Question

A 200 × 150 × 10 angle is to be welded to a steel plate by fillet welds as shown in Figure. If
the angle is subjected to a static load of 300kN, find the length of weld at the top & bottom. The allowable shear stress may be taken as 75 N/mm2

Answer 1

Answer:

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Explanation:

Given static loading =300 kN = 300 \times 10^3×10

3

N ,here dimension is given as 200 \times 150\times 10×150×10 ,

permissible shear stress= 75 \frac{kN}{mm^2}

mm

2

kN

,here a+b=200

s= size of weld = thickness= 10 ,we know that

for single weld

300 \times 10 ^3=0.707 \times s \times l\times \tau300×10

3

=0.707×s×l×τ

on putting value we can get value of l

l= 303.09 mm

l_a+ l_b=565.77l

a

+l

b

=565.77

now distance of centroidal axis from bottom angle

b=55.3 and a= 144.7 so

l_a= \frac{l\times b}{a+b}l

a

=

a+b

l×b

l_a=156.44l

a

=156.44 and l_b= 366.88l

b

=366.88