311ch9

1.

I)Stresses and Strains

In statics we assumedrigidbodies

In strength of materials, we acknowledge that bodies aredeformable , not rigid.

We will study thestressesapplied forces produce in a body and the accompanyingstrains .

A.)Axial Tensile and Compressive

Stresses

Consider a 2 x 4 piece of wood with

force P applied at each end.

Anywhere you cut this bar across its section, in order to satisfyequilibrium , the 800 lb force must act on that section.

F x= 0 = - 800 lb + P A= 0

P A = 800 lb

We assume that the force is distributed evenly throughout the section so that an equal portion of the 800 lb force acts on each square inch of the cross-section

Since we have 8 square makes, the amount of force on each square inch is:

800 lb = 100lb= 100 psi

8in 2 in 2

Which is the definition ofstress :

= stress = unit stress

= average stress

= engineering stress

P = applied force

A= constant cross-sectional area overwhich the stress develops

t= Tensile Stress (produced by

Tensile Forces)

c = Compressive Stress (produced by

Compressive Forces)

Normal Stress = stress acting on a plane perpendicular to (ornormalto) the line of action of the applied force (as in our example).

we will discuss stresses oninclinedplanes in week 4.

B.TENSILE AND COMPRESSIVE

STRESSES - Application

1.)Analyze the capacity of existing

member

P (all)= (all)A

A= cross-sectional area, perpendicular to the direction of the force

P (all)= axial load capacity (max allowableaxial load)

=amount of load the member can

safely carry

(all)= allowable axial stress=amount of stress which is judged acceptable for the given material

Example:2 x 4 wood with tensile forceapplied. Find theaxial load capacityof the 2 x 4.

A = 3.5 x 1.5 = 5.25in 2 , (all)= 400 psi

(all) =400 psi

P t(all) = (all)A=(400 psi)(5.25in 2 ) = 2,100 lb

2.Design a member to support a known

A (req)=P_

A(req)= required cross-sectional area ofthe axially loaded member beingdesigned.

P =applied axial load

(all)=allowable axial stress

Example: Find what diameter steel rod is required to support a 2100# tension force.

P = 2100 lb

(all)= 24,000 psi (= lb/in 2 )

A (req)= P= 2,100 lb=.0875 in 2

(all)24,000 lb/in 2

A (req)= .0875 in 2= (r)2= (d req )2

(d req ) 2 =.0875 in 2 (4)=0.1114 in 2

d req=(0.1114 in 2 )1/2= .333 =3/8

Therefore a3/8 diameter steel rod supports as much tensile force as 2 x 4.

Note :App. A,B,C*,D* give cross-sectional areas for different steel shapes. E gives cross-sectional areas for lumber.

3.)Cautions concerning design and

analysis:

a.) Some materials have a higher allowable compressive stress than tensile stress ( t(all)< c(all) )

Examples:Wood, concrete, cast iron

(refer to App. F, G, 722, 723)

b.)Thelengthof a member effects how much compressive force it can sustain.Will discuss in depth in Week 14 (columns).

C all< T all

For now, we will assume compressive members are short enough to ignore effect of length.

c).Holes for bolts etc. must be subtractedfrom the gross cross-sectional area toobtain the net cross-sectional area(A net ).

d.) Design Manuals have been publishedby Steel (AISC), Wood (AITC, NDS),Concrete (ACI), and Aluminumassociations.

e.) Assuming even force distribution over the cross-sectional area is usually valid, butnot always .

C.STRESS ON NET AREA

Ex.2 x 4 with bolt holes for 2-1/2 bolts

Determine allowable tensile load

T all= t,allAnet

t(all)= 400 psi

A net= A - Aholes

= (3.5 x 1.5) - (2)(9/16)(1.5)

= 3.56 in 2

T all= (400 psi)(3.56 in 2)= 1425 lb

Note :3.5 and 1.5 are dresseddimensions(see APP.E)

2 x 4 is actually 1.5 x 3.5

2 x 8 is actually 1.5 x 7.25

D.) Bearing Stress ( p )

-A contact pressure between separatebodies.

-A type of compressive stress

Example:Wood rafter in a building

E.) Shear Stress

Normal stress(tensile and compressive) acts in a directionperpendicularto the surface on which it acts.It is produced by a force whose line of action is perpendicular to the surface on which the stress is produced.

Shear stressacts in a directionparallelto the surface on which it acts.It is produced by a force whose line of action is parallel to the surface on which the stress is produced.

Example:Nail through boards

Example: Find the shear stress in the 1 diameter bolt shown.

V = 4,000 lb

A = d 2=(1) 2= .785 in 2

v=4,000 lb= 5,096 psi

.785 in 2

F.)TENSILE AND COMPRESSIVE

STRAINS AND DEFORMATIONS

In picking a member size, we not only

need to know it can handle the load

without failing due to too much stress

(i.e. fracture or break), we must also

be sure it will not deflect (or deform)

excessively.

Example:Dock with wooden ladder for

a footbridge.

This is an example of deformation or

deflection due to bending stress which

we will cover later.

Similarly, when a steel rod is in Tension,

it will deform, but it is not as noticeable.

= deformation = the amount a body is

lengthened by a Tensile force and

shortened by a compressive force.

To permit comparison with acceptable

values, the deformation is usually

converted to a unit basis, which is the

strain .

= strain (= unit strain)

= deformation that occurs over length L

L = original length of member

Example:

Example:3/8 cable, 100 long stretches

1 before freeing a Jeep which

is stuck in the mud.

Find - the strain in the cable.

L = 100 (12/1) = 1200

= 1= 0.0008333 in/in

Well come back to see if this is will break the cable.

G.)Shear Strain SKIP !

The effect of shear on a body is to cause the body to rotate thru an angle.

SKIP !

tan = v _ _

= shear strain (radians)

v= shear deformation (in)

L = Distance over which shear deformation

occurs (in)

SKIP !

Sinceis always a very small angle, and

for small angles:

Then: = v/L

Reviewof Stress and Strain

Axial Stress and Strain

Shear Stress and Strain

= v/L -SKIP !

H.) The Relationship Between Stress

and Strain

As you apply load to a material, the strain increases constantly (or proportionately) with stress.

Example: In a tension test you apply a gradually increasing load to a sample.You can determine the amount of strain ( that occurs in a sample at any given stress level ( .

(ksi) (in/in)

0 0.000

3 0.001

6 0.002

9 0.003

12 0.004

Since the stress is proportional to the strain, ratio of stress to strain isconstant .

(ksi) (in/in) (ksi)

3 0.001 3000

6 0.002 3000

9 0.003 3000

12 0.004 3000

This constant ratio of stress to strain is called theModulus of Elasticity (E).

The Modulus of Elasticity is always the same for a given material (see p. 617). We call it amaterial constant .

Knowing E for a given material and :

1.) We can find how much stress is in thematerial if we know the strain:

2.) We can find how much strain is in thematerial if we know the stress:

CAUTION !

If the tension test continues, the stress will reach a level called theProportional Limit ( PL).If the stress is increased above PL,the strain will increase at a higher rate.

Strain is directly proportional to stress as long as the stress does not exceed the proportional limit ( PL ) of the material. That is:

=E= constant, ONLYif PL

Anytime you use E to find stress or strain, you must check to make sure PL.

Ex. Given:Previous Truck cable strain

Find: Stress in the steel cable

L = 1200

= 1= 0.0008333 in/in

E( as long as PL )

E= 30,000,000 psi (for steel)

E = 30,000,000 psi (.0008333 in/in)

= 24,990 psi (pretty high)

CHECK:is < PL?

= 24,990 psi< PL= 34,000 psi(OK)

For Shear Stress and Strains:

G = v = Modulus of Rigidity

Substituting=Pand=

into E = Assuming< pL )

E =P/A

Note:E, G are constant

Solving for

=PLIF ,< PL

Example 2:

How much will a 1 diameter by 150 long steel bar stretch if a 1000 lb axial tension load is applied to it ?

=P=1000 lb= 1,273 psi < Y 36,000 psi

A (0.5) 2

= = __1,273psi___= 0.00004244 in/in

E30,000,000 psi

L 0.00004244 in/in)(150) =0.006366

311ch9

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