Wall Brace Buckling 1

[reesecc66]

Want to make sure I'm going about this correctly.
I have a wall leaning at angle 'phi' at a weight of 25,000#. I'd like to put a bracing column up to support it at angle 'theta'. The brace will be welded to the wall and then bolted into the concrete through a baseplate. I was going to design my column for buckling and my bolts and baseplate weld for shear. Wanted to see if my forces were correct. Thanks in advance.

 

[BAretired]

Doesn't look right to me.
A = Bottom of wall, B = top of wall
C = Bottom of brace, D = top of brace (and intersection with AB)
E is below D on ground level
F is point on CD which is nearest point A (i.e. AFC is a right angle)

height of wall AB = h
brace length = b
weight of wall = W

Overturning Moment (OTM) about A = (Wh/2)sin(phi)
Height DE = b.sin(theta)
AE = b.sin(theta)tan(phi)
EC = b.cos(theta)
AC = AE + EC = b[cos(theta) + sin(theta).tan(phi)]
AF = AC sin(theta) = b.sin(theta)[(cos(theta) + sin(theta).tan(phi)]

Brace force F = OTM/AF = (Wh/2(sin(phi)/b.sin(theta)[(cos(theta) + sin(theta).tan(phi)]


For some reason, I couldn't get the Extended characters to work properly so I wrote the Greek letters as text.




BA

 

[BAretired]

If you want to include wind on the wall, add w.h^2/2 to the OTM.

w is the uniform wind load on the wall

BA

 

[BAretired]

Doesn't look right to me.
A = Bottom of wall, B = top of wall
C = Bottom of brace, D = top of brace (and intersection with AB)
E is below D on ground level
F is point on CD which is nearest point A (i.e. AFC is a right angle)

height of wall AB = h
brace length = b
weight of wall = W

Re-writing using T for theta and P for phi:
Overturning Moment (OTM) about A = (Wh/2)sinP
Height DE = b.sinT
AE = b.sinT.tanP
EC = b.cosT
AC = AE + EC = b(cosT + sinT.tanP)
AF = AC sinT = b.sinT(cosT + sinT.tanP)

Brace force F = OTM/AF = (Wh/2)sinP/b.sinT(cosT + sinT.tanP)


BA

 

[SlideRuleEra]

I'll offer a graphical approach to the problem for review, see the attached marked up sketch:

W[sub]T[/sub] acts through the wall's center of gravity, not necessarily the attachment point for the brace.
Sum moments about the wall's base to get the magnitude of R. Of course, R is perpendicular to the wall.
F[sub]C[/sub] has to act along the length of the brace.
V keeps the brace from sliding.
The trigonometry to get the answer should be straightforward.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[BAretired]

Something amiss there, SRE. Your vectors do not look quite right.

BA

 

[BAretired]

If the distance between A and C in the attached diagram is specified, reactions at A and C can easily be found. The load W is 1 unit from A and 4 units from C so the vertical reactions are 0.8W and 0.2W at A and C respectively.

so Fc = Vc/sinT
and Hc = Vc/tanT
and Ha = Hc (not true if wind is included)

BA

 

[SlideRuleEra]

Thanks, BA, that helps a lot.
Do believe that I have gotten the vector's corrected (see attached file).
R remains perpendicular to the wall.
F[sub]C[/sub] continues to act along the length of the brace.
A new vector, called M, acts along the height of the wall. If the OP is designing a real brace, the value of M is the sliding force the connection between the brace and the wall has to resist.

As a check of my diagram, I printed your drawing, computed the angles, and scaled the lengths. After a lot of trigonometry, got the same reactions and other values you do.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[BAretired]

SlideRuleEra,
I can see how the vectors M and R relate to Fc. Is the force Fc found graphically from the Wc vector? Offhand, I can't see how that could be done.

In any case, there is not enough information given by the OP to solve the problem. Without specifying a few additional parameters such as wall height and brace length, there are infinitely many answers.

BA

 

[reesecc66]

Thanks guys, very helpful and much appreciated.

Here's a little bit more info and my numbers.

W: 30,000#
AC: 22ft
DC: 36ft
AB: 67ft
AD: 33.5ft
T: arccos[(DC^2-AD^2+AC^2)/(2.DC.AC)] = arccos[(36^2-33.5^2+22^2)/(2.36.22)] = 65.47*
DE: DCsin(T) = 32.75ft
P: arccos(DE/AD) = arccos(32.75/33.5) = 12.15*
AF: ACsin(T) = 22sin(65.47) = 20ft
OTM: (W(AB)/2)sin(P) = (30(67)/2)sin(12.15) = 221.5k-ft
Fc: OTM/AF = 221.5/20 = 11kips

I am a little confused at how you got Fc. I'm not sure how AF factors in.

Thanks again.

 

[SlideRuleEra]

BA - Yes, I did calculate F[sub]C[/sub] using a semigraphical solution. Did sneak in a method from my Mechanical Engineering background (Kinematics and Dynamics of Machinery). Recognized that the problem has an exact ME analog - a slider-crank mechanism:

For reference, I have attached the marked up working copy of your helpful sketch. The letters below are from your sketch, not the above image.

Calculated the angles using your block gird. Phi = 26.6 degrees, Theta = 40.6 degrees.

Using a scale, measured the dimensions for the "wall":
Wall height (AB) = 11 units
Center of Gravity = 5.50 units.
Location of Brace (AD) = 8.30 units.

I considered point A to be an "instant center of rotation". Temporarily ignored DC and replaced it with vector R, perpendicular to the wall. Summed moments about A to solve for R.

After that, it was pretty much "plug and chug" using geometry and trigonometry. Got answers accurate to about 0.6%+, compared to your exact solution... not to bad for a semigraphical approach. BTW, I am not recommending the above way of solving the problem (with the possible exception of determining M). Just wanted to see if I could do it. Your elegant solution is much better.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[BAretired]

OTM: (W(AB)/2)sin(P) = (30(67)/2)sin(12.15) = 221.5k-ft
Fc: OTM/AF = 221.5/20 = 11kips

I am a little confused at how you got Fc. I'm not sure how AF factors in.

OTM = 211.5 foot-kips
Fc = OTM/AF = 10.6 kips

Fc is the axial force in the brace. It has a moment about point A of Fc.AF which must be equal to the overturning moment about A.

When you consider only vertical load, it may be easier to calculate Fc from the reaction at C but if you want to add wind perpendicular to the wall, it is easier to use OTM/AF.

Assume there is a wind pressure of w acting normal to the wall in the sloped position. Then OTM(wind) = wL[sup]2[/sup]/2 where L is 67 feet. So the brace force due to wind is wL[sup]2[/sup]/(2*20) = 112.2w.

If braces are spaced 10 feet apart and wind pressure is 20 psf, w = 200#/'
So Fc from wind alone is 22,445# or 22.5k, substantially more than the force due to the dead load of the wall.

Another point to consider is the strength and deformation of the wall. It has a cantilever length of 33.5'. Are you sure you want to attach the brace at mid-height of the wall?

BA

 

[BAretired]

SlideRuleEra,
I got used to solving trusses using the graphical method when I started in Toronto in 1955. The plan checkers at that time insisted on a force vector diagram for each truss to be included on the drawings. It was a good way of solving for member forces and the accuracy was plenty good enough for engineering purposes.

Even now, when I want a quick and dirty solution to a problem, I scratch out a vector diagram on whatever is available and measure the length of the resulting vectors.

BA

 

[reesecc66]

BA, thanks for the explanation. Also, I just looked at my calcs and did get the 211.5 for OTM and accidentally wrote down 221.5. No biggie though.

As far as the cantilever goes, the "wall" is technically a bowed out banana shaped wall. The peak of the bow is right about h/2 and is connected at the base and the top of h. I was assuming a fully leaning wall as a worst case scenario and after running the calcs had more than enough built into my section.