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Torsion vs Shear Cracks Stirrups Resistances Behavior

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Structural
Jul 16, 2018
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In diagonal shear crack.. when this happens.. the stirrups were being engaged so you still have shear capacity by Vc + Vs.. Is it not torsion cracks are similar where after it occurs.. torsion is resisted by the stirrups? But in ACI 318 there is this passage "At torsional cracking, however, a large twist occurs under an essentially constant torsional moment, resulting in a large redistribution of forces in the structure (Collins and Lampert 1973; Hsu and Burton 1974)"

Were they referring to beam without any torsion stirupps or reinforcement such that a large twist can still occur or does this happen even with stirrups? How? In diagonal shear crack, you don't have happening because the stirrups resisted them.

Let's review the concept of Torsion cracks from Mcgregor textbook:

"Pure Torsion

When a concrete member is loaded in pure torsion, shearing stresses, and principal stresses develop as shown in Fig. 7-8a and b. One or more inclined cracks develop when the maximum principal tensile stress reaches the tensile strength of the concrete. The onset of cracking causes failure of an unreinforced member. Furthermore, the addition of longitudinal steel without stirrups has little effect on the strength of a beam loaded in pure torsion because it is effective only in resisting the longitudinal component of the diagonal tension forces.

A rectangular beam with longitudinal bars in the corners and closed stirrups can resist increased load after cracking. Figure 7-12 is a torque-twist curve for such a beam. At the cracking load, point A in Fig. 7-12, the angle of twist increases without an increase in torque as some of the forces formerly in the uncracked concrete are redistributed to the reinforcement. The cracking extends toward the central core of the member, rendering the core ineffective. Figure 7-13 compares the strengths of a series of solid and hollow rectangular beams with the same exterior size and increasing amounts of both longitudinal and stirrup reinforcement [7-4]. Although the cracking torque was lower for the hollow beams, the ultimate strengths were the same for solid and hollow beams having the same reinforcement, indicating that the strength of a cracked reinforced concrete member loaded in pure torsion is governed by the outer skin or tube of concrete containing the reinforcement.

After the cracking of a reinforced beam, failure may occur in several ways. The stirrups, or longitudinal reinforcement, or both, may yield, or, for beams that are overreinforced in torsion, the concrete between the inclined cracks may be crushed by the principal compression stresses prior to yield of the steel. The most ductile behavior results when both reinforcements yield prior to crushing of the concrete."
 
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In diagonal shear crack.. when this happens.. the stirrups were being engaged so you still have shear capacity by Vc + Vs.. Is it not torsion cracks are similar where after it occurs.. torsion is resisted by the stirrups? But in ACI 318 there is this passage "At torsional cracking, however, a large twist occurs under an essentially constant torsional moment, resulting in a large redistribution of forces in the structure (Collins and Lampert 1973; Hsu and Burton 1974)"

Were they referring to beam without any torsion stirupps or reinforcement such that a large twist can still occur or does this happen even with stirrups? How? In diagonal shear crack, you don't have happening because the stirrups resisted them.

It's kind of the same concept as flexural cracking moment in a beam/slab: once cracking happens.....then reinforcement takes over. You ignore torsion until you reach that limit. (I.e. about a quarter of the cracking torsion.) The reinforcement doesn't play much of a role until that happens.
 
If that's the case.. how do you explain this passage ""At torsional cracking, however, a large twist occurs under an essentially constant torsional moment, resulting in a large redistribution of forces in the structure (Collins and Lampert 1973; Hsu and Burton 1974)" Well. If there is still stirrups resisting it.. how can a large twist occur?
 
If there is still stirrups resisting it.. how can a large twist occur?

Quite simply because prior to cracking, you have the entire (uncracked) cross section resisting it and the stirrups are not engaged. If you look at the postcracking stiffness equation, i.e. GK[sub]cr[/sub], it's a portion of the cross section involved.....and typically the ratio of cracked to uncracked stiffnesses are quite low. Usually, I am use to seeing GK[sub]cr[/sub]/GK[sub]g[/sub] values between 0.25 and 0.05. (Typically on the lower end of that.)

 
Ok thank you. what do you think about the ratio of cracked and uncracked stiffness for the diagonal inclined shear crack failure?
 
what do you think about the ratio of cracked and uncracked stiffness for the diagonal inclined shear crack failure?

I'm afraid I don't understand your question.
 
I mean in the shear diagonal inclined failure mode that is not torsion failure.. what is the difference between post cracking stiffness and pre cracking stiffness of the beam? Is it somewhat still similar? At diagonal inclined cracking (not Torsion related), there is still Vc or aggregate interlock engaged with stirrups holding them. So can you say in diagonal shear crack, the stiffness before and after cracking is still the same (in contrast to Torsion crack where the stiffness before and after cracking differs by much maybe because in Torsion crack, aggregate interlock is lost..)
 
I mean in the shear diagonal inclined failure mode that is not torsion failure.. what is the difference between post cracking stiffness and pre cracking stiffness of the beam? I mean in the shear diagonal inclined failure mode that is not torsion failure.. what is the difference between post cracking stiffness and pre cracking stiffness of the beam? Is it somewhat still similar? At diagonal inclined cracking (not Torsion related), there is still Vc or aggregate interlock engaged with stirrups holding them. So can you say in diagonal shear crack, the stiffness before and after cracking is still the same (in contrast to Torsion crack where the stiffness before and after cracking differs by much maybe because in Torsion crack, aggregate interlock is lost..)

I'm still not 100% certain I follow you. Flexural stiffness and torsional stiffness (both before and after cracking) are well defined numerically. The cracking induced in torsion occurs in a spiral on all faces (as opposed to shear). While the shear cracking probably ultimately contributes to getting you to the cracked level (for torsion).....it's hard to say how much. There might be a way to back out of the codes with some number.....but it would be fairly meaningless.
 
I wasn't referring to combined torsion and shear failure.. but shear failure alone.. specifically diagonal inclined shear failure. How do you characteristize the prior and post stiffness of beam in diagonal inclinde shear failure (without taking into account torsion so try to remove the torsion failure in our discussion now). Maybe there is similiar stiffness since at shear diagonal failure, aggregate interlock is engaged (by Vc) and dowel action and compression block in addition to stirrups?
 
I wasn't referring to combined torsion and shear failure.. but shear failure alone.. specifically diagonal inclined shear failure. How do you characteristize the prior and post stiffness of beam in diagonal inclinde shear failure (without taking into account torsion so try to remove the torsion failure in our discussion now).

I don't know. I would think the stiffnesses would be close because it doesn't take too much slip to engage the re-bar. (In many tests I've seen, we are talking less than 0.1 inches.) For unreinforced sections, a good percentage (of the shear) is carried in the compression zone....so there probably wouldn't be much change there either.
 
Going back to pure torsion. Is your "slip" got to do with the concept of "angle of twist" ... were you talking about the angle of twist being related to the stiffness such that in cracked concrete, the angle of twist is larger in value hence stiffness poorer? What is the technical word for your "slip" if it is not the "angle of twist"?
 
Going back to pure torsion. Is your "slip" got to do with the concept of "angle of twist" ... were you talking about the angle of twist being related to the stiffness such that in cracked concrete, the angle of twist is larger in value hence stiffness poorer? What is the technical word for your "slip" if it is not the "angle of twist"?

Slip may play a role in "angle of twist"......but as far as I know: not a significant one. The rotational stiffness of a concrete beam after a certain level of torsional cracking is overwhelmingly controlled by the fact you have a reinforced core tube of a beam resisting a torsion where before it was the entire (uncracked) cross section (i.e. K[sub]cr[/sub] vs. K[sub]g[/sub]).
 
Do you have any reference for your comment? It seems to conflict with my reference by Mcgregor, he said the "once torsional cracking has occurred, the concrete in the center of the member has little effect on the torsional strength of the cross section and hence can be ignored" (see also figure in the attachment):

"This model for the torsional strength of beams combines the thin-walled tube analogy from Fig. 7-6 with the plastic-truss analogy for shear presented in Section 6-4. This gives a mechanics-based model of the behavior that is easy to visualize and leads to much simpler calculations than the skew bending theory.

Both solid and hollow members are considered as tubes. Test data for solid and hollow beams in Fig. 7-13 suggest that, once torsional cracking has occurred, the concrete in the center of the member has little effect on the torsional strength of the cross section and hence can be ignored. This, in effect, produces an equivalent tubular member.

Torsion is assumed to be resisted by shear flow, q, around the perimeter of the member as shown in Fig. 7-15a. The beam is idealized as a thin-walled tube. After cracking, the tube is idealized as a hollow truss consisting of closed stirrups, longitudinal bars in the corners, and compression diagonals approximately centered on the stirrups, as shown in Fig. 7-15b. The diagonals are idealized as being between cracks that are at an angle generally taken as 45° for reinforced concrete.

The derivation of the thin-walled tube/plastic space truss method as used in the ACI Code was presented and compared to tests in [7-10]. In this book, the analogy is referred to as the thin-walled tube analogy because this is the terminology used in the derivation of Eq. (7-9) in mechanics of materials textbooks. The walls of an equivalent tube for a concrete member are actually quite thick, being on the order of one-sixth to one-quarter of the smaller side of a rectangular member.
 
 https://files.engineering.com/getfile.aspx?folder=d53326a5-65e5-4bfc-8b3f-f991b444ce61&file=torsion_dynamics.JPG
Yes, it is more accurate to think of it as a tube rather than a "core" (after cracking).
 
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