In [12]:

```
import math
#Initilization of variables
#Consider Equilibrium of joint A
#As there are no Load applied at A members AC and AB have nothing to Balance
#So they are null members
F_AB=0
F_AC=0
#Consider Equilibrium of joint B
#Applying the summation of horizontal forces we get
F_DB=4*(cos(45*pi*180**-1))**-1
#Applying the summation of vertical forces we get
F_BC=F_DB*sin(45*pi*180**-1)
#Consider Equilibrium of joint B
#Applying the summation of vertical forces we get
F_CE=4*(sin(45*pi*180**-1))**-1
#Applying the summation of horizontal forces we get
F_DC=F_CE*cos(45*pi*180**-1)
#Result
print"The Forces in Each members are as follows:F_AB",F_AB,"KN"
print" :F_AC",F_AC,"KN"
print" :F_DB",round(F_DB,2),"KN(compression)"
print" :F_BC",round(F_BC,2),"KN(Tension)"
print" :F_CE",round(F_CE,2),"KN(Tension)"
print" :F_DC",round(F_DC,2),"KN (compression)"
```

In [34]:

```
import math
#Initilization of variables
#Taking moment at Pt A we get
R_B=100*8*4**-1
#Applying the summation of vertical forces we get
R_AV=-R_B
#Applying the summation of horizontal forces we get
R_H=100
#joint B
#Applying the summation of vertical forces we get
F_CB=R_B
#Applying the summation of horizontal forces we get
F_AB=0 #As there is no force to balance in horizontal direction
#joint A
#Applying the summation of horizontal forces we get
F_AC=R_H*(cos(45*pi*180**-1))**-1
#Applying the summation of vertical forces we get
F_AD=200-F_AC*sin(45*pi*180**-1)
#joint C
#Applying the summation of vertical forces we get
F_EC=200-F_AC*cos(45*pi*180**-1)
#Applying the summation of horizontal forces we get
F_DC=F_AC*cos(45*pi*180**-1)
#joint D
#Applying the summation of horizontal forces we get
F_DE=F_DC*(cos(45*pi*180**-1))**-1
#DF and EF are null members at this joint as each member individually has nothing to balance
F_DF=0
F_EF=0
#Result
print"The Forces in Each members are as follows:F_AB",F_AB,"KN"
print" :F_CB",F_CB,"KN(compressive)"
print" :F_AC",round(F_AC,2),"KN(Tensile)"
print" :F_AD",F_AD,"KN(Tensile)"
print" :F_EC",F_EC,"KN(Compressive)"
print" :F_DC",F_DC,"KN(compressive)"
print" :F_DE",round(F_DE,2),"KN(Tensile)"
print" :F_DF",F_DF,"KN"
print" :F_EF",F_EF,"KN"
```

In [5]:

```
import math
#Initilization of variables
#taking moment at pt A we get
R_D=(90*6+120*3)*9**-1 #Reaction at Pt D
#Joint D
#Applying the summation of vertical forces we get
F_GD=100*(sin(60*pi*180**-1))**-1
#Applying the summation of horizontal forces we get
F_DC=F_GD*cos(60*pi*180**-1)
#Joint G
#Applying the summation of vertical forces we get
F_GC=F_GD
#Applying the summation of horizontal forces we get
F_FG=F_GD*cos(60*pi*180**-1)+F_GC*cos(60*pi*180**-1)
#joint C
#Applying the summation of vertical forces we get
F_FC=(115.5*sin(60*pi*180**-1)-90)*(sin(60*pi*180**-1))**-1
#Applying the summation of horizontal forces we get
F_CB=F_DC+F_GC*cos(60*pi*180**-1)+F_FC*cos(60*pi*180**-1)
#joint F
#Applying the summation of vertical forces we get
F_FB=F_FC
#Applying the summation of horizontal forces we get
F_EF=F_FG+F_FC*cos(60*pi*180**-1)+F_FB*cos(60*pi*180**-1)
#Joint B
#Applying the summation of vertical forces we get
F_EB=(120-F_FB*sin(60*pi*180**-1))*(sin(60*pi*180**-1))**-1
#Applying the summation of horizontal forces we get
F_BA=F_CB+F_FB*cos(60*pi*180**-1)-F_EB*cos(60*pi*180**-1)
#Joint E
#Applying the summation of vertical forces we get
F_AE=F_EB
#Result
print"Forces in Each members are as follows:F_GD",round(F_GD,1),"KN(compression)"
print" :F_DC",round(F_DC,2),"KN(Tension)"
print" :F_GC",round(F_GC,1),"KN(Tension)"
print" :F_FG",round(F_FG,1),"KN(compression)"
print" :F_FC",round(F_FC,1),"KN(compression)"
print" :F_CB",round(F_CB,2),"KN(Tension)"
print" :F_FB",round(F_FB,1),"KN(Tension)"
print" :F_EF",round(F_EF,2),"KN(compression)"
print" :F_EB",round(F_EB,2),"KN(Tension)"
print" :F_BA",round(F_BA,2),"KN(Tension)"
print" :F_AE",round(F_AE,2),"KN(compression)"
```

In [7]:

```
import math
#Initilization of variables
#JOint D
#Applying the summation of vertical forces we get
F_1=6*sin(30*pi*180**-1)**-1
#Applying the summation of horizontal forces we get
F_5=F_1*cos(30*pi*180**-1)
#Joint C
#Resolving forces perpendicular to plane
F_6=10*cos(30*pi*180**-1)
#Resolving forces parallel to plane
F_2=F_1+10*cos(60*pi*180**-1)
#Joint E
#Applying the summation of vertical forces we get
F_7=(8+F_6*sin(60*pi*180**-1))*(sin(60*pi*180**-1))**-1
F_4=F_5+F_6*cos(60*pi*180**-1)+F_7*cos(60*180**-1*pi)
#Resolving forces perpendicular to plane
F_3=F_7*sin(60*pi*180**-1)
#Resolving forces parallel to plane
F_8=F_2+F_7*cos(30*pi*180**-1)
#Result
print"Forces in Each members are as follows:F_1",round(F_1,2),"KN(Tension)"
print" :F_5",round(F_5,2),"KN(compression)"
print" :F_6",round(F_6,2),"KN(compression)"
print" :F_2",round(F_2,2),"KN(Tension)"
print" :F_7",round(F_7,2),"KN(Tension)"
print" :F_4",round(F_4,2),"KN(compression)"
print" :F_3",round(F_3,2),"KN(compression)"
print" :F_8",round(F_8,2),"KN(Tension)"
```

In [31]:

```
import math
#Initilization of variables
BC=6 #m
#Calculations
AB=2*BC*(3**0.5)**-1
#Taking moment about B we get
R_A=-(-2000*3-1000*6)*(12*(3**0.5)**-1)**-1 #reaction at the roller support A
#The resultant of all the three Loads is 4000 N acting at right angle to BC at D
#Resolving it vertically we have
V=4000*sin(60*pi*180**-1)
#Resolving it horizontal we have
H=4000*cos(60*pi*180**-1)
#Applying the summation of vertical forces we get
R_B_v=V-R_A
#Applying the summation of horizontal forces we get
R_B_h=H
R_B=((R_B_v)**2+(R_B_h)**2)**0.5
tan_theta=R_B_v*R_B_h**-1
#Joint B
#Applying the summation of vertical forces we get
F_BD=1000*(3**0.5)*2
#Applying the summation of horizontal forces we get
F_BE=R_B_h+F_BD*cos(30*pi*180**-1)
#Joint D
F_DE=2000 #N
F_CD=F_BD
#Consider equilibrium of truss to the Left of section 2-2
F_CE=R_A*AB*(sin(30*pi*180**-1)*6)**-1
#Joint A
#Applying the summation of vertical forces we get
F_AC=R_A*(sin(60*pi*180**-1))**-1
#Applying the summation of horizontal forces we get
F_AE=F_AC*cos(60*pi*180**-1)
#Result
print"Forces in Each members are as follows:F_BD",round(F_BD,2),"KN(compression)"
print" :F_BE",round(F_BE,2),"KN(Tension)"
print" :F_DE",round(F_DE,2),"KN(compression)"
print" :F_CD",round(F_CD,2),"KN(compression)"
print" :F_CE",round(F_CE,2),"KN(Tension)"
print" :F_AC",round(F_AC,2),"KN(compression)"
print" :F_AE",round(F_AE,2),"KN(Tension)"
```

In [3]:

```
import math
#Calculations
#Taking moment of the Forces about the hinge A
P=1000*2**0.5*1.2*(0.9)**-1
#Let R_AH be the Horizontal component of the reaction at A
R_AH=P-1000*2**0.5
R_A=((R_AH)**2+(1000*2**0.5)**2)**0.5
#Resolving the forces vertically we get
R_AV=1000*2**0.5 #vertical component of the reaction at A
#joint A
#Resolving vertically we get
F_BA=1000*2**0.5*(sin(30*pi*180**-1))**-1
#Resolving horizontally we get
F_AD=2000*2**0.5*3**0.5*2**-1-1000*2**0.5*3**-1 #N
#Joint C
BD=1.2*sin(30*pi*180**-1)
BE=0.6*sin(30*pi*180**-1)
ED=0.6*cos(30*pi*180**-1)
CE=0.9-0.52
theta=arctan(BE*CE**-1)*(180*pi**-1)
F_CB=P*(sin(38.29*pi*180**-1))**-1
#Resolving vertically
F_CD=F_CB*cos(theta*pi*180**-1)
#Joint D
#Resolving horizontally
F_DB=(F_AD-1000*2**0.5)*(cos(60*pi*180**-1))**-1
#Result
print"The Pull in chain is",round(P,2),"N"
print"Force in the each members are as follows:F_BA",round(F_BA,2),"KN(compressive)"
print" :F_AD",round(F_AD,2),"KN(Tensile)"
print" :F_CB",round(F_CB,2),"KN(compression)"
print" :F_CD",round(F_CD,2),"KN(Tensile)"
print" :F_DB",round(F_DB,2),"KN(compressive)"
```

In [7]:

```
import math
#Calculations
theta=arctan(1*2**-1)*(180*pi**-1) #Radian
#Taking moment about A
R_EH=10*8*4**-1
#Horizontal component of reaction at A
R_AH=20 #KN
#Applying the summation of horizontal forces we get
F_AB=20*cos(theta*pi*180**-1)**-1
#Applying the summation of vertical forces we get
R_AV=10*5**0.5*sin(theta*pi*180**-1)
#Vertical Reaction at E
R_EV=0
#Joint C
#Applying the summation of vertical forces we get
F_DC=10*sin(theta*pi*180**-1)**-1
#Applying the summation of horizontal forces we get
F_CB=F_DC*cos(theta*pi*180**-1)
#Joint D
#Applying the summation of vertical forces we get
F_DB=F_DC*sin(theta*pi*180**-1)
#Applying the summation of horizontal forces we get
F_DE=F_DC*cos(theta*pi*180**-1)
#Joint E
#Applying the summation of vertical forces we get
F_EB=R_EV*sin(theta*pi*180**-1)
#Result
print"Forces in Each members are as follows:F_AB",round(F_AB,2),"KN(Tensile)"
print" :F_DC",round(F_DC,2),"KN(compression)"
print" :F_CB",round(F_CB,2),"KN(Tensile)"
print" :F_DB",round(F_DB,2),"KN(Tensile)"
print" :F_DE",round(F_DE,2),"KN(compression)"
print" :F_EB",round(F_EB,2),"KN"
```

In [26]:

```
import math
#Initilization of variables
F_c=20 #KN #Force at C
F_d=5 #KN #Force at D
F_e=15 #KN #Force at E
F_f=10 #KN #Force at F
L_CD=3.6 #m #Length of CD
L_DE=3.6 #m #Length of DE
L_EF=4.8 #m #Length of EF
L_AD=L_BE=3.6 #m #Length of AD & BE
#Calculations
#Let R_A and R_B be the reactions at pts at A and B
#Taking moment at A
R_B=-(-F_f*(L_DE+L_EF)+F_c*L_CD-F_e*L_DE)*(L_DE)**-1
R_A=50-R_B
#Considering section 1-1 through members AB,DB,DE and taking F.B.D of left side of section 1-1
#Taking moment at B
sigma_1=(F_d*L_DE+F_c*(L_CD+L_DE)-R_A*L_DE)*L_AD**-1 #Force i member DE
#Taking moment @ D
sigma_3=(F_c*L_CD)*L_AD**-1 #KN #force in member AB
#Consider triangle DBE
theta=arctan(L_BE*L_DE**-1)*(180*pi**-1)
#Taking moment @ A
sigma_2=(-sigma_1*L_AD+F_c*L_CD)*(L_AD*cos(theta*pi*180**-1))**-1 #Force in member F_DE
#Now considering section 2-2 passing through members AB,AD,CD and taking left hand F.B.D
#Taking moment @C
sigma_5=(R_A*L_CD-sigma_3*L_AD)*L_CD**-1 #Force in member AD
#Taking moment @A=0
sigma_4=F_c*L_CD*L_AD**-1 #Force in member CD
#Result
print"Force in member CD is",round(sigma_4,2),"KN(Compressive)"
print"Force in member AD is",round(sigma_5,2),"KN(Tensile)"
print"Force in member BD is",round(sigma_2,2),"KN(Compression)"
print"Force in member AB is",round(sigma_1,2),"KN(Tension)"
```