Dr. D. B. Wallace

Basic Stress Equations

Internal Reactions:

6 Maximum(3 Force Components & 3 Moment Components)

Normal Force

)ττ(

)σσ(

Shear Forces

z

x

y

PVy

Vx

Torsional Moment)ττ(

)σσ(Bending Moments

z

x

y

TMy

Mx

or TorqueForce Components Moment Components

"Cut Surface" "Cut Surface"

Centroid ofCross Section

Centroid ofCross Section

Normal Force:

Axial Force

z

x

y

P

Centroidσσ

Axial Stress

"Cut Surface" σ =PA

l Uniform over the entire cross section.l Axial force must go through centroid.

Shear Forces:Cross Section

y

Aa

Point of interestLINE perpendicular to V through point of interest

= Length of LINE on the cross section

= Area on one side of the LINE

Centroid of entire cross section

Centroid of area on one side of the LINE

= distance between the two centroids

= Area moment of inertia of entire cross sectionabout an axis pependicular to V.

V

b

Aa

y

I

"y" Shear Force

z

x

y

Vy

"x" Shear Force

z

x

y

Vx

ττ

ττ

τ =⋅ ⋅

⋅V A y

I bab g

Note: The maximum shear stress for common cross sections are:

Cross Section: Cross Section:

Rectangular:

τmax = ⋅3 2 V A Solid Circular: τmax = ⋅4 3 V A

I-Beam or H-Beam: webflange τmax = V Aweb

Thin-walledtube: τmax = ⋅2 V A

Basic Stress Equations Dr. D. B. Wallace

Torque or Torsional Moment:

Solid Circular or Tubular Cross Section:

r = Distance from shaft axis to point of interestR = Shaft RadiusD = Shaft Diameter

JD R

JD D

for solid circular shafts

for hollow shaftso i

=⋅

=⋅

=⋅ −

π π

π

4 4

4 4

32 2

32

e jTorque

z

x

y

T

"Cut Surface"

ττ

τ =⋅T r

J

τπ

τπ

max

max

=⋅

⋅

=⋅ ⋅

⋅ −

16

16

3

4 4

T

DT D

D D

for solid circular shafts

for hollow shaftso

o ie j

Rectangular Cross Section:

Torque

z

x

y

T

Centroid

ττ

Torsional Stress

"Cut Surface" ττ1

2

a

b

Note:a > b

Cross Section:

Method 1:

τ τmax .= = ⋅ ⋅ + ⋅ ⋅12 23 1 8T a b a bb g e j ONLY applies to the center of the longest side

Method 2:

τα

1 21 2

2,,

=⋅ ⋅T

a b

a/b αα1 αα21.0 .208 .2081.5 .231 .2692.0 .246 .3093.0 .267 .3554.0 .282 .3786.0 .299 .4028.0 .307 .414

10.0 .313 .421∞ .333 ----

Use the appropriate αα from the table on the right to get the shear stress at either position 1 or 2.

Other Cross Sections:

Treated in advanced courses.

2

Basic Stress Equations Dr. D. B. Wallace

Bending Moment

"x" Bending Moment

z

x

y

z

x

y

Mx

σσ

σσ

My

"y" Bending Moment

σ σ=⋅

=⋅M y

Iand

M x

Ix

x

y

y

where: Mx and My are moments about indicated axes

y and x are perpendicular from indicated axes

Ix and Iy are moments of inertia about indicated axes

Moments of Inertia:

h

c

b D

c RIb h

h

ZI

c

b h

is perpendicular to axis=⋅

= =⋅

3

2

12

6

ID R

ZI

c

D R

=⋅

=⋅

= =⋅

=⋅

π π

π π

4 4

3 3

64 4

32 4

Parallel Axis Theorem:I = Moment of inertia about new axis

I I A d= + ⋅ 2

centroid

d

new axisArea, A I = Moment of inertia about the centroidal axis

A = Area of the regiond = perpendicular distance between the two axes.

Maximum Bending Stress Equations:

σπ

max =⋅

⋅32

3

M

DSolid Circularb g σmax =

⋅⋅

62

M

b hRectangulara fσmax =

⋅=

M c

I

M

Z

The section modulus, Z, can be found in many tables of properties of common cross sections (i.e., I-beams,channels, angle iron, etc.).

Bending Stress Equation Based on Known Radius of Curvature of Bend, ρρ.

The beam is assumed to be initially straight. The applied moment, M, causes the beam to assume a radius ofcurvature, ρρ.

Before:

After:

M Mρρ

σρ

= ⋅Ey

E = Modulus of elasticity of the beam material

y = Perpendicular distance from the centroidal axis to thepoint of interest (same y as with bending of astraight beam with Mx).

ρρ = radius of curvature to centroid of cross section

3

Basic Stress Equations Dr. D. B. Wallace

Bending Moment in Curved Beam:

Geometry:

r

r

e

rrnσσ

σσo

i

centroidcentroidal

neutral axis

axis

o

i

ynonlinear

stressdistribution

M

c

ci

o

ρρ

rA

dA

e r r

n

area

n

=

= −

z ρ

A = cross sectional area

rn = radius to neutral axis

r = radius to centroidal axis

e = eccentricity

Stresses:

Any Position: Inside (maximum magnitude): Outside:

σ =− ⋅

⋅ ⋅ +M y

e A r ynb g σii

i

M c

e A r=

⋅⋅ ⋅

σoo

o

M c

e A r=

− ⋅⋅ ⋅

Area Properties for Various Cross Sections:

Cross Section rdA

area ρz A

t

hriro

rRectangle

ρ rh

i +2 t

r

ro

i⋅

FHG

IKJln h t⋅

t

hriro

rTrapezoid

ρot i

rh t t

t tii o

i o+

⋅ + ⋅⋅ +

2

3

b gb g

For triangle:set ti or to to 0

t tr t r t

h

r

ro io i i o o

i− +

⋅ − ⋅⋅

FHG

IKJln h

t ti o⋅+2

rHollow Circle

ρa

b

r 2 2 2 2 2⋅ − − −LNM

OQPπ r b r a π ⋅ −a b2 2e j

4

Basic Stress Equations Dr. D. B. Wallace

Bending Moment in Curved Beam (Inside/Outside Stresses):

Stresses for the inside and outside fibers of a curved beam in pure bending can beapproximated from the straight beam equation as modified by an appropriatecurvature factor as determined from the graph below [i refers to the inside, and orefers to the outside]. The curvature factor magnitude depends on the amount ofcurvature (determined by the ratio r/c) and the cross section shape. r is the radiusof curvature of the beam centroidal axis, and c is the distance from the centroidalaxis to the inside fiber.

M

M

CentroidalAxis

r

c

Inside Fiber: σi iKM c

I= ⋅

⋅

Outside Fiber: σo oKM c

I= ⋅

⋅

0

1.0

0.5

1.5

2.0

2.5

3.0

3.5

4.0

1 2 3 4 5 6 7 8 9 10 11

K

i

o

CurvatureFactor

Amount of curvature, r/c

K r

b

c

b

c

b

c

c

b/2

b/3

b/4b/8

b/6B A B A

B A

B A

B A

B A B A

Values of K for inside fiber as at i A

Values of K for outside fiber as at o B

U or T

Round or Elliptical

Trapezoidal

I or hollow rectangular

Round, Elliptical or Trapezoidal

U or TI or hollow rectangular

5