In [4]:

```
#Given
w_varying = 270.0 #N/m, torque
l_crossection = 9.0 #m , length
l_cb = 6.0 #m length
l_ac = 2.0 #m length
#Calculation
w_c = (w_varying/l_crossection) * l_cb #By proportion, load at C is found.
f_resultant_c = 0.5* w_c *l_cb
#Balancing forces in the x direction:
n_c = 0
#Balncing forces in the y direction:
v_c = f_resultant_c
# Balncing the moments about C:
m_c = - (f_resultant_c*l_ac)
#Results
print"The resultant force at C is ",f_resultant_c,"N"
print"The horizontal force at C is ",n_c,"N"
print"The vertical force at C is ",v_c,"N"
print"The moment about C is ",m_c,"Nm"
```

In [3]:

```
#Given
f_d = 225 #N
w_uniform = 800 # N/m
l_ac = 0.200 #m
l_cb = 0.05+0.1 #m
l_bd = 0.100 #m
l_bearing = 0.05 #m
f_resultant = w_uniform*l_cb #120N
l_f_resultant_b = (l_cb/2)+ l_bearing #0.125m
l = l_ac + l_cb + l_bearing + l_bd
#Calculation
m_b = 0 # Net moment about B is zero for equilibrium
a_y = -((f_d*l_bd) - (f_resultant*l_f_resultant_b))/ (l - l_bd) # finding the reaction force at A
# Refer to the free body diagram in Fig.1-5c.
f_c = 40 #N
#Balancing forces in the x direction
n_c = 0
#Balncing forces in the y direction
v_c = a_y - f_c #-18.75N - 40N-Vc = 0
# Balncing the moments about C
m_c = ((a_y * (l_ac + 0.05)) - f_c*(0.025) ) # Mc+40N(0.025m)+ 18.75N(0.250m) = 0
# Result
print'The horizontal force at C =',n_c,"N"
print'The vertical force at C =',v_c,"N"
print'The moment about C =',round(m_c,2),"NM"
```

In [21]:

```
# Given
l_ac = 2 #m.,length
l_cb = 1 #m.
l_ad = 1.5 #m.
#r_a = 0.125 #m.
#r_d = 0.125 #m.
W = 500 # N, force
g=9.81
#Calculation
#Balancing forces in the x direction
n_c = (W*g*(l_ac+l_cb))/(3*2/5.0) # N
#Balncing forces in the y direction
v_c = n_c*(4/5.0) #N
# Balncing the moments about C
m_c = n_c*(3/5.0)-(W*g)
# Result
print'The horizontal force at C = ',-v_c/1000,"KN"
print'The vertical force at C = ',-m_c/1000,"KN"
print'The moment about C = ',-m_c/1000,"KNm"
```

In [31]:

```
# Given
l_ag = 2 #Length of AG is 1m.
l_gd = 1 #Length of GD is 1m.
l_de = 3 #Length of DE is 1m.
f_a = 1500 #Force at A is 1500N.
l_ec = 1.5 #Length of EC is 1m.
l = l_ag +l_gd +l_de
w_uniform_varying = 600 #Nm.
f_ba = 7750 #N
f_bc = 6200 #N
f_bd = 4650 #N
#Calculation
w_resultant = 0.5*l_de*w_uniform_varying
# calling point of action of resultant as P
l_ep = (2/3.0)*l_de
l_ap = l - l_ep
#Free Body Diagram Using the result for Fba, the left section AG of the beam is shown in Fig 1-7d.
# Equations of equilibrium
#Balancing forces in the x direction
n_g = -f_ba * (4/5.0) # N
#Balncing forces in the y direction
v_g = -f_a + f_ba*(3/5.0) #N
# Balncing the moments about C
m_g = (f_ba * (3/5.0)*l_ag) - (f_a * l_ag) #Nm
# Result
print'The horizontal force at G = ',n_g,"lb"
print'The vertical force at G = ',v_g,"lb"
print'The moment about G = ',m_g,"lb-ft"
```

In [34]:

```
# Given
f_a = 50 #N, force
m_a = 70 # Moment at A in Nm
l_ad = 1.25 #Length of AD in m.
l_bd = 0.5 #Length of BD in m.
l_cb = 0.75 #Length of BC in m.
w_l = 2 #Kg/m
g = 9.81 #N/kg- acceleration due to gravity
#Free Body Diagram
import math
w_bd = w_l*l_bd*g #in N.
w_ad = w_l*l_ad*g
# Equations of Equilibrium
#Balancing forces in the x, y and z direction
f_b_x = 0 # N
f_b_y = 0 #N
f_b_z = g + w_ad + f_a #N
# Balancing Moments in the x,y and z direction
m_b_x = - m_a + (f_a*l_bd) + (w_ad*l_bd) + (l_bd/2.0)*g #Nm
m_b_y = - (w_ad*(l_ad/2.0)) - (f_a*l_ad) #Nm
m_b_z = 0 #Nm
v_b_shear = sqrt(f_b_z **2 + 0) #Shear Force in N
t_b = - m_b_y #Torsional Moment in Nm
m_b = math.sqrt(m_b_x **2+ 0) # Bending moment in Nm
# Result
print' The weight of segment BD =',w_bd,"N"
print' The weight of segment AD =',w_ad,"N"
print' The force at B in the Z direction =',f_b_z,"N"
print' The moment about B in the X direction =',m_b_x,"Nm"
print' The moment about G in the Y direction =',m_b_y,"Nm"
print' The Shear Force at B =',v_b_shear,"N"
print' The Torsional Moment at B =',t_b,"Nm"
print' The Bending Moment at B =',m_b,"Nm"
```

In [37]:

```
#Given
netf_b = 18*(10 **3) #N Net force at B.
netf_c = 8*(10**3) #N Net force at C.
f_a = 12 *(10**3) #N Force at A.
f_d = 22* (10**3) #N Force at D.
w = 35.0 #mm Width.
t = 10.0 #mm Thickness.
#calculations
p_bc = netf_b + f_a #N Net force in region BC.
a = w*t #m**2 The area of the cross section.
avg_normal_stress = p_bc/a #Average Normal Stress.
# Result
print'The Average Normal Stress in the bar when subjected to load = ',round(avg_normal_stress,1),"MPa"
```

In [15]:

```
#Given :
m_lamp = 80 #Mass of lamp in Kg.
d_ab = 10 # Diameter of AB in mm.
d_bc = 8 # Diameter of BC in mm.
ab_h = 60 *(math.pi/180.0) # In degrees - Angle made by AB with the horizontal.
w = m_lamp*9.81 #N
a_bc = (math.pi/4.0)*(d_bc**2) #m**2 Area of cross section of rod BC
a_ab = (math.pi/4.0)*(d_ab**2) #m**2 Area of cross section of rod AB
#calculation
import math # Equations of equilibrium: Solving equilibrium equations simultaneously ,
f_bc=395
f_ab=f_bc*(4/5.0)/(math.cos(60*3.14/180.0))
avg_normal_stress_a = f_ab / a_ab #Mpa Average Normal Stress in AB
avg_normal_stress_c = f_bc/ a_bc# Mpa Average Normal Stress in BC
# Displaying results:
print"The Average Normal Stress in AB when subjected to load = MPa",round(avg_normal_stress_a,2),"MPa"
print"The Average Normal Stress in BC when subjected to load = MPa",round(avg_normal_stress_c,2),"MPa"
```

In [42]:

```
#Given
h_above_ab = 2.75 #ft , height
h_below_ab = 0.2
d_a = 0.75 #diameter ft
d_b = 0.1
sp_w = 490
# Equation of Equilibrium
import math
a = math.pi* (d_a**2) # Area of cross section in m**2
p = sp_w * h_above_ab * a
avg_comp_stress = p/a # The average compressive stress in kN/m**2
#Display
print'The average compressive stress = ',avg_comp_stress,"lb/ft**2"
```

In [44]:

```
#Determine Distance of force from AB
#Given
f = 3000 #N Force acting at distance x from AB.
l_ac = 200 #Length of AC in mm.
a_ab = 400 #Cross sectional area of AB in mm**2.
a_c = 650 # area of C in mm**2.
#Calculation
#Equations are
#Fab+Fc=3000
#-3000*x+Fc*200=0
Fc=1857
Fab=3000-Fc
x=Fc*200/3000
#Display
print'Distance of force from AB = ',x,"mm"
```

In [68]:

```
#Given
af =30 #kN Axial force along centroidal axis
t1 = 2 #m thickness of square cross section
t2=4
#Internal loading
import math
Fb=af*t1/((t1+t2)*(4/5.0))
Ax=Fb*(3/5.0)
Ay=af-(Fb*(4/5.0))
Fa=math.sqrt(Ax**2+Ay**2)
Va=Fa/t1
Vb=Fb
Aa=(math.pi/4.0)*((t1/100.0)**2)
Ab=(math.pi/4.0)*(((t1+t2)/200.0)**2)
Ta=Va/Aa
Tb=Vb/Ab
#Display
print'The Average Normal Stress for section b-b = ',round(Ta/1000,0),"Mpa"
print'The Average Shear Stress for section b-b = ',round(Tb/1000,1),"Mpa"
```

In [74]:

```
#Given
f = 6 #kN
d_rod = 0.1 #Diameter of steel rod in mm.
l_bc = 0.125 #Length of side bc in mm.
l_ac=0.15
#Calculation
F=f/2.0
#Appling Force balance
Va=F
Vb=F
a_rod=d_rod*l_ac
a_strut=l_bc*l_ac
#Average shear stress
avg_shear_rod = F/a_rod #for rod in Mpa
avg_shear_strut = (f/2)/a_strut #for strut
#Display
print'The average shear stress for the rod = ',avg_shear_rod,"Kpa"
print'The average shear stress for the strut = ',avg_shear_strut,"Kpa"
```

In [81]:

```
#Given
l_bc = 1 #Length of BC in inc.
l_db = 3 # in.
l_ed = 2 # in.
l_ab = 1.5 # in.
f_diagonal = 600 #lb
#Calculation
a1 = l_bc*l_ab #Area of face AB in mm**2.
a2 = l_ab*l_ed #mm**2.
a3 = l_db*l_ab # mm**2.
#Balancing forces along the x- direction.
f_ab = f_diagonal*(3/5.0) #Force on segment AB in N
V = f_ab
#Balancing forces along the Y direction.
f_bc = f_diagonal*(4/5.0) #Force on segment BC in N.
avg_comp_ab = f_ab/a1 # N/mm**2
avg_comp_bc = f_bc/a2 # N/mm**2
avg_shear = f_ab/a3 # N/mm**2
#Display
print a1
print'The average compressive stress along AB = ',avg_comp_ab,"psi"
print'The average compressive stress along BC = ',avg_comp_bc,"psi"
print'The average shear stress along EDB = ',avg_shear,"psi"
```

In [87]:

```
#Given
shear_allow = 3 #ksi, stress
tensile_allow = 5 #ksi, stress
l_C1 = 3 #in
l_C2 = 2 #in
l_ac=8 #inch
Tauallow=8 #ksi
#Calculation
import math
Fab=(tensile_allow*(3/5.0)*(l_C1+l_C2)+shear_allow*l_C1)/(l_ac)
Cx=-shear_allow+tensile_allow*(4/5.0)
Cy=tensile_allow*(3/5.0)+l_C1
Fc=math.sqrt(Cx**2+Cy**2)
A=(Fc/2.0)/Tauallow
d=2*math.sqrt(A/math.pi)
#Result
print" The diameter of rod is = ",round(d,3),"inch"
print"Wse a pin of diameter =",3/4.0,"inch"
```

In [93]:

```
#Given
P= 20.0 #kN, force
d_hole = 40.0 #mm
normal_allow = 60.0 #MPa, pressure
shear_allow = 35.0 #MPa, pressure
#Diameter of Rod
import math
area1 = (P*10**3)/(normal_allow*10**6) #Area in m**2
d = ((math.sqrt((4*area1)/math.pi))*1000) # Area = (math.pi\4)d**2
#Thickness of disc
V = P
area2 = (V*10**3)/(shear_allow*10**6) #Area in m**2
thickness = (area2*10**6)/(d_hole*math.pi)# A = pi*d*t
#Result
print"The diameter of rode = ",round(d,2),"mm"
print"The thickness of disc = ",round(thickness,2),"mm"
```

In [96]:

```
#Given
import math
bearing_allow = 75.0 #MPa, pressure
tensile_allow = 55.0 #MPa, pressure
d_shaft = 60.0 #mm
r_shaft = d_shaft/2.0 #mm
area_shaft = math.pi*(r_shaft**2) #Area = pi*r**2
d_collar = 80 #mm
thick_collar = 20 #mm
#Calculation
r_collar = d_collar/2.0 #mm
area_collar = math.pi*(r_collar**2) #Area = pi*r**2
#Normal Stress
P1 = (tensile_allow* area_shaft)/3.0 #Tensile stress = 3P/A.
P1_kN = P1/1000.0
#Bearing Stress
bearing_area = area_collar-area_shaft
P2 = (bearing_allow*bearing_area)/3
P2_kN= P2/1000.0
#Result
print"The load calculated by Normal Stress ",round(P1_kN,1),"kN"
print"The load calculated by Bearing Stress ",round(P2_kN,0),"kN"
```

In [102]:

```
#Given
d_ac= 20.0 #mm
area_al = 1800.0 #mm**2, area
d_pins = 18.0 #mm
st_fail_stress = 680.0 #MPa, pressure
al_fail_stress = 70.0 #MPa, pressure
shear_fail_pin = 900.0 #MPa
fos = 2.0 #Factor of safety
l_ab = 2.0 #m
l_ap = 0.75 #m
#Calculation
import math
area_pins = math.pi*(d_pins/2.0)**2
area_ac = math.pi*(d_ac/2.0)**2 #Area = (math.pi\4)d**2
st_allow= st_fail_stress /fos #MPa
al_allow = al_fail_stress/fos #MPa
pin_allow_shear = shear_fail_pin/fos #MPa
#Rod AC
f_ac = (st_allow*area_ac)/1000.0
P1 = ((f_ac*l_ab)/(l_ab-l_ap))
#Block B
f_b =(al_allow*area_al)/1000.0
P2 = ((f_b*l_ab)/l_ap)
#Pin A or C
V = (pin_allow_shear*area_pins)/1000.0
P3 = (V*l_ab)/(l_ab-l_ap)
#Result
print"The load allowed on rod AC ",round(P1,0),"kN"
print"The load allowed on block B ",P2,"kN"
print"The load allowed on pins A or C ",round(P3,0),"kN"
```