Analysis and Design of V-Shaped Beams

Beams that have V-shapes (in the plan view) are commonly found in the corners of residential and commercial buildings. According to Hassoun and Al-Manaseer (2008), such beams can be analysed and designed using strain-energy principles especially when the beam is fixed at both ends.

In some scenarios, engineers may be tempted to design v-shaped beams as two cantilevers meeting at a point, but this is not strictly the case. There will be a need to assess other internal stresses such as torsion, and to understand the actual nature of the distribution of internal forces.

For a v-shaped beam subjected to a uniformly distributed load w (kN/m), the internal forces in the members can be obtained as follows;

(The bending moment at the centre of the beam Mc is given by;
Mc = (wL2)/6 Ã— [sin2Î¸/(sin2Î¸ + Î»cos2Î¸)

Where;
Î» = EI/GJ
L = half the total length of the beam AC
Î¸ = Half the angle between the two sides of the v-shape beam.

The torsional moment at the centreline of the section is given by;
TC = (MC/sinÎ¸) Ã— cosÎ¸ = MCCotÎ¸

At any section N along the length of the beam at a distance x from the centreline C,

MN = MC â€“ wx2/2
TN = TC = MCCotÎ¸ (constant torsional moment)

At the supports, let x = L
MA = MC â€“ wL2/2

Worked Example

A v-shaped beam at the corner of a building has a depth of 400mm and a width of 225 mm. The plan view of the beam is shown below. It is to support an ultimate uniformly distributed load of 30 kN/m inclusive of the factored self-weight. Design the beam according to the requirements of EC2. fck = 30 MPa, fyk = 500 MPa, Concrete cover = 35 mm

Solution

For fck = 30 MPa,
Modulus of elasticity Ecm = 31476 MPa
Shear modulus G = Ecm/2(1 + v) = 31476/2(1 + 0.2) = 13115 MPa
Moment of inertia I = bd3/12 = (225 Ã— 4003)/12 = 12 Ã— 108 mm4
Polar moment of inertia J = 985033091.649413 mm4
Î» = EI/GJ = (31476 Ã— 12 Ã— 108)/(13115 Ã— 985033091.649413) = 2.9237

Mc = [wl2sin2Î¸/6(sin2Î¸ + Î»cos2Î¸) = (30 Ã— 2.52 Ã— 0.75)/[6 Ã— (0.75 + 2.9237 Ã— 0.25)] = 140.625/8.88555 = 15.755 kNm
MA = MB = MC â€“ wl2/2 = 15.755 â€“ (30 x 2.52)/2 = -77.995 kNm
Torsional moment = TA = MCcotÎ¸ = 15.755 x 0.5773 = 9.095 kNm
Shear at support = VA = VB = 30 Ã— 2.5 = 75 kN