ME 270 Class Notes - Part 3

Shear-Force & Bending-Moment Diagrams

Distributed Loads

Learning Objectives

1) To evaluate the shear-force and bending-moment diagrams for systems

with discrete loads.

2) To do an engineering estimate of these quantities.

Beam Sign Convention

Distributed load - An upward load is positive

Shear Force - A positive internal shear force causes a clockwise

rotation of beam segment. (i.e., it pushes a left-

facing cross-section upward or a right-facing cross-

section downward).

Bending Moment - A positive internal moment causes compression in

the top fibers of the segment (i.e., clockwise on a

left-facing cross-section or counter-clockwise on a

right-facing cross-section).

Procedure

1. Determine support reactions

2. Specify beam sections origin (left end) to between each discrete load

(force or moment). Be sure V and M are shown acting in the positive

sense.

3. Sum forces vertically to determine V

4. Sum moments at sectioned end to determine M. (This eliminates V

from the moment equation).

Shear-Force & Bending-Moment Diagrams

Graphical Methods

Learning Objectives

1) To evaluate the shear-force and bending-moment diagrams for systems

with discrete loads.

2) To do an engineering estimate of these quantities.

Beam Sign Convention

Distributed load - An upward load is positive

Shear Force - A positive internal shear force causes a clockwise

rotation of beam segment. (i.e., it pushes a left-

facing cross-section upward or a right-facing cross-

section downward).

Bending Moment - A positive internal moment causes compression in

the top fibers of the segment (i.e., clockwise on a

left-facing cross-section or counter-clockwise on a

right-facing cross-section).

Procedure

1. Determine support reactions

2. Specify beam sections origin (left end) to between each discrete load

(force or moment). Be sure V and M are shown acting in the positive

sense.

3. Sum forces vertically to determine V

4. Sum moments at sectioned end to determine M. (This eliminates V

from the moment equation).

Normal Stress and Axial Strains

Normal Stress - The stress acting normal to a

surface along an axis of a member.

= Normal Stress = 0

lim = A

F dF

A dA

AVGF= , dA = AA

y z

AVG

F Axial Load = =

A Cross-Sectional Area

where the units are given by:

AVG 2 2

lbs N = = psi or = Pa

in m

Note:

AVG AVG > 0 (Tension) ; < 0 (Compression)

Axial Strain ( ) – The elongation (or shortening) of a part

along an axis.

0

- xAxial Strain = lim

x

x

x

L* - L L*Avg. Axial Strain = = - 1

L L

L* - 1 ,

L

εx =

(εx )AVG =

εave =

Hooke’s Law

Hooke’s Law – The nearly linear relationship between

normal stress and axial strain in the proportional region

(typically under low strains)

where

E = Modulus of Elasticity or Young’s Modulus, which is a

measure of the stiffness of a material (E has units of psi or

Pa).

Elongation is the axial direction causes contraction in

transverse directions.

where

= Poisson's ratio (dimensionless)

= Eε

εy = εx = - εx

Stress-Strain Regions:

Proportional Region (A-B): Nearly linear relationship between

σ and ε, whose slope is E. Hooke’s

Law is valid up to the proportional

limit (σPL).

Elastic Region (A-C): Similar to the proportional region

except between B and C the

material is still behaving elastically

up to the upper yield point (σγP)u,

but is not linear (Hooke’s Law is

not valid).

Yielding Regions (C-E): Plastic deformation begins. A

slight material “softening” occurs

from C-D, down to the lower yield

point (σγP)l. Perfectly plastic

behavior occurs from D-E as the

material strain increases without an

increase in stress.

Strain Hardening Region (E-F): Strain hardening occurs from E-

F up to the “ultimate stress”

(σU).

Necking Region (F-G): Above σU, the material “necks

down” resulting in a significant

reduction in cross-sectional area

and ultimately in fracture (G) at the

fracture stress (σF).

General Stress-Strain Curves

for Various Materials

Design for Allowable Stress

In Axially-Loaded Members

Structural Design Strategies

Design for strength

Design for stiffness (avoidance of large deformations)

Design for ductility (avoidance of fracture)

Design for strength (two possibilities)

Yield strength (σYP)

Ultimate Strength (σU)

Factor of Safety (FS)

To avoid failure FS > 1

Typical FS values 1.3 < FS < 3

Failure LoadFS =

Allowable Load

For yield failure,

For ultimate failure,

allow

FS = YP

YPallow =

FS

allow

FS = U

Uallow =

FS

Shear Stress Due to Pure Torsion

in Circular Members

Local Shear Stress at radius ( )

where T = applied Torque

ρ = radius at some location in the shaft

J = polar area moment of inertia of the shaft

cross-section

The maximum shear stress occurs at the outer perimeter

of the shaft when (ρ = r)

Two Common Circular Configurations:

Solid Shaft

Tubular Shaft

T( )

J( )

MAX

Tr

J

Stress Distribution in a Solid Shaft

where r = radius of shaft

and J = 4π

r2

4

MAX 3

T(ρ) 2(ρ) = = (ρ)

Tr 2T = =

J πr

T

J r

Stress Distribution in a Tubular Shaft

where ro = outer radius of tubular shaft

ri = inner radius of tubular shaft

J = 4 4π

(r - r )2

o i

4 4

MAX 4 4

T(ρ) 2(ρ) = = (ρ)

- r

Tr 2Tr = =

J π - r

o i

o o

o i

T

J r

r

Power Transmission

where P = Power (work performed per unit of time)

T = Torque

W = Angular speed of shaft

Power often measured in horsepower (hp)

Power can also be expressed as a function of frequency

(f)

where f = frequency (in cycles/sec or Hz)

Note: ω = 2πf

P = T

ft-lb1 hp = 550

s

P = 2 f T

Shaft Design

When the Power (p) and frequency (f) of a shaft are

known, the torque developed can be determined

Knowing the torque (T) and the allowable shear stress

(τAllow), the size of the shaft’s cross-section can be

determined (assuming small strains in the linear elastic

range).

A common geometric (design) parameter is

Remember, J = 4π

r 2

Solid Shaft

J = 4 4

o i

π r - r

2 Tubular Shaft

T = P / 2 f

Allow

T =

τ

J

r

Stresses Due to Pure Bending

Pure Bending

Sign Convention

Pure Bending in a Beam

When V = 0 and M ≠0 => Pure Bending

Below Pure Bending occurs between B and C.

Stress and Strain Distributions Due to Pure Bending

x x

1 = ε (y) = (y)ε

x x x

E = (y) = E(ε ) = (y)

Stress Distribution

where M = resultant internal moment about the neutral

axis of the cross-section

y = perpendicular distance from the neutral axis

I = second area moment of inertia of the cross-

sectional area about the neutral axis

E = Young’s modulus of elasticity

Second Area Moment of Inertia

x x

- M(y) = ( ) =

Iy

A

2I = y dA

xx x

- M(y) = (y) = =

E E Iε

Maximum Normal Stress

1. Only normal stress σx exist at the cut.

2. Neutral axis passes through centroid.

3. Normal stresses vary linearly in the y-direction.

4. Normal stresses are constant in the z-direction.

5. Normal stress is zero at neutral axis.

6. Max normal stress exists at most outer surface of

beam.

MAXx MAX

M y =

I

Second Area Moment of Inertia By Integration

Recall

Similar to centroids by integration except there is a y2

(rather than a y) term in the integrand.

2

xA

2

yA

2

o x yA

I = y dA

I = x dA

J = r dA = I + I

A

2I = y dA

Stresses Due to Combined

Transverse Shear and Bending

Previously, we considered pure shear and pure bending

separately.

For combined transverse shear and bending

where: V = shear force at cross-section

A* = cross-sectional area above the element

y* = centroid of the area above the element

I = centroidol second area moment for the entire

cross-section

t = depth of the beam at the location of the

stress element of interest

For a rectangular cross-section beam of height (h) and

width (t), (remember 31

I = th12

),

V A y =

I t

Combining equations gives,

hA* = ( - y) t ,

2

1 hy* = ( +y) .

2 2

3

h 1 hV - y t ( + y)

2 2 2

th t

12

=

22

2

6V h  = - y

Ah 4

Note:

1. τ distribution is quadratic

2. τ = 0 at y = ± h/2

3. τ = max at the neutral axis (y = 0)

MAX

3V =

2A

where:

V = shear force at cross-section

A = cross-sectional area