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1. Stress- Strain curve for 6061-T6 aluminum alloyData was chosen to obtain a reasonably shaped curve.

6061-T6 Aluminum Alloy

Stress-Strain Curve

0

50

100

150

200

250

300

350

0 4 8 12 16 20

Stress (MPa)

S t r a i n ( % )

2. Annealed Pure Copper

E (GPa) G (GPa) !(unitless) "y (MPa) "t (MPa) Ductility

(%EL)

n K

(MPa)

Copper 110 46 0.34 69 200 45 0.44 530

a) E = "y / # therefore # = "y / E = 69MPa / 110,000MPa = 0.000627 unitless

# = (lf – lo) / lo therefore lf = #* lo + lo

lf = (0.000627*100mm)+100mm = 100.0627mm (0.0627mm stretched)

b) ! = -#y / #z therefore #y = -#z* ! = -0.000627*0.34 = -0.0002132

r f = #* r o + r o = (-0.0002132*5mm)+5mm = 4.999mm

c) %EL = % plastic strain at fracture = 45% or .45

Total #f = plastic strain + elastic strain = 0.45 + 0.000627 = 0.450627

#T = ln(1+#f ) = ln(1+0.450627) = 0.3720

"T = K #Tn = 530MPa * 0.37200.44 = 343 MPa

(NOTE that this is much higher than given value of 200 MPa above)

d) "T, f = "f (1+#f ) therefore "f = "T,f /(1+#f )

"f = = 343 MPa / (1+0.450627) = 237 MPa

This answer is questionable because 237 MPa is greater than the given tensilestrength of 200 MPa, and we expect fracture strength in engineering stress terms

to be below tensile strength. The fact that we have utilized Eqn 7.19 in a casewhere necking is likely helps explain this discrepancy.

Tensile Strength = 310 MPa

Failure Strain = 0.175 = 17.5%

Slope = Modulus = 69GPa

Yield Strength = 276MPa

2

3. 4 Legs

lo = 400mm

A (cross section) = 50mm * 25mmP = 6,300,000Nll (leg length under load) = 397.6mm (completely elastic strain)

a) The table legs are loaded in compression. The legs, which are shortened

under loading, return to their original length when the load is removed; this

indicates that all the strain is elastic.

b) The cross section area will increase under compression; after the load is

released the dimensions will return to their original values.#z = (lf – lo) / lo = (397.6mm-400mm) / 400mm = -0.006

! = -#y / #z therefore #y = -#z * ! = 0.006 * .30 = 0.0018

#y = (lf – lo) / lo therefore lf = #y* lo + lo

lf = (0.0018*25mm)+25mm = 25.045mm

! = -#x / #z therefore, #x = -! * #z = -0.30 *-0.006 = .0018

lf = (0.0018*50mm)+50mm = 50.09mm

c) " = P/A = 6,300,000N / 4x(25.045*50.09)mm2

= 1255.5 MPa

(NOTE multiplication by 4 in denominator to account for 4 legs…)"z > 1255.5 MPa

d) E = " / # = 1255.5 MPa / 0.006 = 209 GPa

4.a) Section 7.20 discusses two ways of incorporating a margin of error into

engineering structural calculations for greater safety: design stress and safestress. The design stress method assumes know conditions, and overestimates

them to produce a conservative estimate. The safe stress method assumesknow material properties and underestimates them to produce a conservative

estimate. Design stress is usually preferred since it is based on the anticipatedmaximum applied stress instead of the yield strength of the material.

b) Case 1- Safe Stress- because the yield strength of the material is known andwind is a factor effecting the yield strength

Case 2- Design Stress- because the strength of the bed is unknown and must be estimated (determination of an applied stress)

c) In both cases, the stresses are altered to provide a margin of error. Therequired truck bed strength will be overestimated, while the allowable wind

speed will be underestimated.

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5.a) For the brass specimen, the stress-strain behavior for which is shown in Figure

7.12, the tensile strength is 450MPa. From Figure 7.31, the hardness for

brass corresponding to this tensile strength is about 125HB or 70HRB. b) The steel alloy (Figure 7.33) has a tensile strength of about 1970 MPa. This

corresponds to a hardness of about 560HB or ~55HRC from the line

(extended) for steels in Figure 7.31.c) Hardness tests are non-destructive, quick, and simple; they are often used for

quality control.

d) Strength-hardness correlations are material dependent. No overall

relationship exists between the two values (strength and hardness).

6.

)(

0

2

0)9.09.11(

nP

fs e

P P E E

!

=

+!=

" "

Plug in values at P= 0 to find

MPa

e

GPa E

E

n

220

220

90

))0(9.0)0(9.11(90

0

))0((

0

0

2

0

=

=

=

+!=

!

"

"

Then find (n) by plugging in 25% values

60 = 220e("n(.25))

n = 5.197

Next solve for 50% values

E = 90(1"1.9(.5)+0.9(.5)2)= 24.75GPa

# fs= 220e

("5.197*0.5)=16.36 MPa

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7.

Relaxation Modulus vs. Temp for PMMA

1

10

100

1000

10000

0 20 40 60 80 100 120 140 160

Temperature C

R e l a x a t i o n M o d u l u s M P a

0.01 hrs

10 hrs

The glass transition temperature will be 105 to 110 C

The melt temperature will be around 140 or 150 C

The 0.01 hour time yields larger values of Er than the 10 hour time, since less relaxationhas been allowed to occur at the lower time values. This also influences the T[g] and

T[m] estimates to some degree.

8.

Use Equation 8.2, $r = " cos(%) cos(&), and the closely related Eqns 8.3 & 8.4.

a) For " =12 MPa, % = 60° and & = 35° Eqn 8.3 gives a resolved shear stress of

$r = 4.91 MPa. Since this is less than the critical stress, no yielding will occur.

b) To reach the critical resolved shear stress, Eqn 8.4 shows that an applied tensile

stress of 15.1 MPa would be needed.

c) If the slip plane normal angle were reoriented to 0°, all possible slip directionswould be at angles of 90° [cos(&)=0], giving a resolved shear stress of zero. At

the opposite extreme, a slip plane normal angle of 90° [cos(%)=0] for any possible

slip direction would also give a resolved stress of exactly zero. The max resolvedshear stress would occur when both % & & = 45°, when $r would equal 6 MPa.

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9) Refer to Figs 8.15 and 8.19…

a) The minimum yield strength of 25 MPa shown is at the point where d-1/2

=0. This

will be "0 = 25 MPa. The slope of the yield strength line represents k Y and can bevisually estimated as k Y = 12.5 MPa•mm

1/2(or 0.395 MPa•m

1/2).

b) As suggested by the method above, "0 represents the minimum strength level of

the material, which would occur at d-1/2

=0, or d=! (the strength when there are no

grain boundaries present, because there is only a single grain).c) A grain size of 2.0 x 10

-3mm gives d

-1/2=22.36 mm

-1/2. Given the constants from

part (a), this should correspond to a strength of approximately 305 MPa.d) The brass in Fig 8.19 shows an undeformed yield strength of approximately 175

MPa (when no additional dislocation strengthening would exist). Here, thestrength level would be entirely due to grain size and alloying (composition)

effects. If the composition is the same as in Fig 8.15, then the grain size should be d

-1/2=12 mm

-1/2or d=0.0069 mm.

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