A model for a twostory building is shown Determine the force

A model for a two-story building is shown. Determine the force supported by each member of the truss. T_A B = _______kip T_B D = _______kip T_A D = _______kip T_A C = _______kip T_C D = _______kip T_D F = _______kip T_C F = _______kip T_C E = _______kip

Solution

The reaction forces at equlibrium at the supports E & F,

sigma Fv: Re + Rf = 22 ------ eqn 1

sigma Fh: He + Hf = 7 ------- eqn 2

Assuming, sigma Me = 0

6 x 5 + 5 x 5 - (4 x 54 + 3 x 27) + Rf x 5 = 0

Rf = 48.4 kip

Re = 22 - 48.4 = (-) 26.4 kip

At joint E:

vertical force sigma Fe = 0

Tce = Re = (-) 26.4 kip (tensile)

At point F:

theta = arctan(36/27) = 53.13 deg

equilibrium forces at F, sigma horizontal Fh= 0 and Fv=0

sigma Fv = 0

Rf - Tcf cos 53.13 - Tdf = 0

Tcf cos 53.13 - Tdf = 48.4  -------- eqn 3

At point B:

Tab = 4 kip , and Tbd = 5 kip

At point D:

Tcd + Tad sin 53.13 = 3 ----- eqn 4

Tdf - Tcd cos 53.13 - Tbd = 6

Tcd = (Tdf - Tbd - 6)/ cos 53.13

Tcd = 1.66Tdf - 18.33 ----------- eqn 5

Eqn eqn 4 & 5 and rearrange,

1.66 Tdf + 0.79Tad = 21.33 ---------- eqn 6

At point A:

sigma Fv, Tac + Tad cos 53.13 = 5 ----- eqn 7

sigma Fh, Tab + Tad sin 53.13 = 0

using Tab = 4 kip solve for Tab in the above eqn we get,

Tad = (-) 4 / sin 53.13 = (-) 5 kip

Using Tad = (-) 5 kip solve for Tdf in eqn 6,

Tdf = (21.33 + 3.95) / 1.66 = 15.22 kip

Solve for Tcd in eqn 4 using Tad = (-) 5 kip

Tcd = 3 - Tad sin 53.13 = 3 + 5 sin 53.13 = 6.99 kip

Using eqn 3 find Tcf, Tdf = 15.22 kip

Tcf cos 53.13 - Tdf = 48.4

Tcf = 1.66 (48.4 + 15.22) = 105.60 kip

solve for Tac in eqn 7 using Tad = (-) 5kip

Tac + Tad cos 53.13 = 5

Tac = 5 - Tad cos 53.13 = 5 + 3 = 8

Tac = 8 kip