311 ch11

1.

III .)Stress Considerations

A.)Poissons Effect - Uniaxial Load

Poisson's Ratio( ) is a known

material constant that tells you howmuchstrainoccurs in a direction

perpendicularto an applied load.

We already know how to find strains and deformations in the directionparallelto the load axis.

x = x where x =P x/ A x

x = x L x.O

To find the strains and deformations in a directionperpendicularto the loading.

= - y / x .25 (for steel)

.33 (for aluminum)

y= - x

y= - xbecause x= x

y= y L yO

Example:Find the deformation in the x & y-directions of the steel bar showndue to the 20 kip Uniaxial load.

x =P x/ A x=20 k____=8.0 ksi(< pl)

( 2.5)(1.0)

x= x=(8.0 ksi)__=2.67x10 -4in/in

E 30,000 ksi

x = x L x.O = (2.67x10 -4in/in)(12.0)

x= 3.20x10 -3in.(longer)

y= - x= - (0.25)(8.0 ksi)

E 30,000 ksi

y = - 6.67x10 -5in/in

y= y L y.O = (- 6.67x10 -5in/in)(2.5)

y= -1.67x10 -4in.(shorter)

B.)Poissons Effect - Biaxial Loads

The strain in the y-direction now has two parts:

y1= - x is caused by P x

y2= y is caused by P y

y= y- x is the total strain inEE the y-direction dueto biaxial loading

Similarly, The strain in thex-directionis found to be.

x= x- y the total strain in the

EEx-direction due to

biaxial loading

Example:Find the deformation in the x & y-directions of the of the same steel bar if a load in the y-direction is added.

Note: pl= 34 ksi

x =P x/ A x=20 k=8.0 ksi(< pl)

( 2.5)(1.0)

y =P y/ A y=60 k=5.0 ksi(< pl)

( 1.0)(12.0)

x= x - y=(8.0 ksi)-(0.25)(5.0 ksi)

EE 30,000 ksi30,000 ksi

x=2.25x10 -4in/in

x = x L x.O = (2.25x10 -4in/in)(12.0)

x= 2.70x10 -3in.(longer)

( x was 3.20x10 -3in. without y-dir. load)

y= y- x= 5.0 ksi-(0.25)(8.0 ksi)

E E 30,000 ksi30,000 ksi

y = 1.00x10 -4in/in

y= y L y.O = ( 1.00x10 -4in/in)(2.5)

y= 2.50x10 -4in.(longer)

( y was-1.67x10 -4in. before adding the y-load)

The bar got longer and wider

(and thinner)

z= - y- x=- (0.25)(5.0 ksi)-(0.25)(8.0 ksi)

E E 30,000 ksi30,000 ksi

z =- 1.08x10 -4in/in

z= z L z.O = ( -1.08x10 -4in/in)(1.0)

z= -1.08x10 -4in.(thinner)

B.) Thermal Effects

1.) Thermal Expansion

-most materials expand when heated

-the amount ofdeformationcaused

by a temperature change is

proportionalto the change in

temperature.

That is, if raising the temperature 10 o F

lengthens a part by 0.10,

then raising the temperature 20o F will

lengthen the same part by 0.20.

T =L ( T)

T= Change in length caused by a temp change (in)

coefficient of thermalexpansion,in/in

L= Original length of member (in)

T= Change in temperature ( o F)

Example:

Mackinac Bridge is 5 miles long ( + ).

What is the change in length from 3 p.m. when the temp.= 80 o F to midnight when temp.= 50 o F. (A36 Steel)

T = L ( T)

steel 6.5 X 10 -6 in/in(App G)

L= (5mi)(5280 ft/mi)(12in/ft)

=316,800in

T= 50 o F - 80 o F= -30 o F

T =(6.5x10-6in/in/ o F)(316,800in)(-30 o F)

= -61.8 in= -5.15 ft .

2.)Thermal Stress

If you restrain a member when you change its temperature, internal stresses will be produced which are called thermal stresses.

Thermal stress is determined by picturing a member that is expanded when heated and determining the force (P) which would be required to bring in back to its original length.

PL= L( T)

P = T)E

T= - T)E

=Thermal stress in a restrainedmember due to a change intemperature (psi). ( for T PL )

E= Modulus of Elasticity (psi)

C.) Composite Members(SKIP)

Examples:

Concrete slab reinforced with steel

Carbon fiber and plastic snow skis

Kevlar and fiberglass boats

Concrete slab on a metal deck

1.) Relationship Between Load and Stress

We know:

a.) E A= E B because two different

materials.

Example:E A= 30,000 ksi (steel)

E B=3,000 ksi(conc)

b.) A = B- because the two materials

move (deform)together

c.) A= B

A = B - because=

E A E B E

A =E A B = n B, where n = E A

E B E B

We also know:

d.)P = P A+ P Bbecause each material

carries a part of the load

P = A A A+ B A BbecauseP = A

P = (n B )A A+ B A Bbecause A= n B

P = B (nA A+ A B )

P = A (A A+ A B /n)because B= A /n

2.)To find thestressesdue to a known

axial load (P) in a member (A) made of two known materials (E A , E B ).

a.) first find Bfrom:

P = B (nA A+ A B )

b.) then find Afrom:

A =n B

3.)To find theallowable axial load(P) on a member (A A, A B ) made of two known

materials (E A , E B , A,all, B,all ).

a.) find Afrom:

A =n B,all

b.) if A< A,all, then find P all from:

P all= B,all (nA A+ A B )

c.) if A> A,all, then find P all from:

P all= A,all (A A+ A B /n)

D.) Stress Concentration:

1.)At a section with a hole in wepreviouslyassumed the stress isuniformlydistributed across the net section.

A net = A - A hole

2). The actual distribution is notuniform, but is much higher than avg near the hole and less than avgaway from the hole.

t,max= stress next to the hole

t,max=k(P )

k = stress concentration factor

3.)Stress Concentration Factor (k)

k is determined from empirical curves.

(see Fig.11-12, p.285)

k is based on the ratio r/d:

r= the radius of the hole

d = the net width of the member

3.)Stress Concentration Factor (k)

4.) When to consider stress concentration in design.

a.)If a member isductile andsubjected to fairlystatic loads , it is assumed plastic deformation will provide a uniform stress distribution,use avgin design.

b.) If the member is made of abrittle

material or is subjected torepetitive loads , you must consider the higher stress( t,max ) which occurs near the hole.

4.) When to consider stress concentration in design.

Repetitive (cyclical) stress Examples:

- rotational machinery:

- axles, gears, sprockets, drive shafts

- others

- number of cycles:

III.) Stress Considerations (Review)

A.) Poissons Ratio

trans/ long y/ x

1.)Uniaxial Loads

2.)Biaxial Loads

x x- y

B.) Thermal Effects

1.) Thermal Expansion

t= L( T)

2.) Thermal Stresses

t= E( T)

C.) Composite Members -SKIPPED

n = E A/ E B= Modular Ratio

A = n B

P = B(nA A+ A B)

D.) Stress Concentration

t,max = k avg = k(P/A net )

E.) Stresses on Inclined Planes

1.) We are interested in finding the

stresses on planes other than the plane which is normal to the axial load because failure does not always occur on the normal plane.

a.) An axially loadedductilematerialwill usually fail along a plane which isat a 45 degree angle to the load.

b.) An axially loadedbrittlematerialwill usually fail along a plane which isnormal to the load.

If A is the area of plane DB,

and the thickness is constant,

Thenthe area of plane DC is:

A/ cos

The load perpendicular (normal)

to plane DC is:

P n= Pcos

The load parallel (shear)

to plane DC is:

P v= Psin

The stress perpendicular (normal)

to plane DC is:

n=P n__

n = Pcos

n= P(cos

The stress parallel (shear)

to plane DC is:

v=P v__

v = Psin

v= Psin cos

v= (P/2A)sin2

The maximum normal stress is when the plane is inclined at 0 degrees:

n= P(cos P

The minimum normal stress is when the plane is inclined at 90 degrees:

n= P(cos90 0

The maximum shear stress is when the plane is inclined at 45 degrees:

v=P(sin(2x45)) = P

The minimum shear stress is when the plane is inclined at 0 or 90 degrees:

v= P(sin(2x90) = P(sin(2x0) = 0

III.) Stress Considerations (Ch. 11Review)

A.) Poissons Ratio

trans/ long y/ x

1.)Uniaxial Loads

2.)Biaxial Loads

x x- y

B.) Thermal Effects

1.) Thermal Expansion whenmember is freeto move

t= L( T)

2.) Thermal Stresses when memberis restrained

t= E( T)

C.) Stress Concentrations t,max= k avg = kP A

D.) Stresses on Inclined Planes

n= P(cos

n,max=P, when = 0

v= Psin cos

v,max= P/2A , when = 45

311 ch11

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