Project

challenge on internal hinge.

A steel beam was designed as simply supported beam of span 20m. the applied load on the

beam is 15kn/m. however it was realized that a 20m beam cannot be shipped at once with

an upper limit of 10m. so the contractor suggested to split the beam in 2 pieces of 10m each.

the contractor also recommended to have a fixed support at one end and roller at other end.

the contractor also suggested to have an internal hinge at the connection of two pieces.

determine if the contractor's suggestion is acceptable or not.

soln:

let us consider a fixed support at A, internal hinge at B and roller support at C.

consider a span BC subjected to UDL of 15kn/m of span 10m.

there will be vertical shear force at B and vertical reaction at C.

taking the summation of moment about B,we get

Rc*10 - wl^l/2 = 0

10Rc - 15(10)^2/2 = 0

10 Rc = 750

Rc = 75KN.

now take the summation of vertical forces, we get

Vb + Rc = 150

Vb + 75 = 150

Vb = 75KN.

now consider the span AB subjected to UDL of 15KN/m, fixed support at A and internal hinge at B.

at A there will be one vertical reaction and one moment and at B ,there will be one vertical force.

now take the summation of moment about A , we get

Ma - 75 * 10 - 15 * (10)^2/2 = 0

Ma - 750 - 750 = 0

Ma = 1500KN

now take the summation of vertical forces,

Ra + Vb = 15 * 10

Ra + 75 = 150

Ra = 75KN

Ship work like this
in your own course.