MAE 343 - Intermediate Mechanics of
Materials
Tuesday, Sep. 14, 2004
Textbook Section 4.4
Overview of Structural Analysis for
Beams and Shafts
Overview of Loads ON and IN
Structures / Machines
Structural Analysis
Applied LoadsForces &Moments
Body forces &couples
Reaction Forces
& Moments (at supports)
From FBD of
entire structure
Surface forces&couples
Flow Lines from
Applied to Reaction
Forces
Resultants on
Cutting Surface
From FBD of
Part of Structure
Equilibrium Eqs.
(3 in 2D, 6 in 3D)
Concentrated
FBD of Each
Component from
Forces through
Contact Surfaces
Distributed /
Pressures
Internal Forces
& Moments
Depend on Location&
Orientation of
Cutting Plane
Equilibrium Eqs.
(3 in 2D, 6 in 3D)
Shear Force &
Bending Moment
Diagrams
Overview of Various Stress Patterns
Stresses- from Distribution of
Internal Forces over Given
Cross-Section
Uniaxial Tension or
Compression of
Straight Bars or
Lap Joints
Pure Bending
of Long Beams
Beams Subjected
to Transverse Forces
Torsion of Round
or Prismatic Shafts
Uniform Distributions
of Stresses
Neutral Axis at
Centroid of Cross-section
Transverse Shear Stresses
Result (Add-up) in
Shear Force in the
Cross-Section
Moment of differential
torsional stress about
centroid results in the
internal torque in cross-section
Normal Stress-Straight Bars
Sigma=F/A
Linear Distribution
over from Neutral Axis
Sigma=(My)/I
(Normal Axial Stresses)
Distribution from
Neutral Axis Depends on
Shape of Cross-section
Tau(y)=(V*MomentArea)/(I*Z(y))
Circular Cross-section:
Linear distribution, with
Max. at the outer edge:
Tau=(Tr)/J
Maximum may not be
at neutral axis
Tau_max=(coef.)Tau_average
Prismatic Shafts -Torsion
Leads to Warping
Membrane Analogy:
Tau_max=T/Q, Theta=(TL)/KG
Direct Shear in Lap Joints
Tau =P/A
The Resultant of
Bending Stresses
is the Bending Moment
in the Cross-section
Maximum at top
or bottom, and zero
at neutral axis
Zero at top
and bottom edges
Bending of Asymmetric Beams:
IF Plane of Transverse Forces
passes through the SHEAR
CENTER, NO Torsion occurs
Power-Torque-RPM Relations:
hp=(Tn)/63,025
kw=(Tn)/9549
Solution of Example Problem 4.3
• Step 1 –Construct shear&moment diagrams
• Step 2 –Find dmin=0.90in to resist
allowable=35,000psi at section where Mmax=2500in-lb
• Step 3 –Find dmin=0.56in to resist
allowable=20,200psi for maximum direct average
shear stress
• Step 4 –Find dmin=0.65in to resist
allowable=20,000psi for maximum transverse shear
stress
Example 4.2- Calculation of
Transverse Shear Stresses
• Hollow rectangular cross-section (Channel)
– Moment of Inertia about neutral axis, Izz=8.42 in4
• Max. transverse shear stress at neutral axis
– Direct application of “area moment” method
• xy(max)=0.34V, xy(ave)=0.2V, so that
• xy(max) =1.7 xy(ave)=1.7(F/A)….Table 4.3
– Divide irregular section into several regular parts
• Transverse shearing stress distribution, xy(y)
Summary of Solutions to Textbook
Problems – Problem 4.10
• Module support D6AC steel beam with two small hole
and cracks
– Ignore for now stress concentrations&fracture mechanics
– Beam subjected to four-point bending
• Uniform bending moment at mid-span, M=PL/3 =281.25 kN-m
• No transverse shear stress in central span, V =0
• Check for possible failure by yielding
– Max. bending stress at:
• The tip of bottom crack, yct =10.3cm, x(ct)=(M*yct)/Izz=556 MPa
• At the outer fibers: x(max)=(281.25x103)(12.5x10-2)/5.21x10-5=675 Mpa
– From Table 2.1, Syp=1570MPa, so that the “safety factor” is equal
to 1570/675=2.3  NO yielding
Summary of Solutions to Textbook
Problems – Problem 4.15
• Short tubular cantilever bracket, AISI 1020 steel (CD)
– Critical points at the wall:
• Top and bottom fibers for bending
• At neutral axis for transverse shearing stress
• Calculate maximum stresses
– Axial bending stress: x=48,735 psi
– Transverse shear: use Table 4.3 to obtain yz(ts)=2(F/A)=48,890psi
• Yielding failure mode for uniaxial tensile stress
– No failure since Table 3.3 shows Sy=51,000psi>48,735 psi
• Multiaxial failure theory is necessary for yz(ts)
Example 4.5- Calculation of Stresses in
Channel-Section Cantilever Beam
• Determine maximum bending stress
– z(max)=48000(2.0)/4.74 =20,250 psi, Ixx=4.74in4
• Max. transverse shear at neutral axis
– Area moment methods yields: zy(max)= 7720 psi
• Flow of torsional shearing stresses
– Locate SHEAR CENTER by using Table 4.5: e=0.56 in
– Find torsional moment since the plane of “P” is located at a
distance a=1.26 in. from the shear center: T=Pa =10,080 in-lb
– Find horizontal shear forces that form resisting couple, T = Rd
– Maximum torsional shearing stresses in the flanges:
• zx(max)= R/Af= 2739/((1.72)(0.32)) = 4976 psi
• Could be eliminated by translating the plane of “P” by 1.26in to the left