CTC / MTC 222Strength of Materials

Final Review

Final Exam

• Wednesday, December 12, 10:15 -12:15• 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

Course 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.

Chapter 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• Poisson’s ratio, modulus of elasticity in

tension and modulus of elasticity in shear

Direct Stresses

• Direct Normal Stress , • σ = 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

Direct Stresses

• Bearing Stress, σb • σb = 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

Chapter Two – Design Properties

• Basic Definitions• Yield point, ultimate strength,

proportional limit, and elastic limit• Modulus of elasticity and how it

relates strain to stress• Hooke’s Law• Ductility - ductile material, brittle

material

Chapter 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

Chapter Four – Axial Deformation and Thermal Stress

• Axial strain ε, • ε = δ / L , where δ = total deformation, and L =

original length• Axial deformation, δ

• δ = F L / A E

• If unrestrained, thermal expansion will occur due to temperature change

• δ = α x L x ∆T

• If restrained, deformation due to temperature change will be prevented, and stress will be developed

• σ = E α (∆T)

Chapter Five – Torsional Shear Stress and Deformation

• For a circular member, τmax = Tc / J• T = applied torque, c = radius of cross section, J = polar

moment of inertia

• Polar moment of Inertia, J• Solid circular section, J = π D4 / 32• Hollow circular section, J = π (Do

4 - Di4 ) / 32

• Expression can be simplified by defining the polar section modulus, Zp= J / c, where c = r = D/2

• Solid circular section, Zp = π D3 / 16• Hollow circularsection, Zp = π (Do

4 - Di4 ) / (16Do)

• Then, τmax = T / Zp

Chapter Six – Shear Forces and Bending Moments in Beams

• Sign Convention• Positive Moment M • Bends segment concave upward

compression on top

Relationships 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.

Relationships 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.

Chapter Seven – Centroids and Moments of Inertia of Areas

• Centroid of complex shapes can be calculated using:• AT ̅Y̅ = ∑ (Ai yi ) where:

• AT = total area of composite shape• ̅̅Y̅ = 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 ̅Y̅ = ∑ (Ai yi ) / AT• Perform calculation in tabular form• See Examples 7-1 & 7-2

Moment 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, ̅Y̅ from

some reference axis• Compute the moment of inertia of each part with respect to

its own centroidal axis, Ii

• Compute the distance, di = ̅Y̅ - 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 di

2)

• See Examples 7-5 through 7-7

Chapter 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)• σmax = 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

• σtop = M ct / I • σbot = M cb / I

Section Modulus, S

• Maximum Stress due to bending • σmax = M c / I • Both I and c are geometric properties of the section

• Define section modulus, S = I / c

• Then σmax = M c / I = M / S• Units for S – in3 , mm3

• Use consistent units• Example: if stress, σ, 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

Chapter 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 ̅y̅ , 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

• ̅y̅ = distance to the centroid of AP from the overall centroidal axis• Units of Q are length cubed; in3, mm3, m3,

Shear 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

Chapter Fifteen – Pressure Vessels

• If Rm / t ≥ 10, pressure vessel is considered thin-walled

• Stress in wall of thin-walled sphere• σ = p Dm / 4 t

• Longitudinal stress in wall of thin-walled cylinder• σ = p Dm / 4 t • Longitudinal stress is same as stress in a sphere

• Hoop stress in wall of cylinder• σ = 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