A 20 m long beam is simply supported at the left end and 5m from the right end. It is to be used as a bridge to carry a moving uniform highway load of 8 kN/m and a concentrated load of 12 kN. Compute the maximum span positive moment due to the highway loads.   0kN - m   460kN - m   350kN - m   225kN - m

The beam is subjected to a service dead load of 2.5k/ft for the entire length and a service live load of 20 kips at 4 ft from left end. The self weight of the beam b = 12in h=24 in has to be added to the dead load for the calculations. How much is the reaction for the design load at point A and paint B?

3.11 (B). A beam, 12 m long, is to be simply supported at 2m from each end and to carry a u.d.i. of 30 kN/m together with a 30 kN point load at the right-hand end. For ease of transportation the beam is to be jointed in two places, one joint being situated 5 m from the left-hand end. What load (to the nearest kN) must be applied to the left- hand end to ensure that there is no B.M. at the joint (i.e. the joint is to be a point of contraflexure)? What will then be the bes position on the beam for the other joint? Determine the position and magnitude of the maximum B.M. [114 kN, 1.6 m from r.h. reaction; 4.7 m from 1h. reaction; 43.35 kN m.) present on the beam.

A simply supported beam spans 8 m. the cross section width is 300mm, the effective depth is 500mm. It is required to carry a uniformly distributed load of 150 kN/m at ULS. Assume fe=30 MPa and fuk=500 MPa. Design the beam for both bending and shear.