The signboard truss is designed to support a horizontal wind load of 800 lb. A separate analysis shows that 5 8 of this force is transmitted to the center connection at C and the rest is equally divided between D and B. Calculate the forces in members BE and BC.

BE = 559lb T

BC = 300lb T

Explanation:

Base on the figure attached, let us first find the load on joint B, C and D

FC = 5/8 * 800 = 500lb

FD = 13/28 * 800 = 150lb

FB is also equal to 150lb

The angle alpha α at joint D will be:

Tan α = 6/12

α = tan ^-1 (0.5)

α = 26.6

by using resolution of forces, resolve the force at that point to horizontal and vertical component.

Horizontal component

Fd + DE sin α = 0

DE sinα  = -Fd

DE = -Fd/sinα

DE = -150/26.6

DE = - 335.4lb

Vertical component

-CD - DEcosα = 0

CD = -DEcosα

CD = -(-335.4)cos26.6

CD = 300lb

So let us do the same at joint C and joint E

At joint C

Horizontal component

Fc + CE = 0

CE = -Fc

CE = -500lb

Vertical component

CD - BC = 0

BC = CD

BC = 300lb T

At joint E

The horizontal component will be:

-CE - DEsinα - BEsinα + EFsinα = 0

-(-500) - (-335.4)sin26.6 - BEsin26.6 + EFsin26.6 + EFsin 26.6

650 - BEsin26.6 + EFsin26.6 = 0

Make BE the subject of the formular

BEsin26.6 = 650 + EFsin26.6

BE = (650 - EFsin26.6)/ sin26.6

BE = 1453.2 + EF ................ (1)

The vertical component

DEcosα - BEcosα - EFcosα = 0

substitute all the parameters

- 335.4cos26.6 - BEcos26.6 - EFcos26.6 = 0

Substitute BE in the equation 1 into the equation above

-300 - (1453.2 + EF)cos26.6 - EFcos26.6 = 0

-1599.7 - 1.789EF = 0

EF = -1599.7 / 1.789

EF = -894.2lb

Substitute EF in equation 1

BE = 1453.2 - 894.2

BE = 559lb T

Therefore the forces in members BE and BC

BE = 559lb T

BC = 300lb T

this is true

explanation:

i looked it up