1.

• II)Properties of Materials

A.) Tension Test

Stress ), ksi Strain ( ), in/in PL Y U E = =slope 2.

B. Mechanical Properties of Materials

1.) Ultimate Strength ( U ) - Themaximum stress a material willwithstand before fracture.

2.) Yield Strength ( Y ) - Themaximum stress a material will withstand before deforming permanently.

3.

3.) Proportional Limit ( PL ) - Themaximum stress a material willwithstand before stress-strainrelationship becomes non-linear.

4.) Modulus of Elasticity - the ratio of stress over strain in thelinearregion of the stress-strain curve.

4.

5.) Ductility

Ductile Material - will undergo plasticdeformation before failing

Stress Strain Ductile Material 5.

6.)Brittleness

Brittle Material - will fail without any plastic deformation (opposite of ductile)

Stress Strain Brittle Material 6. C.)Allowable Stress, Actual Stress,Factor of Safety 7.

1.) The Actual Stress,is the amount of stress in a member due to the applied load.

= P/A

2.) Allowable stress allof a material is the maximum stress that may be considered safe for a given structural member.

= P/A< all

8.

2.) Allowable Stress (continued)

The allowable stress is determined by applying a Factor of Safety (FS) to the yield strength orultimate strength.

all= Y or all= U

F.S.F.S.

FS varies from 1.5 to 3 in typical design situations.

9.

- FS depends on the

a.) Material - properties & variability

b.) Importance of the product

c.) Uncertainty of the load

- FS ensures the yield strength will not be

exceeded.

- FSallows for flaws in the material &

workmanship.

10.

Example:Suspending yourself from a 300 high building with a cable.

How big mustthe cable be to hold you?

Assume the cable has a solid circular cross-section and is made ofA36 steel .

Y= 36,000 psi(p617)

F.S. = 4.0( safe enough?)

P = 200 lb(your weight)

11.

all= Y_ =36,000 psi= 9,000 psi

FS 4.0

In structural design, we wantto find the smallestcross-sectional area(A) that will safely hold a known force (P).

The smaller the area (A) is, the higher the actual stress ( ) will be.

= P

A

12.

The absolute smallestArea (A) we can safely use is when the actual stress is equal the allowable stress.

200 lb= 9,000 psi

A

A= 200 lb __ = 0.0222 in 2

9,000 psi

d= [4(0.0222 in 2 ) 1/2= 0.168 in

13.

So the bare minimum diameter is 0.168

Readily available sizes are in diameter increments of 1/16.

2/16 = 0.125 (too small)

3/16 = 0.188 > 0.168 (OK)

14. 15. Spreadsheet for HW # 10-4 256.7577 0.21 53.9 0.42 10.8 322.017 0.2 64.4 0.4 12.9 388.3064 0.18 69.9 0.36 14 433.7244 0.16 69.4 0.32 13.9 559.5056 0.116 64.9 0.232 13 907.7293 0.055 49.9 0.11 10 2212.747 0.01805 39.9 0.0361 8 6490.265 0.006 38.9 0.012 7.8 18182.03 0.00215 39.1 0.0043 7.83 31952.07 0.00125 39.9 0.0025 8 31531.65 0.00095 30.0 0.0019 6 30723.15 0.00065 20.0 0.0013 4 33283.41 0.0003 10.0 0.0006 2 0 0.0 0 0 (ksi) (in) (kips) E Strain Stress Deformation Load 16.

Example:Hyd. Piston Rod Stainless Steel

y= 40 ksiFS = 1.5

A= ( 10,000 lb )(1.5)= 0.375 in 2

(40,000 psi)

D =[ (4 )( .375 in 2 )/ ] 1/2= 0.69 = 3/4

10000lb 10000 lb