Question

An ISJC 200 channel section has the following details: width of flange 70 mm, depth of channel 200 mm, the thickness of flange t_f = 7.1 mm, a moment of inertia I_xx = 1161.2 cm⁴: The distance of shear centre from the centre of the web will be
1. 16.82 mm
2. 18.58 mm
3. 22.87 mm
4. 27.87 mm

Answer 1

Correct Answer - Option 4 : 27.87 mm

Concept:

Shear Centre:

It is the point (inside or outside a section) through which the applied shear force produces no torsion or twist of the member

The location of shear centre is given for different sections as:

1) For channel section:

\({\rm{e}} = \frac{{{{\rm{h}}^2} × {{\rm{t}}_1} × {{\rm{b}}^2}}}{{4 × {{\rm{I}}_{{\rm{xx}}}}}}\)

where,

h = Distance between the centre lines of the flanges

t₁ = Thickness of the flange

b = Length of the flange upto the centreline of the web

I_xx = Moment of inertia of the channel about axcis of symmetry (X-X axis)

e = Horizontal distance of applied vertical downward shear force from the centre of the web

2) For Unequal I-Section:

\({\rm{e}} = \frac{{{{\rm{h}}^2} × {{\rm{t}}_1} × \left( {{\rm{b}}_2^2 - {\rm{b}}_1^2} \right)}}{{4 × {{\rm{I}}_{{\rm{xx}}}}}}\)

where,

t1 = Thickness of the flange of I-Section

b₂ = Longer length of the I-Section from the vertical centre line of the web

b2 = Shorter length of the I-Section from the vertical centre line of the web

3) For sections having symmetry about both the axis, the shear centre coincides with the centroid, wheras for sections having symmetry about one axis, the shear centre lies on the axis of symmetry but does not coincides with centroid.

Calculation:

Given,

t₁ = 7.1 mm, b = 70 mm

h = Distance between the centre lines of the flanges

h = 200 - (7.1/2) - (7.1/2) = 192.9 mm

Ixx = 1161.2 cm4 = 1161.2 × 10⁴ mm⁴

\({\rm{e}} = \frac{{{{\rm{h}}^2} × {{\rm{t}}_1} × {{\rm{b}}^2}}}{{4 × {{\rm{I}}_{{\rm{xx}}}}}}\)

\({\rm{e}} = \frac{{{{\left( {70} \right)}^2} × {{\left( {192.9} \right)}^2} × 7.1}}{{4 × 1161.2 × {{10}^4}}}\)

e = 27.87 mm