Statics Problem 4/49: Method of Sections

[ErLupo]

Homework Statement

Determine the force in member BE of the loaded
truss.

See the attached picture.

Homework Equations

Sum of the Moments = 0
Sum of the Forces = 0

The Attempt at a Solution

[/B]
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.

 

[SteamKing]

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?)

 

[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!

 

[SteamKing]

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

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

To solve the right triangle with hypotenuse BE:

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

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.

 

[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.

 

[SteamKing]

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?

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

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.

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,

Sum moments about A to get:

-12L - 56L +40Dy = 0

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.

 

[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!

 

[SteamKing]

Glad everything worked out!