- Area Ab is the area over which the load is
transferred - For flat surfaces in contact, Ab is the area of
the smaller of the two surfaces - For a pin in a close fitting hole, Ab is the
projected area, Ab Diameter of pin x
material thickness
7Chapter Two Design Properties
- Basic Definitions
- Yield point, ultimate strength, proportional
limit, and elastic limit - Modulus of elasticity and how it relates strain
to stress - Hookes Law
- Ductility - ductile material, brittle material
8Chapter Three Direct Stress
- Basic Definitions
- Design stress and design factor
- Understand the relationship between design
stress, allowable stress and working stress - Understand the relationship between design
factor, factor of safety and margin of safety - Design / analyze members subject to direct stress
- Normal stress tension or compression
- Shear stress shear stress on a surface, single
shear and double shear on fasteners - Bearing stress bearing stress between two
surfaces, bearing stress on a fastener
9Chapter Four Axial Deformation and Thermal
Stress
- Axial strain e,
- e d / L , where d total deformation, and L
original length - Axial deformation, d
- d F L / A E
- If unrestrained, thermal expansion will occur due
to temperature change - d a x L x ?T
- If restrained, deformation due to temperature
change will be prevented, and stress will be
developed - s E a (?T)
10Chapter Five Torsional Shear Stress and
Deformation
- For a circular member, tmax Tc / J
- T applied torque, c radius of cross section,
J polar moment of inertia - Polar moment of Inertia, J
- Solid circular section, J p D4 / 32
- Hollow circular section, J p (Do4 - Di4 ) /
32 - Expression can be simplified by defining the
polar section modulus, Zp J / c, where c r
D/2 - Solid circular section, Zp p D3 / 16
- Hollow circularsection, Zp p (Do4 - Di4 ) /
(16Do) - Then, tmax T / Zp
11Chapter Six Shear Forces and Bending Moments in
Beams
- Sign Convention
- Positive Moment M
- Bends segment concave upward ? compression on
top
12Relationships Between Load, Shear and Moment
- Shear Diagram
- Application of a downward concentrated load
causes a downward jump in the shear diagram. An
upward load causes an upward jump. - The slope of the shear diagram at a point (dV/dx)
is equal to the (negative) intensity of the
distributed load w(x) at the point. - The change in shear between any two points on a
beam equals the (negative) area under the
distributed loading diagram between the points.
13Relationships Between Load, Shear and Moment
- Moment Diagram
- Application of a clockwise concentrated moment
causes an upward jump in the moment diagram. A
counter-clockwise moment causes a downward jump. - The slope of the moment diagram at a point
(dM/dx) is equal to the intensity of the shear at
the point. - The change in moment between any two points on a
beam equals the area under the shear diagram
between the points.
14Chapter Seven Centroids and Moments of Inertia
of Areas
- Centroid of complex shapes can be calculated
using - AT Y ? (Ai yi ) where
- AT total area of composite shape
- Y distance to centroid of composite shape
from some reference axis - Ai area of one component part of shape
- yi distance to centroid of the component part
from the reference axis - Solve for Y ? (Ai yi ) / AT
- Perform calculation in tabular form
- See Examples 7-1 7-2
15Moment of Inertia ofComposite Shapes
- Perform calculation in tabular form
- Divide the shape into component parts which are
simple shapes - Locate the centroid of each component part, yi
from some reference axis - Calculate the centroid of the composite section,
Y from some reference axis - Compute the moment of inertia of each part with
respect to its own centroidal axis, Ii - Compute the distance, di Y - yi of the
centroid of each part from the overall centroid - Compute the transfer term Ai di2 for each part
- The overall moment of inertia IT , is then
- IT ? (Ii Ai di2)
- See Examples 7-5 through 7-7
16Chapter Eight Stress Due to Bending
- Positive moment compression on top, bent
concave upward - Negative moment compression on bottom, bent
concave downward - Maximum Stress due to bending (Flexure Formula)
- smax M c / I
- Where M bending moment, I moment of inertia,
and c distance from centroidal axis of beam to
outermost fiber - For a non-symmetric section distance to the top
fiber, ct , is different than distance to bottom
fiber cb - stop M ct / I
- sbot M cb / I
17Section Modulus, S
- Maximum Stress due to bending
- smax M c / I
- Both I and c are geometric properties of the
section - Define section modulus, S I / c
- Then smax M c / I M / S
- Units for S in3 , mm3
- Use consistent units
- Example if stress, s, is to be in ksi (kips /
in2 ), moment, M, must be in units of kip
inches - For a non-symmetric section S is different for
the top and the bottom of the section - Stop I / ctop
- Sbot I / cbot
18Chapter Nine Shear Stress in Beams
- The shear stress, ? , at any point within a beams
cross-section can be calculated from the General
Shear Formula - ? VQ / I t, where
- V Vertical shear force
- I Moment of inertia of the entire cross-section
about the centroidal axis - t thickness of the cross-section at the axis
where shear stress is to be calculated - Q Statical moment about the neutral axis of the
area of the cross-section between the axis where
the shear stress is calculated and the top (or
bottom) of the beam - Q is also called the first moment of the area
- Mathematically, Q AP y , where
- AP area of theat part of the cross-section
between the axis where the shear stress is
calculated and the top (or bottom) of the beam - y distance to the centroid of AP from the
overall centroidal axis - Units of Q are length cubed in3, mm3, m3,
19Shear Stress in Common Shapes
- The General Shear Formula can be used to develop
formulas for the maximum shear stress in common
shapes. - Rectangular Cross-section
- ?max 3V / 2A
- Solid Circular Cross-section
- ?max 4V / 3A
- Approximate Value for Thin-Walled Tubular Section
- ?max 2V / A
- Approximate Value for Thin-Webbed Shape
- ?max V / t h
- t thickness of web, h depth of beam
20Chapter Fifteen Pressure Vessels
- If Rm / t 10, pressure vessel is considered
thin-walled - Stress in wall of thin-walled sphere
- s p Dm / 4 t
- Longitudinal stress in wall of thin-walled
cylinder - s p Dm / 4 t
- Longitudinal stress is same as stress in a sphere
- Hoop stress in wall of cylinder
- s p Dm / 2 t
- Hoop stress is twice the magnitude of
longitudinal stress - Hoop stress in the cylinder is also twice the
stress in a sphere of the same diameter carrying
the same pressure