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 Crosss-Section Neutral Axis (Table 4.2)

• Choose x-y-z Ref. System Centered at Centroid

• Normal Bending Stress– Zero at neutral axis– Max. at top and/or bottom surface

• Transverse Shear Stress– Zero at top and bottom edge– Max. at neutral axis unless cross-

section is narrower elsewhere– Z(y=y1)=width of cross-section where

shear stress is calculated

A

A

dA

ydA

y

zz

zx I

yM

AyyyZI

xVyy

zzxy )(

)()(

11

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

– Relationships between power, speed, torque of rotating shaft, based on fact that 1 hp=33,000ft-lb/min, and 1 kw = 60,000N-m/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)

dArJwhereJ

Tr 232

4DJ

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)

4.4,,,

4.4,,,

4

3max

TableinKKG

TL

TableinQQ

T

Bending of Usymmetric BeamsShear 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