Answer:
77362.56 J
163730.28571 J
Explanation:
A = Area = 25 mm²
l = Length = 300 mm
K = Constant = 
= Heat transfer factor = 0.75
= Melting factor = 0.63
T = Melting point of low carbon steel = 1760 K
Volume of the fillet would be

The unit energy for melting is given by

Heat would be

Heat required to weld is 77362.56 J
Amount of heat generation is given by

The heat generated at the welding source is 163730.28571 J
Answer:
Check attachment for better understanding
Explanation:
Given that
a= 0.75m
b=1.31m
c= 2.2m
Weight of pole is 26.90
Then, Fg = Weight = 26.90
Using Equilibrium of forces
ΣFy = 0
U — D — Fg = 0
U — D = Fg
U — D = 26.9
To calculate U,
We will take moment about point A.
ΣMa = 0
Let the clockwise moment be positive and anti-clockwise be negative
Fg(a+b) — U(a) = 0
26.9(1.31+0.75) —0.75U = 0
26.9(2.06) = 0.75U
0.75U = 55.414
U = 55.414/0.75
U = 73.89 N
To calculate D,
U — D = 26.9
73.89—D =26.9
73.89—26.9 = D
D = 46.99N