A beam of I section 600mm deep and 200 mm wide has flanges 25 mm thick...
Solution:
Given:
Depth of the section, d = 600 mm
Width of the section, b = 200 mm
Thickness of the flange, tf = 25 mm
Thickness of the web, tw = 20 mm
Shear stress in the web, q = ?
To find: Shear stress in the flange at the junction.
Assumptions:
1. The section is symmetric about the y-axis.
2. The section is subjected to pure bending.
3. The distribution of the shear stress is linear across the depth of the section.
Calculation:
1. Calculation of the shear force in the section:
The shear force, V at the section can be calculated using the formula:
V = (w x L)/2
Where,
w = load per unit length
L = span of the beam
As the load on the beam is not given, let's assume a uniformly distributed load of 10 kN/m.
L = 1 m (Assuming a simply supported beam)
V = (10 x 1)/2 = 5 kN
2. Calculation of the shear stress in the web:
The maximum shear stress in the web occurs at the junction of the web and flange.
Shear stress in the web, q = (V x tw)/(b x tf)
q = (5 x 1000 x 20)/(200 x 25) = 40 N/mm^2
3. Calculation of the shear stress in the flange:
The distribution of the shear stress is linear across the depth of the section. Hence, the shear stress in the flange can be calculated using the following formula:
Shear stress in the flange, σf = (q x (d/2 - tf))/((d/2))
σf = (40 x (600/2 - 25))/(600/2) = 30 N/mm^2
Answer:
Hence, the shear stress in the flange at the junction is 30 N/mm^2.
A beam of I section 600mm deep and 200 mm wide has flanges 25 mm thick...
T=(VQ)/Ib, T(shear stress) is indirectly proportional to b.
so T2/T1=b2/b1
T2=(b2/b1)×T1
where b2=width of web, b1= width of flange
given T1=q
T2=(20/200)×q
=0.1q