Ain Shams University

Faculty of Engineering

New Program

assignment rd3

Presented to: Dr. Nahed Abd El-Salam

Presented by: Ahmed Hassan Ibrahim

Mostafa sherif Ibrahim

S.MANF

Problem (1)

Given: two points σt =222 MPa at ε = 0.05 and σt = 303MPa at ε = 0.15

- Power law : σt=Kεn

1st point : 303=K(0.15)n, 2nd point 222=(K)(0.05)n

Applying the ratio → σ1

σ2

= (ε1

ε2)

𝑛 → ln both side ln (

𝜎1

𝜎2) = 𝑛 𝑙𝑛 (

ε1

ε2)

𝑛 =ln(

𝜎1

𝜎2)

𝑙𝑛(ε1

ε2)

=ln(

303

222)

𝑙𝑛(0.15

0.05)

= 𝟎. 𝟐𝟖𝟑 n=0.283

Substitution in At the 1st point to get K

: K= σt

εn=

303

0.150.283 = 518.333 𝑀𝑃𝑎 K=518.333 MPa

At ε =0.30 σt=Kεn = (518.333)(0.3)0.283 = 368.6669 𝑀𝑃𝑎

- σt = σo (1-𝑒𝑥𝑝−𝐴ε)

Applying the ratio → σ1

σ2

= (1−𝑒𝑥𝑝−𝐴ε1)

(1−𝑒𝑥𝑝−𝐴ε2) →

303

222= 1.36486

Using trial and error: A = 25.2 A=25.2

Substitution in At the 1st point to get σo ,

σo =σt

1−𝑒𝑥𝑝−𝐴ε = 303

1−𝑒𝑥𝑝−(25.2)(0.15) = 310.0767868 𝑀𝑃𝑎

At ε =0.30 σt = 310.076(1-𝑒𝑥𝑝−(25.2)(0.3)) = 309.91 𝑀𝑃𝑎

The problem show that there are considerable differences between values of σ .

Problem (2)

Required : Express the uniform elongation, e, in terms of the constants, A, and Y.

σt = Y + Aε → 𝑑𝛔

𝑑𝛆= 𝐴 → Y + Aε = A → ε =

𝐴−𝑌

𝐴= 1 −

𝑌

𝐴

the engineering strain : e = exp(ε) -1 = exp(1- 𝑌 𝐴⁄ ) − 1

Problem (3)

Required: value of n

the exponent, n, is the

slope of the line:

n =∆𝑠𝑡𝑟𝑒𝑠𝑠

∆𝑠𝑡𝑟𝑎𝑖𝑛 =

log(𝑠𝑡𝑟𝑒𝑠𝑠1𝑠𝑡𝑟𝑒𝑠𝑠2

)

log(𝑠𝑡𝑟𝑎𝑖𝑛1

𝑠𝑡𝑟𝑎𝑖𝑛2)

= log(

300200

)

log(0.16

0.016)

= 0.176

n=0.176

Problem (4)

Required: expressions σt = K (εo + ε) n - At ultimate:

𝑑𝝈

𝑑𝜺= 𝝈𝒕 → in this model:

𝑑𝝈

𝑑𝜺= 𝒏𝑲(𝜺𝐨 + 𝜺)𝒏−𝟏

𝒏𝑲(𝜺𝒐 + 𝜺)𝒏−𝟏 = 𝑲(𝜺𝒐 + 𝜺)𝒏

𝒏 = (𝜺𝒐 + 𝜺)

- 𝝈𝒕𝒆𝒏𝒔𝒊𝒍𝒆 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉 = 𝑲 (𝜺𝐨 + 𝜺)(𝜺𝒐 + 𝜺)

Problem (5)

Given : σt = 150 + 185ε

Required : percent uniform elongation, material’s tensile strength .

- From problem 2 : e at ultimate = exp(ε) -1 = exp(1- 𝑌

𝐴) − 1

General formula σt = Y + Aε → Y=150, A=185

e= exp(1- 𝑌 𝐴⁄ ) − 1 = exp(1 − 𝑌

𝐴) − 1 = exp (1 −

150

185) − 1

= 0.20826 = 20.826 % e=20.8%

- From problem 2: ε at ultimate = 1 −𝑌

𝐴= 1 −

150

185= 0.1891891892

Sub. Value of ε in σt = 150 + 185εu

σt = 150 + 185(0.1892) = 185.002 MPa

σt= 185.002 MPa

y = 50.982ln(x) + 409.32R² = 0.9259

1

10

100

1000

0.008 0.04 0.2 1

tru

e s

tre

ss

true strain

true stress strain

Log. (true stressstrain)

Problem (6)

Given: two points σt = 278MPa at ε = 0.08, and σt = 322MPa at ε = 0.16

- n & K

Power law : σt=Kεn

1st point : : 278=K(0.08)n, 2nd point 322=(K)(0.16)n

Applying the ratio → σ1

σ2 = (

ε1

ε2)

𝑛 ln both side→ ln (

𝜎1

𝜎2) = 𝑛 𝑙𝑛 (

ε1

ε2)

𝑛 =ln(

𝝈𝟏

𝝈𝟐)

𝑙𝑛(𝛆𝟏

𝛆𝟐)

=ln(

𝟐𝟕𝟖

𝟑𝟐𝟐)

𝑙𝑛(𝟎.𝟎𝟖

𝟎.𝟏𝟔)

= 0.2119758054 n= 0.212

Substitution in At the 1st point to get K

K= σt

𝛆n=

278

0.080.212= 474.886 𝑀𝑃𝑎 K=474.886 MPa

At ε =0.20 σt=Kεn = (474.886)(0.2)0.212 = 337.6 𝑀𝑃𝑎

approximation σt = K (εo + ε)n

Applying the ratio → σ1

σ2= (

𝜺𝒐+𝜺𝟏

𝜺𝒐+𝜺𝟐)

𝑛 → ln both side ln (

𝜎1

𝜎2) = 𝑛 𝑙𝑛 (

𝜺𝒐+𝜺𝟏

𝜺𝒐+𝜺𝟐)

Given εo = 0.01

𝑛 =ln(

𝜎1𝜎2

)

𝑙𝑛(𝜺𝒐+𝜺𝟏𝜺𝒐+𝜺𝟐

)=

ln(322

278)

𝑙𝑛(0.01+0.16

0.01+0.08)

= 0.231 n=0.231

Substitution in At the 1st point to get K

𝐾 =σ1

(𝜺𝒐 + 𝜺𝟏)𝒏=

278

(0.01 + 0.08)0.231= 484.866 𝑀𝑃𝑎

At ε =0.20 σt = K (εo + ε) n = 484.866(0.01+0.2)0.231 =338.107 MPa

Note : The problem show that there are small differences between values of n.

Problem (7)

Given : 𝜺𝒐 = 0.03 , 𝑛 = 0.18

- Power law : σt=Kεn

At ultimate 𝜺𝒖 = 𝒏

𝒏𝒏 =σ𝑡

𝐾 → 𝟎. 𝟏𝟖𝟎.𝟏𝟖 =

𝛔𝒕

𝑲 → (1)

σt=K(εo+ε)n → σ𝑡

𝐾= (εo + εu)n →

σ𝑡

𝐾= (εo + n)n → (2)

from 1 & 2

0.180.18 = (0.03 + n)n → calculation n = 0.2775

- n at ε = 0.05 and 0.15

0.180.18 = (0.03 + 0.05)n → ln both sides :0.18 ln(0.18)= n ln(0.03+0.05)

n=𝟎.𝟏𝟖 𝐥𝐧(𝟎.𝟏𝟖)

𝐥𝐧(𝟎.𝟎𝟑+𝟎.𝟎𝟓)= 0.103

0.180.18 = (0.03 + 0.15)n → ln both sides :0.18 ln(0.18)= n ln(0.03+0.15)

n=𝟎.𝟏𝟖 𝐥𝐧(𝟎.𝟏𝟖)

𝐥𝐧(𝟎.𝟎𝟑+𝟎.𝟏𝟓)= 0.1627

Problem (8)

Given: three points s = 133.3 MPa at e = 0.05, s = 155.2 MPa at e = 0.10 and

s = 166.3 MPa at e = 0.15

Engineering strain Engineering stress true strain true stress

0.05 133.3 MPa 0.04879 139.965 MPa 0.1 155.2 MPa 0.09531 170.72 MPa

0.15 166.3 MPa 0.139762 191.245 MPa - Power law : σt=Kεn

Sub. In points : 139.965 = K (0.4879)n , 170.72 = K (0.9531)n

Applying the ratio → σ1

σ2 = (

ε1

ε2)

𝑛 → ln both side ln (

𝜎1

𝜎2) = 𝑛 𝑙𝑛 (

ε1

ε2)

𝑛 =ln(

𝜎1

𝜎2)

𝑙𝑛(ε1

ε2)

=ln(

𝟏𝟑𝟗.𝟗𝟔𝟓

𝟏𝟕𝟎.𝟕𝟐)

𝑙𝑛(𝟎.𝟎𝟒𝟖𝟕𝟗

𝟎.𝟎𝟗𝟓𝟑𝟏)

= 𝟎. 𝟐𝟗𝟔𝟔

Substitution in At the 1st point to get K

K= σt

εn=

139.965

0.048790.29669= 342.9087𝑀𝑃𝑎

Check by using the 3rd point :

σt=Kεn → 191.245 = (342.908)(0.139762)0.29669

∴ 𝑡ℎ𝑒 𝑑𝑎𝑡𝑎 fit Power law

- the strain at the start of necking equals n

n = εu = 0.2966

ε =ln(e+1) → eat necking = expε -1 = 0.345398

Problem (9)

Given : σ = 100 MPa, e = 0.20

Engineering strain Engineering stress true strain true stress 0.2 100 MPa 0.18232 120 MPa

True strains for equivalent amounts of deformation in tension and

compression are equal except for sign.

At a true strain of −0.182 in compression, the engineering strain would be

ecomp = exp(−0.18) − 1 = −0.1667, and the engineering stress would be

scomp = σ/ (1 + e)= −120 MPa/(1 − 0.1667) = - 144 MPa.

Problem (10)

The shape of the engineering stress–strain curve in compression can

be predicted from the true stress–strain curve in tension, assuming

that absolute values of true stress in tension and compression are the

same at the same absolute strain values.

Each point σt, ε on the true stress-true strain curve corresponds to a point σ, e on the engineering

stress-strain curve

0

100

200

300

400

500

600

0 0.1 0.2 0.3 0.4 0.5 0.6

stre

ss (

Mp

a)

strain

engineering stress

true stress