Chapter Outline

Lecture Slides

Chapter 3

Load and Stress Analysis

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Free-Body Diagram Example 3–1 Equilibrium

 A system that is motionless, or has constant velocity, is in

equilibrium.
 The sum of all force vectors and the sum of all moment vectors
acting on a system in equilibrium is zero.

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 Free-Body Diagram Example 3–1 Free-Body Diagram Example 3–1

Fig. 3–1

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Free-Body Diagram Example 3–1 Free-Body Diagram Example 3–1

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 Shear Force and Bending Moments in Beams Sign Conventions for Bending and Shear

 Cut beam at any location x1

 Internal shear force V and bending moment M must ensure
equilibrium

Fig. 3−2

Fig. 3–3
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Distributed Load on Beam Relationships between Load, Shear, and Bending

 Distributed load q(x) called load intensity

 Units of force per unit length

 The change in shear force from A to B is equal to the area of the

Fig. 3–4 loading diagram between xA and xB.
 The change in moment from A to B is equal to the area of the
shear-force diagram between xA and xB.

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EXAMPLE 3–2
Derive the loading, Shear-Moment Diagrams Example 3–2 (continued)
shear-force, and
bending-moment
relations for the beam
of Fig. 3–5a.

Figure 3–5
(a) Loading diagram for
a simply-supported
beam.
(b) Shear-force diagram.
(c) Bending-moment
diagram.

Fig. 3–5
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Moment Diagrams – Two Planes Combining Moments from Two Planes

 Add moments from two

planes as perpendicular
vectors


Fig. 3–24
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Fig. 3–24
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 Stress Stress element

 Normal stress is normal to a surface, designated by 

 Tangential shear stress is tangent to a surface, designated by 
 Normal stress acting outward on surface is tensile stress
 Normal stress acting inward on surface is compressive stress
 U.S. Customary units of stress are pounds per square inch (psi)
 SI units of stress are newtons per square meter (N/m2)
 1 N/m2 = 1 pascal (Pa)

 Represents stress at a point

 Coordinate directions are arbitrary
 Choosing coordinates which result in zero shear stress will
produce principal stresses

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Cartesian Stress Components Cartesian Stress Components

 Defined by three mutually orthogonal surfaces at a point within  Defined by three mutually orthogonal surfaces at a point within
a body a body
 Each surface can have normal and shear stress  Each surface can have normal and shear stress
 Shear stress is often resolved into perpendicular components  Shear stress is often resolved into perpendicular components
 First subscript indicates direction of surface normal  First subscript indicates direction of surface normal
 Second subscript indicates direction of shear stress  Second subscript indicates direction of shear stress

Fig. 3−8 (a) Fig. 3−7

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 Cartesian Stress Components Plane-Stress Transformation Equations

 In most cases, “cross shears” are equal  Cutting plane stress element at an arbitrary angle and balancing
stresses gives plane-stress transformation equations
 Plane stress occurs when stresses on one surface are zero

Fig. 3−8
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Fig. 3−9
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Principal Stresses for Plane Stress Extreme-value Shear Stresses for Plane Stress
 Differentiating Eq. (3–8) with respect to  and setting equal to  Performing similar procedure with shear stress in Eq. (3–9), the
zero maximizes  and gives maximum shear stresses are found to be on surfaces that are
±45º from the principal directions.
 The two extreme-value shear stresses are
 The two values of 2p are the principal directions.
 The stresses in the principal directions are the principal stresses.
 The principal direction surfaces have zero shear stresses.
 Substituting Eq. (3–10) into Eq. (3–8) gives expression for the
non-zero principal stresses.

 Note that there is a third principal stress, equal to zero for plane
stress.
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 Maximum Shear Stress Elastic Strain

 There are always three principal stresses. One is zero for plane  Hooke’s law
stress.
 There are always three extreme-value shear stresses.

 E is Young’s modulus, or modulus of elasticity

 Tension in on direction produces negative strain (contraction) in
 The maximum shear stress is always the greatest of these three. a perpendicular direction.
 Eq. (3–14) will not give the maximum shear stress in cases  For axial stress in x direction,
where there are two non-zero principal stresses that are both
positive or both negative.
 If principal stresses are ordered so that 1 > 2 > 3,
 The constant of proportionality n is Poisson’s ratio
then max = 1/3
 See Table A–5 for values for common materials.

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Elastic Strain Elastic Strain

 For a stress element undergoing x, y, and z, simultaneously,  Hooke’s law for shear:

 Shear strain  is the change in a right angle of a stress element

when subjected to pure shear stress.
 G is the shear modulus of elasticity or modulus of rigidity.
 For a linear, isotropic, homogeneous material,

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 Uniformly Distributed Stresses Normal Stresses for Beams in Bending

 Uniformly distributed stress distribution is often assumed for  Straight beam in positive bending
pure tension, pure compression, or pure shear.  x axis is neutral axis
 For tension and compression,  xz plane is neutral plane
 Neutral axis is coincident with the
centroidal axis of the cross section
 For direct shear (no bending present),
Fig. 3−13

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Normal Stresses for Beams in Bending Normal Stresses for Beams in Bending

 Bending stress varies linearly with distance from neutral axis, y  Maximum bending stress is where y is greatest.

 I is the second-area moment about the z axis

 c is the magnitude of the greatest y
 Z = I/c is the section modulus

Fig. 3−14
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 Assumptions for Normal Bending Stress Example 3–5

 Pure bending (though effects of axial, torsional, and shear

loads are often assumed to have minimal effect on bending
stress)
 Material is isotropic and homogeneous
 Material obeys Hooke’s law
 Beam is initially straight with constant cross section
 Beam has axis of symmetry in the plane of bending
 Proportions are such that failure is by bending rather than
crushing, wrinkling, or sidewise buckling
 Plane cross sections remain plane during bending

Fig. 3−15
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Dimensions in mm
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Example 3–5 (continued) Example 3–5 (continued)

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 Example 3–5 (continued) Example 3–5 (continued)

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Two-Plane Bending Example 3–6

 Consider bending in both xy and xz planes

 Cross sections with one or two planes of symmetry only

 For solid circular cross section, the maximum bending stress is

Fig. 3−16
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 Example 3–6 (continued) Example 3–6 (continued)

Fig. 3−16
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Example 3–6 (continued) Shear Stresses for Beams in Bending

Fig. 3−17

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 Transverse Shear Stress Transverse Shear Stress in a Rectangular Beam

Fig. 3−18

 Transverse shear stress is always accompanied with bending

stress.
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Maximum Values of Transverse Shear Stress Significance of Transverse Shear Compared to Bending
 Example: Cantilever beam, rectangular cross section
 Maximum shear stress, including bending stress (My/I) and
transverse shear stress (VQ/Ib),

Table 3−2

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 Significance of Transverse Shear Compared to Bending Example 3–7
 Critical stress element (largest max) will always be either
◦ Due to bending, on the outer surface (y/c=1), where the transverse
shear is zero
◦ Or due to transverse shear at the neutral axis (y/c=0), where the
bending is zero
 Transition happens at some critical value of L/h
 Valid for any cross section that does not increase in width farther away
from the neutral axis.
◦ Includes round and rectangular solids, but not I beams and channels

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Fig. 3−20
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Example 3–7 (continued) Example 3–7 (continued)

Fig. 3−20(c)
Fig. 3−20(b)
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 Example 3–7 (continued) Example 3–7 (continued)

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Example 3–7 (continued) Example 3–7 (continued)

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 Torsion Torsional Shear Stress

 Torque vector – a moment vector collinear with axis of a  For round bar in torsion, torsional shear stress is proportional to
mechanical element the radius 
 A bar subjected to a torque vector is said to be in torsion
 Angle of twist, in radians, for a solid round bar
 Maximum torsional shear stress is at the outer surface

Fig. 3−21
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Assumptions for Torsion Equations Torsional Shear in Rectangular Section

 Equations (3–35) to (3–37) are only applicable for the  Shear stress does not vary linearly with radial distance for
following conditions rectangular cross section
◦ Pure torque  Shear stress is zero at the corners
◦ Remote from any discontinuities or point of application of  Maximum shear stress is at the middle of the longest side
torque  For rectangular b x c bar, where b is longest side
◦ Material obeys Hooke’s law
◦ Adjacent cross sections originally plane and parallel remain
plane and parallel
◦ Radial lines remain straight
 Depends on axisymmetry, so does not hold true for
noncircular cross sections
 Consequently, only applicable for round cross sections

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 Power, Speed, and Torque Power, Speed, and Torque

 Power equals torque times speed  In U.S. Customary units, with unit conversion built in

 A convenient conversion with speed in rpm

where H = power, W
n = angular velocity, revolutions per minute

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Example 3–8 Example 3–8 (continued)

Fig. 3−22 Fig. 3−23

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 Example 3–8 (continued) Example 3–8 (continued)

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Example 3–8 (continued) Example 3–8 (continued)

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 Example 3–8 (continued) Example 3–9

Fig. 3−24
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Example 3–9 (continued) Example 3–9 (continued)

Fig. 3−24

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Fig. 3−24
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 Example 3–9 (continued) Example 3–9 (continued)

Fig. 3−24
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Example 3–9 (continued) Stress Concentration

 Localized increase of stress near discontinuities

 Kt is Theoretical (Geometric) Stress Concentration Factor

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 Theoretical Stress Concentration Factor Stress Concentration for Static and Ductile Conditions

 Graphs available for  With static loads and ductile materials

standard configurations ◦ Highest stressed fibers yield (cold work)
 See Appendix A–15 and ◦ Load is shared with next fibers
A–16 for common ◦ Cold working is localized
examples
◦ Overall part does not see damage unless ultimate strength is
 Many more in Peterson’s
exceeded
Stress-Concentration Fig. A–15–1
Factors ◦ Stress concentration effect is commonly ignored for static
loads on ductile materials
 Note the trend for higher
Kt at sharper discontinuity
radius, and at greater
disruption

Fig. A–15–9
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Techniques to Reduce Stress Concentration Example 3–13

 Increase radius
 Reduce disruption
 Allow “dead zones” to shape flowlines more gradually

Fig. 3−30

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 Example 3–13 (continued) Example 3–13 (continued)

Fig. A−15−1
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Example 3–13 (continued)

Fig. A−15−5
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