5 m o tnc dion of the beam at B Problem 3: Two beams support a distributed load

Question

5 m o tnc dion of the beam at B Problem 3: Two beams support a distributed load of 28 kN/m as shown. Beam (1) is supported by a fixed support at A and by a simply supported beam (2) at D. In the unloaded condition, beam (1) touches, but exerts no force on, beam (2). Beam (1) has a depth of 400 mm, a moment of inertia of I1 130 x 10 mm, a length L 3.6 m, and an elastic modulus of E 200 GPa. Beam (2) is a timber beam 175-mm wide and 300-mm deep. The elastic modulus of the timber beam is E2 12 GPa, and its length is L2-5 m. Use superposition to determine the deflection at point B. 4 Problem 4: The stresses shown in the figure act at a point in a stressed body. Using the equilibrium equation approach (do not use the 110 MPa

Explanation / Answer

The first beam is fixed at one end (A) and supported at other end B

As per the theory of deflection of beams we use formulae to calculate deflections for 2 beams independently

Principle of superposition is applied at the end to calculate effective deflection

Beam 1 )

Bending Moment formula

w = Continuous load (Unit - N/m)

M= R1 - wx                 

where

R1 = 3wl/8 = 3(28,000)(3.6) / 8 = 37800 N

and Shear Force at B = R1 - wx = 37800 - (28000)(3.6) = -63000 N

Bending Moment at end B = 37800 - (28000)(3.6) = -63000 Nm2

Deflection at end B is wx( l3-3lx2+2x3) / 48EI = 0

There is zero deflection at end B due to uniform continuous load

Now for Beam 2 we see that it is a simply supported beam with concentrated load at the centre point B

The deflection at point B = Pl3 / 48EI = ( - 63000)( 2.53 ) / 48 (12 GPa) ( .000393750000 )

= 4.340 mm

using superposition principle the total deflection is = 0 + 4.34 = 4.34 mm at point B

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