Transcript Document

MAE 343 - Intermediate Mechanics of
Materials
Thursday, Sep. 9, 2004
Textbook Section 4.4
Transverse Shear Stress in Beams
Torsional Shear Stress
Main Steps of Beam Bending
Analysis
• Step 1 – Find Reactions at External Supports
– Free Body Diagram (FBD) of Entire Beam
– Equations of Force and Moment Equilibrium (3 in 2D)
• Step 2 – Shear and Bending Moment Diagrams
– Cutting Plane and FBD of Part of the Beam
– Use Equilibrium Eqs. to Express Internal Forces in Terms of
Position Variable, “x”
• Step 3 – Stress Distributions at Critical Sections
– Linear Distribution of Bending (Normal) Stresses
– Transverse Shear Stress Distribution in Terms of “Area Moment”
Summary of Stress Analysis Procedure
in Symmetrical Beams
• Find Centroid of CrosssSection
Neutral Axis
(Table 4.2)
• Choose x-y-z Ref. System
Centered at Centroid
• Normal Bending Stress
y
A
 dA
A
– Zero at neutral axis
– Max. at top and/or bottom surface
• Transverse Shear Stress
 ydA
Mzy
x 
I zz
V ( x)
 xy ( y  y1 ) 
yA
I zz Z ( y  y1 )
– Zero at top and bottom edge
– Max. at neutral axis unless crosssection is narrower elsewhere
– Z(y=y1)=width of cross-section
where shear stress is calculated
yA = Moment of area above y=y1
with respect to neutral axis
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
TORSIONAL SHEAR
• Circular Cross-sections (Fig. 4.7)
– Cross-sectional planes remain plane and parallel if shaft is straight,
torque about longitudinal axis, etc.
– Maximum shear stress is always at the outer fibers
4

D
Tr

where J  r 2 dA J 
32
J
– Relationships between power, speed, torque of rotating shaft,
based on fact that 1 hp=33,000ft-lb/min, and 1 kw = 60,000Nm/min: hp=(Tn)/63,025, or kw=(Tn)/9549, where n=shaft speed in
rev/min, T =torque in in-lb or Newton-meter

• Noncircular Cross-sections: Complex solutions because of
“warping” of cross-sectional surfaces (Fig. 4.8)
MEMBRANE ANALOGY for Torsion
Analysis of Noncircular Bars
• Based on similar governing differential equations and
boundary conditions for:
• the contoured deflection surface of a pressurized membrane and the
• distribution of torsional shearing stresses in a twisted bar
• Analytically and Experimentally Verified Observations:
– Tangent to contour line coincides with direction of torsional shear
at the corresponding point of cross-section
– Max. slope of membrane at any point is proportional to
corresponding magnitude of shear stress
– Volume enclosed by datum base-plane and contoured surface is
proportional to the torque applied on the twisted bar
• Critical points of maximum shear identified by
visualizing deformed membrane (Fig. 4.9)
T
 max  , Q, in 3 , Table 4.4
Q
TL

, K , in 4 , Table4.4
KG
Bending of Usymmetric Beams
Shear Center
• Torsion occurs if the plane of applied transverse
force is not a plane of symmetry
• Shear Center (Center of Twist) of Cross-Section
–
–
–
–
–
No torsion if plane of transverse forces passes through it
If one axis of symmetry, it lies on it, but not at centroid
If two axes of symmetry, it lies at their intersection point
If there is a point of symmetry, it lies at the centroid
See Table 4.5 for Shear Center location in cross-sections
MAE 343-Intermediate Mechanics of Materials
Homework No.3 - Thursday, Sep. 9, 2004
• Textbook problems required on
Thursday, Sep. 16, 2004:
– Problems 4.17 and 4.19
• Textbook problems recommended for
practice before Sep. 16, 2004:
– Problems 4.16, 4.18 and 4.20