Consider the diagram of a simply supported beam shown in Figure P10 . This beam has different dimensions to in the previous question. A triangular distributed load is applied to the beam, increasing from 0 kN/m at point B (at the right edge of the beam) , to a magnitude of w 2 = 9 kN/m at mid-span of the beam. The span of this beam from point A to point B, L = 12.0 m . The beam is in equilibrium. Determine the magnitude of the equivalent point load (F EQ ) due to the triangularly distributed load. Select the correct value of F EQ (in kN). Figure P10. Simply supported beam with a triangular distributed load.

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Consider the diagram of a simply supported beam shown in Figure P10 . ...

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Consider the diagram of a

simply supported beam shown in

Figure P10. This beam has different dimensions to in the previous question.

A triangular distributed load is applied to the beam, increasing from

0 kN/m at point B (at the right edge of the beam), to a magnitude of w₂ = 9 kN/m at mid-span of the beam. The span of this beam from point A to point B, L = 12.0 m

. The beam is in equilibrium.

Determine the

magnitude of the equivalent point load (F_EQ)

due to the triangularly distributed load.

Select the correct value of F

_EQ
(in kN).

Figure P10. Simply supported beam with a triangular distributed load.

F_EQ = 27.00 kN

100%

F_EQ = 108.00 kN

F_EQ = 216.00 kN

F_EQ = 54.00 kN

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