# Statics Problem 4/49: Method of Sections

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1. Mar 14, 2015

### ErLupo

1. The problem statement, all variables and given/known data
Determine the force in member BE of the loaded
truss.

See the attached picture.

2. Relevant equations
Sum of the Moments = 0
Sum of the Forces = 0

3. The attempt at a solution

Sum moments about A to get:

-12L - 56L +40Dy = 0
Dy=1.7L

Sum the forces in the y direction:

Ay + Dy - L - 2L = 0
Ay = 1.3L

Cut through BC, BE, and FE at "Q"

Then sum the forces in the y direction:

-BEsin(θ) + Ay - L = 0
BE=.3L/sin(θ)

I'm really at a loss at how to get theta. The main problem I have is determining the horizontal distance between B and F. I am unable to determine what that distance is. If I had that distance I believe I could use the Law of sines and subsequently find theta.

Last edited by a moderator: Apr 30, 2017
2. Mar 15, 2015

### SteamKing

Staff Emeritus
The key is solving triangle ABF.

Can you figure the vertical distance between points A and B? (Hint: what angle does the line AB make with the horizontal?)

3. Mar 15, 2015

### ErLupo

Wouldn't that just be;

Let x = vertical distance.

12/sin30 = y/sin60

y = 20.78.

That would make the hypotenuse = 24 (Pythagorean Theorem).

Then to solve for the hypotenuse BF of the small 15 degree triangle:

BF/sin15 = 24/sin120

BF = 7.17

Then to solve for the two legs joined at a right angle of the small 15 degree triangle:

Let BFx = horizontal distance between B and F.

BFx = 7.17*cos75°
BFx = 1.86

BFy = 7.17*sin75°
BFy = 6.93

To solve the right triangle with hypotenuse BE:

Horizontal distance between B and E
BEx = 16 - 1.86
BEx = 14.14

Now do inverse tangent to find the angle which is the same as theta:

θ = tan^-1(6.93/14.14)
θ = 26.1°

However, when I use that in my previous equation, BE=.3L/sin(θ), I come up with .682L which is off from the answer of .787L.

Any thoughts?
Thank you!

4. Mar 15, 2015

### SteamKing

Staff Emeritus
I agree with your calculations up to this point. Good job.

The calculation above is where the error crept into your calculations.

Notice that the frame is symmetrical about the vertical axis.
The length BC = 16 feet.

In order to find the length FE, you must use the symmetry of the frame to find the length of FE w.r.t. the length of BC.
Your calculation assumed FE = BC - BEx, which is slightly in error, because it corrects for only one side of the frame.

5. Mar 15, 2015

### ErLupo

Yes, FE = BC - 2BEx, correct?

So that would make FE = 12.29

But if I were to use a right triangle with hypotenuse BE wouldn't I just add BEx which would be 14.14?

Even if I only use the actual length of FE (12.29) and a combination of law of sines and cosines to get the unknowns I still end up with an angle 26.1 degrees.

6. Mar 15, 2015

### SteamKing

Staff Emeritus
Just as point F is not located directly under point B along a vertical line, neither is point E located directly beneath point C.

BC = 16 ft.
FE' = 16 - 1.86 = 14.14 ft. (horizontal distance between B and E)

BF = 7.17 ft.

BF cos 15° = 6.93' (vertical distance between B and F)

∠FBE + 15° + θ = 90°

∠FBE = tan-1(14.14 / 6.93) - 15° = 48.9°

θ = 26.1°

I agree with your calculation of θ.

Looking back into your original calculations,

Notice that the moment arms for the loads L and 2L don't quite line up with the distances measured to points B and C, respectively.

L and 2L are applied at points F and E instead.

7. Mar 15, 2015

### ErLupo

Ah. Yes. This makes sense. That was my mistake.

When I use the changed distances to the moment arms I get Dy = 1.65L and Ay = 1.34L.

When the truss is cut through Q and the vertical forces are summed I got:

Ay - BEsin(θ) - L = 0

BE = (1.34L - L)/(sin(26.1°))

BE = .787L

Which matches the back of the book!

Thank you so much for the help!

8. Mar 15, 2015

### SteamKing

Staff Emeritus
Glad everything worked out!