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In [1]:

    
import sys
import platform
from math import sqrt, pi

print('Python version: ', sys.version)
print('Running on: ', sys.platform)









    



Python version:  3.6.3 |Anaconda, Inc.| (default, Oct 27 2017, 12:14:30) 
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)]
Running on:  darwin

Given: A steel bar with a cross-section of 130 mm² and a length of 50.8 mm is axially loaded in tension with a force of 57.0 kN. The modulus of elasticity of the steel is 210 GPa.

Find: What is the axial stress in the bar?

Solution:

F = 57.0 kN

F = 5700 N

A₀ = 130 mm²

σ = ?

$$ \sigma = \frac{F}{A_0} $$



In [2]:

    
F = 57000
A0 = 130
stress = F/A0
print(stress)









    



438.46153846153845

Find: What is the resulting strain?

Solution:

σ = 438.46 MPa

E = 210 GPa

E = 210000 MPa

ε = ?

$$ E = \frac{\Delta\sigma}{\Delta\epsilon} $$



In [3]:

    
E = 210000
# E = stress/strain
strain = stress/E
print(strain)









    



0.0020879120879120877

Find: What is the change in length of the bar?

Solution:

ε = 0.0020879

L₀ = 50.8 mm

ΔL = ?

$$ \epsilon = \frac{\Delta L}{L_0} $$$$ \Delta L = \epsilon L_0 $$



In [4]:

    
L0=50.8
deltaL =strain*L0
print(deltaL)









    



0.10606593406593405

Find: What is the final length of the bar?

Solution:

L₀ = 50.8 mm

ΔL = 0.10607 mm

L_f = ?

$$ L_f = L_0 +\Delta L $$



In [5]:

    
Lf = L0 + deltaL
print(Lf)









    



50.90606593406593

2. Given: An aluminum rod will be loaded in tension with a force of 350 kip.

Find: What is the minimum diameter required to have a factor of safety of 1.9 agaist yield, if the yield stress is 40 ksi?

Solution:

SF = 1.9

σ_y = 40 ksi

d = ?



In [6]:

    
SF =1.9
stress_y = 40
stress_app = stress_y/SF
print('Stress: ',stress_app)
F = 350
A0 = F/stress_app
print('Cross-sectional Area: ', A0)
d = sqrt(4*A0/pi)
print('Diameter, d = ', d)









    



Stress:  21.05263157894737
Cross-sectional Area:  16.625
Diameter, d =  4.600826820390231

3. A composite rod is axially loaded. It is ridigidly mounted at the wall and a single load P of 50 kN is applied at the end.

Material	Area (mm²)	Length (mm)	Elastic Modulus, E (GPa)	Yield Strength, σ_y (MPa)
Aluminum	400	600	70.0	240
Brass	300	800	105	410

a. What is the stress in the Aluminum?



In [7]:

    
A_Al=400
F = 50000
stress_Al= F/A_Al
print(stress_Al)

b. What is the stress in the Brass?



In [8]:

    
A_Br = 300
stress_Br = F/A_Br
print(stress_Br)









    



166.66666666666666

c. What is the final length of the entire composite rod once the load P is applied?



In [9]:

    
E_Al = 70000
strain_Al = stress_Al/E_Al
print(strain_Al)









    



0.0017857142857142857



In [10]:

    
L0_Al = 600
dL_Al = strain_Al*L0_Al
print(dL_Al)









    



1.0714285714285714



In [11]:

    
E_Br = 105000
strain_Br = stress_Br/E_Br
print(strain_Br)









    



0.0015873015873015873



In [12]:

    
L0_Br = 800
dL_Br = strain_Br*L0_Br
print(dL_Br)









    



1.2698412698412698



In [13]:

    
Lf = L0_Al + dL_Al + L0_Br + dL_Br
print(Lf)









    



1402.3412698412696

d. What is the factor of safety against yeild for the brass?



In [14]:

    
stress_y_Br = 410
SF_Br = stress_y_Br / stress_Br
print(SF_Br)



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