Strut and tie question for beam with both uniform and concentrated loads

[cliff234]

I am design a deep beam per ACI 318-08, Appendix “A”, and I have a question. I searched through the archives of this forum but did not see discussion of the question that I have. Attached is an illustration showing a beam configuration similar to the actual beam that I am designing. The beam has a span/depth ratio greater than 4, but it is still classified as a deep beam because there is a significant concentrated load located closer than twice the beam depth to one of the supports.

My question is this, how do I compute the strut force? Is the strut force calculated based only on the 1,143k concentrated load, or is the strut force calculated based on the 2,000k reaction? If the concentrated load did not exist, then the beam would not have to be designed using a strut-and-tie (because L/d>4). But because of the concentrated load, I need to look at a strut-and-tie model. Is it reasonable to assume that the strut forms solely due to the concentrated load and I can compute the strut force based only on that 1,143k load? The difference is significant – a strut force of 1,414k versus 2,828k. A 2,828k strut force will require significantly more tie steel. The difference primarily affects the quantity of tie steel and anchorage of that tie steel in the nodal zone. I will design the bottom bars in the beam for a moment due to the uniform load plus a tie force due to the concentrated load. I will make sure that the beam has enough shear strength support the 2,000k reaction. I will provide headed anchors on the ends of the bottom (tie) bars, but would only need to consider development of those bars within the nodal zone to resist a tie force resulting from the strut force due to the 1,000k concentrated load.

Carrying this logic further, what if the concentrated load was only 100k (versus 1,000k)? Would I have to design the strut for a force of 141k or 141k + 1,414k = 1,555k? It seems odd that a relatively small concentrated load would suddenly require me to calculate a big strut force – with most of that force coming from a uniform load that would otherwise not be creating a strut.

Again - I am not short-changing design of the beam for shear strength, nor am I short-changing flexural strength. What I am questioning is the need to consider the uniform load increasing the strut force, since that uniform load, where it the only load on the beam, would otherwise not create a strut. The consequence of the increased strut force is a significantly greater quantity of tie steel.

Thanks!

 

[MacGruber22]

I am thinking superposition of Bernoulli (linear strain) and non-linear strain effects is valid, because material linearity is still presumed; and thus the area of steel required across your strut is resisting a force somewhere between 1,414k and 2,000k. I am not 100% sure, but I think that you can design your beam for typical flexure and shear based on the distributed load - then calculate the additional steel required as a result of the STM d-region developed by the point load. Your logic regarding reducing the point load, yet not appreciably reducing the strut force is exactly what I was thinking.

It just doesn't seem correct that you throw all of the your reaction load into the STM d-region. Now, if your span to depth ratio was less than 4, than the answer would be to use the sum total of the distributed and concentrated load for the the strut. Or, maybe I am wrong and the concentrated load messes all of that up within the d-region no matter what. I am curious if someone can answer confidently and with a reference. The 318-14 seems to be void of guidance.

"It is imperative Cunth doesn't get his hands on those codes."

 

[IDS]

It's not a case of the strut developing or not developing, it's two different ways of analysing the same loads, so I'd say you should either design for the total load using strut and tie, or you should design for the total load using flexure and shear (if the span and load arrangement allows this), but not combine the two.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[MacGruber22]

The University of Maryland:

1. Isolate D-regions
2. Complete the internal stresses on the
boundaries of the element
3. Subdivide the boundary and compute
the force resultants on each sub-length
4. Draw a truss to transmit the forces from
boundary to boundary of the D-region
5. Check the stresses in the individual
members in the truss

http://best.umd.edu/publications/stm.pdf

Sub-dividing might get your strut force to a more reasonable value

"It is imperative Cunth doesn't get his hands on those codes."

 

[cliff234]

Thanks for the feedback. I am still confused. An interesting slide in the referenced University of Maryland presentation was one that said this,

1. The successful use of the STM requires an understanding of basic member behavior and informed engineering judgment
2. In reality, there is almost an art to the appropriate use of this technique
3. The STM is definitely a design tool for thinking engineers, not a cookbook analysis procedure
4. The process of developing an STM for a member is basically an iterative, graphical procedure

In other words, there is no one right answer. That is usually the case in structural engineering. I just want to be sure that I am not completely missing something. It seems odd that tie forces have to be calculated in bottom reinforcing steel that would have otherwise been neglected were it not for the concentrated load near the end of the span. It is as if we have to treat the bottom bars as unbonded reinforcing steel anchored within the nodal zone. That makes sense for a tie force countering a strut load resulting from the concentrated load. It does not make sense for what seems to be a fictitious strut force from the uniform load. I'll have to think about this some more. Thanks.

 

[BAretired]

It makes no difference to the area of bottom steel which way you do it but it makes a difference where you anchor it.

The moment at the concentrated load is 2000(5) - 50(5)²/2 = 9375'k. This creates a tension in the bottom steel of 9375/5 = 1875k at the location of the concentrated load.

The uniform load in the last 5' of beam is 50(5) = 250k. Thinking of the last 5' as a trolley, the uniform load in the end 5' can be thought of as two 125k concentrated loads, one at each end of the trolley, increasing the effective concentrated load by 125k.

Whether you use the strut and tie model or the conventional model, the force required in the bottom steel is 1875k but using the strut and tie model, you anchor it at the end of the beam rather than relying on bond.

BA

 

[cliff234]

I like the trolley analogy BAretired. That adds clarity. Thanks!

 

[BAretired]

You're welcome cliff234.

BA