Ch2-notes

Stress and Strain from Axial LoadingBeer, Johnston, DeWolf

G. Milano, PECh. 2, of 8

Today’s lab exercise was about measurements and the definition of

accuracy and precision. The first lab experiment you’ll perform (or

observe) will be Tensile Testing. A specimen could be a circular steel bar

of a particular diameter. The size of the specimen will be known assuming

that someone else measured it or you may be required to measure the

original length and diameter. The specimen will be inserted into a machine

having grips at both ends. (See p.51) While one end is held fixed, the

machine will begin to move slowly exerting a “pulling” force on the other

end of the bar. This “pulling” force is applied along the longitudinal axis of

the bar, therefore considered an AXIAL TENSILE LOAD. The “pulling”

force causes the bar to be in TENSION. The TENSION on the bar causes

a “stretching” effect. The “stretch” will cause the bar to grow longer and

thinner. The change in the overall length (?L) is called deformation, δ.

This change in the length (?L) compared to its original length (L) is called

the strain, ? . But there is also a change in the diameter or width of the bar

that will be measured and computed as a deformation in the transverse

direction. This we’ll discuss later.

The data you’ll collect will be the amount of the load as it increases and the

changes in this load just before and at exactly the time the specimen fails

(breaks). This load will be divided by the cross-sectional area of your

specimen, therefore giving you the STRESS, σ = P/A. Your data will also

provide you with the changing lengths of the bar, therefore giving you the

STRAIN, ? = ?L / L.

Now comes the interesting part. Prove the theory about the relationship

between STRESS and STRAIN. You already know that relationships



between variables are always determined by graphing. The independent

variable is the STRESS since the controlling variable is the applied load, P.

As a result of this applied load, the length changed, resulting in a

dependent variable, STRAIN. The plot will be a STRESS-STRAIN curve.

You will plot the STRESS on the vertical axis and the STRAIN on the

horizontal axis. The plot will begin to show a linear relationship between

these two variables. During this phase, the straight line plot producing a

linear relationship produces an expression for the direct relationship

between STRESS and STRAIN. This linear relationship offers a slope.

And the slope defines the “proportionality constant” also known as a

“modulus”.

slope = ?y / ?x = ? σ / ?? = constant

This proportionality constant is known as the Modulus of Elasticity, E. This

region is known as the elastic region because the bar can return to its

original shape and size if the load is removed.

Strain

Stre

ss



You will then notice something erratic about the curve as the load is

increased. Observe closely what happens to the specimen in the tensile

machine. Watch the thinner area at the center of the bar. (See p.52)

Remember that we assume the stress is uniformly distributed at any cross-section of the specimen.

Refer to the figures at the top of p.49 and keep in mind the following:

§ STRESS is a relationship between load (P) and the cross-

sectional area (A), σ = P/A (lbf / sq.in.)

§ STRAIN is the relationship between the deformation (δ) or change

in length (?L) and the original length (L), ? = ?L / L, (in / in)

So … increase the diameter of the cross-sectional area of the member and

you’ll need to increase the applied load accordingly to produce the same

stress, regardless of the length of the member.

Also … increasing the length of the member will increase the deformation

accordingly, regardless of the area of the member.

Now, put this all together to define the elastic range as the Modulus of

Elasticity.

E = σ / ? = (P/A) / (?L/L) = PL / Aδ = E

This Modulus of Elasticity (or constant of proportionality) is a physical

property and therefore, its value depends on the material.

Steel 29,000 kips/sq.in. - 30,000 kips/sq.in.Aluminum 11,000 kips/sq.in. - 12,000 kips/sq.in.Copper 15,000 kips/sq.in. - 17,000 kips/sq.inWood 2,000 kips/sq.in.

The units for E are generally in magnitudes of 106 psi or 109 MPa



You can predict the deformation by rearranging the expression for the

Modulus of Elasticity, δ = PL / AE

Watch you units! Refer to p. 50 for an example of dimensions.

Back to the experiment …

As you watch the specimen closely, you will notice the thinning area near

the midsection. As with any pliable material (think about taffy), the material

in this region takes on an “hourglass” appearance. If you were to consider

a line or plane tangent to this hourglass surface in the region of the

thinning, you would watch this tangent approach a 45° angle. This thinning

is also known as “necking”. [the thin area between the top and bottom or

the “neck” between the head and the body!] This angled plane is called an

OBLIQUE plane and shares both NORMAL STRESSES and SHEAR

STRESSES.

45°



Now go back to Section 1.11 in Chapter 1 and review Stresses on an

Oblique Plane under Axial Loading.

As the oblique angle increases, the SHEAR STRESSES prevail. Thus, the

member really fails due to SHEAR STRESSES at this point.

The oblique plane is seen as the

red line. The applied load, P, is

along the normal axis parallel to

the longitudinal axis of the

member.

The “normal force” used to

determine the normal stress is

now perpendicular to the oblique

plane.

Therefore, the normal force and the shear force are functions of the applied

load and the angle between the normal axis and an axis normal to the

oblique plane.

Also, the cross-sectional area is now that area projected onto this oblique

plane.

F = P cos ? σ = F/A? = P cos ? / A?

V = P sin ? τ = V/A? = P sin ? / A?

BUT … it is important to look at the geometry of this projected area on the

oblique plane. Treat the oblique area as the hypotenuse and the original

cross-sectional area a leg of the right triangle. Therefore, A = A? cos?

Transverse plane

Normal axis or plane

P

F

V, shear force

?

Oblique area, A?



So, rearranging, A? = A / cos?

Now, substitute for the oblique area in terms of the original cross-sectional

area and the normal and shear stresses are:

σ = F/A? = P cos ? / A? = P cos ? / (A / cos ?) = P cos² ? / A

τ = V/A? = P sin ? / A? = P sin ? / (A / cos ?) = P sin ? cos ? / A

Remember to sketch a FBD and work out the geometry before simplysubstituting into the formulas.

These are the formulas that must be taken into consideration at the time of

failure when analyzing your stress-strain diagram. This will be part of your

discussion and conclusion in explaining how your test results meet the

objective of the theory.

Not all materials behave the same. It is very important to be aware of the

material of your specimen. Some are pliable, called ductile, and some are

brittle. The brittle materials will result in a blunt break transverse to the

normal axis. It is interesting to compare the stress-strain relations for

ductile materials and brittle materials. You may wish to investigate the

physical properties of some materials before your next lab experiment.

There are numerous websites that offer plenty of information on the

physical properties of metals and other materials. What about ceramics

and glass? What about wood or timber? Is all steel the same?

For metals, the ASTM is very helpful. ASTM = American Standard of

Testing Metals.



To determine the ductility of a material, you can calculate the percent

elongation.

% elongation = { (Lf – Lo) / Lo } x 100

where Lo is the original testing length of the specimen and Lf is the

measured length just before breaking.

Or you can calculate the percent reduction in the cross-sectional area.

% reduction in area = { (Ao – Af) / Ao } x 100

where Ao is the original cross-sectional area before applying the tensile

load and Af is the minimum area at breaking or failure.

HOOKE’S LAW and YOUNG’S MODULUS

For linear relations, you know that:

y is proportional to x

For the variables in this case:

Stress is proportional to Strain, or σ ˜ ?

The exact expression for the linear relation comes from the value of the

slope, or the proportionality constant, σ = E ? known as Hooke’s Law

And E = the coefficient of proportionality, also known as Young’s Modulus.

The largest value of Stress to satisfy this relation is called the proportional

limit. This value can also be considered the YIELD POINT on the stress-

strain diagram. The word “yield” refers to the maximum point where the

material will exhibit behavior analogous to being elastic.

SUMMARY:



Hooke’s Law: σ = E ?

Young’s Modulus: E = σ / ? = (P/A) / (?L/L) = PL / Aδ = E

Normal Stress: σ = P/A

Strain: ? = ?L / L

Deformation δ = PL / AEor Deflection:

DEFORMATION of COMPOSITE MEMBERS under AXIAL LOADING

Since the axial force is transferred through the entire length, it is said that

the components of the composite will SHARE the force, but each will

exhibit its own separate deformation. Therefore, the total deformation is

equal to the sum of the parts.

Assuming the same material andthe fact that the force will be thesame:δtotal = δ1 + δ2 + δ3

δtotal = PL1 + PL2 + PL3

A1E A2E A3E

Therefore,δtotal = P L1 + L2 + L3 = (P/E) ∑ ( Li / Ai ) E A1 A2 A3

Some examples!

Next week: Effect of Temperature Changes and Poisson’s Ratio

L1, d1 L2, d2 L3, d3

P

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