Tension in a cable supporting a hinged beam

[adlewis90]

I have tried to complete the following question by taking moments about point A, but to no avail. Could anybody help me with where i am going wrong. Do I need to include perpendicular distance?

The I-beam A-B shown below is supported by a cable at B and a pin at A. The cable is inclined at an angle, θ = 21° to the longitudinal axis of the beam. The beam supports a load, m, of 12 kN at a distance mx = 1.5 m from B.

If the mass of the beam is 95 kg/m and its total weight can be assumed to act through the centre of gravity of the beam, calculate:

a) the cable tension, T

b) the resultant force acting on the pin at A.

Given:

Ax = 0.12 m, Beam Depth, D = 0.5 m, Beam Length, L = 5 m, Gravity, g = 9.81 m/s2



Part a)

Weight of beam = 4.65975 kN

By taking moments about A, T = 26.09±0.2 kN

Part b)

To determine the resultant at A, calculate the horizontal and vertical reactions AH and AV respectively:

AH = 24.36±0.2 kN

AV = 7.31±0.2 kN

Therefore, the resultant (using Pythagoras), R = 25.43±0.2 kN
Incorrect

Here is my solution to the cable tension

(4.88)(T)(sin21)+(111.834x0.06)-(4.5479x10^3)(2.44)-(12x10^3)(3.38)=0

1.7488(T)=+6.7098+11.0968x10^3+40.56x10^3

1.7488(T)=51650.0902

(T)= (51.65x10^3)/1.7488 =29.53kn

 

[Simon Bridge]

I have tried to complete the following question by taking moments about point A, but to no avail. Could anybody help me with where i am going wrong. Do I need to include perpendicular distance?

You need to show your working. i.e. how did you compute the moments?
It helps to do the algebra with the variables first and then plug the numbers in.
If you didn't take into account the beam "depth" D, you should give that a try.
Use the model answer to clue you into how far out you are.

 

[adlewis90]

Sorted the question before looking on here, was simply solved by extending the length of the cable to where it intersected the centreline of the beam and using this point in the x-axis as the distance. Thanks for your help