# Chapter 17:Welded Joints¶

## Problem no.17.1,Page no.379¶

In :
import math

#Initilization of Variables

b=12 #cm #width of steel plates
t=1 #cm #thickness of steel plates
sigma=75 #MPa #stress

#Calculations

#The maximum Load which the plate can carry
P=b*t*10**-6*sigma*10**6 #N

#size of weld is equal to thickness of plate
S=t #cm

#P=2**0.5*l*S*sigma

#After substituting values and simplifying above equation, we get,
l=((P)*(2**0.5*S*sigma)**-1) #cm

#add 1.25 to allow start and stop of weld run
L=l+1.25 #cm

#The stress concentration factor for transverse fillet weld is 1.5

sigma_2=sigma*1.5**-1 #MPa #Permissible tensile stress

#P=2**0.5*l_2*S*sigma_2

#After substituting values and simplifying above equation, we get,
l_2=((P)*(2**0.5*S*sigma_2)**-1) #cm

l_3=l_2+1.25 #cm

#Result

Length of weld for static loading 9.74 cm


## Problem no.17.2,Page no.380¶

In :
import math

#Initilization of Variables

d=6 #cm #diameter of rod
L=40 #cm #Length of steel plate
sigma=180 #MPa #Allowable stress

#Calculations

A=pi*4**-1*d**2 #cm**2 #Area of rod
M=P*10**3*L #Ncm

F=M*A**-1 #N/cm #Force per unit cm of weld at top and bottom

V_s=P*10**3*(pi*d)**-1 #N/cm #vertical shear

S=R*(sigma)**-1*10**-2 #cm #size of weld

#Result
print"size of weld is",round(S,2),"cm"

size of weld is 0.94 cm


## Problem no.17.3,Page no.380¶

In :
import math

#Initilization of Variables

b=12 #cm #width of plate
S=t=1 #cm #thickness of plate
sigma_s=60 #MPa #shear stress

#Calculations (part-1)

#P=2**0.5*l*S*sigma_s

l=((P*10**3)*(2**0.5*S*sigma_s*10**-4*10**6)**-1) #cm

#add 1.25 cm to start and stop weld run

L=l+1.25 #cm #length of weld

#Calculations (part-2)

#stress concentration factor for parallel fillet weld is 2.7

sigma_s_2=sigma_s*2.7**-1 #MPa #permissible shear stress

#P=2**0.5*l_2*S*sigma_s_2

l_2=((P*10**3)*(2**0.5*S*sigma_s_2*10**-4*10**6)**-1) #cm

l_3=l_2+1.25 #cm #length of weld

#Result

Length of weld Under static Loading is 7.143 cm


## Problem no.17.4,Page no.381¶

In :
import math

#Initilization of Variables

sigma_t=100 #MPa #tensile stress

#Calculations

#For equal stress in the welds A and B, the load shared by the fillet welds will be proportional to size of weld

#t_a=0.7*s #Effective throat thickness of weld A in upper plate
#s=size of weld

#t_b=1.05*s #Effective throat thickness of weld B in lower plate

#For weld A
#P_1=l*t*sigma_t

#After substituting values and simplifying above equation, we get,
#P_1=84000*s #N     (equation 1)

#P_2=l*t_2*sigma_t

#After substituting values and simplifying above equation, we get,
#P_2=126000*s #N     (equation 2)

#After adding equation 1 and 2, we get,
#P=210000*s           (equation 3)

#Now equating total forces of the fillets to load on the plates
s=P*10**3*210000**-1 #cm

#Result
print"size of end fillet is",round(s,2),"cm"

size of end fillet is 0.81 cm


## Problem no.17.5,Page no.381¶

In :
import math

#Initilization of Variables

L_1=30 #cm #length of longitudinal weld
L_2=16 #cm #length of transverse weld
#t=0.7*s #Effective thickness of weld
sigma_t_1=100 #MPa #working stress for transverse welds
sigma_t_2=85 #MPa #working stress for longitudinal welds

#Calculations

#For transverse weld
#P_1=L_1*t*10**-4*sigma_t_1*10**6

#After substituting values and simplifying above equation, we get,
#P_1=112000*s #N

#For longitudinal weld
#P_2=L_2*t*10**-4*sigma_t_2*10**6

#Total force of resistance of weld
#P=P_1+P_2 #N

#P=290500*s #N

#Now equating total forces of resistance to pull of the joint
s=P*10**3*290500**-1 #cm

#Result
print"size of weld is",round(s,3),"cm"

size of weld is 0.516 cm


## Problem no.17.6,Page no.382¶

In :
import math

#Initilization of Variables

P=200 #KN #Load carried by the angle
S=0.6 #mm #size of weld
b=4.46 #cm #Distance of centre of gravity of the angle from the top shorter leg
a=10.54 #cm #Distance of centre of gravity of the angle from the top edge of the angle
sigma_s=102.5 #MPa #shear stress
#l_1=Length of the top weld
#l_2=length of the bottom weld
#L=l_1+l_2 #cm #total length weld

#Using the relation
#P=L*0.7*S*sigma_s

#After substituting values and simplifying we get
L=(P*10**3)*(0.7*S*sigma_s*10**-4*10**6)**-1 #cm  (equation 1)

#Using the relation
l_1=(L*b)*(a+b)**-1 #cm

#substituting this value in equation 1 we have,
l_2=L-l_1 #cm

#Result
print"Distance of centre of gravity of the angle from the top edge of the angle",round(l_2,2),"cm"

Distance of centre of gravity of the angle from the top edge of the angle 32.64 cm


## Problem no.17.7,Page no.383¶

In :
import math

#Initilization of Variables

sigma_s=75 #N/mm**2 #shear stress
e=12 #cm
r_1=2.5 #cm

#Calculations

#A=(2*S*l)*(2)**0.5
#sigma_s=P*A**-1 #MPa #shear stress

#After substituting values and simplifying we get
#sigma_s=16.97*S**-1 #MPa

#I_g=S*l*(3*b**2+l**2)*(6)**-1 #cm**4 #Polar moment of Inertia of weld

#After substituting values and simplifying we get
#I_g=180.833*S #cm**4
r_2=((8*2**-1)**2)+((5*2**-1)**2)**0.5 #cm #max radius of weld

#sigma_s_2=P*e*r_2*I_g**-1 #MPa #shear stress due to bending moment

cos_theta=r_1*r_2**-1

#Now using the relation
#sigma_s=(sigma_s_1**2+sigma_s_2**2+2sigma_s_1*sigma_s_2*cos_theta

S=(2363.8958*5625**-1)**0.5 #cm #size of the weld

#Result
print"size of the weld",round(S,3),"cm"

size of the weld 0.648 cm