REINFORCED-CONCRETE STRUCTURES
3.3 DEVELOPMENT LENGTH
2. Between
b dw 兹ƒ⬘c and Vs⬉ 4兹ƒ⬘c wb d
stirrup spacing must be less than or equal to d / 2 .
3. If Vs⬎ 4b dw 兹ƒ⬘c but Vs ⬍ 8b dw 兹ƒ⬘c the maximum stirrup spacing is d / 4
The absolute limit is
Vs⫽ Vs,max⫽ 8b dw 兹ƒ⬘c
Practical Note. Stirrup sizes are normally #3, #4, and #5. Anything larger is for a megaproject. As mentioned earlier, do not use it unless you are de-signing a bridge.
The best ƒyfor stirrups is 40,000 psi. At this strength, stirrups bend well.
For harder steel, ƒy ⫽ 60,000 or higher, stirrups tend to be brittle and might crack around hooks. Then you have development length for the stirrups, which might fail. Minimum spacing: 3 in.
3.3 DEVELOPMENT LENGTH
By now it should be obvious to the reader that there is a potential danger of bars pulling out. The force in the bars can be much larger and closer to the support of simply supported beams than anticipated by a conventional engi-neering approach—a shift in the bending moment diagram.
When the 1963 ACI code made strength design the preferred method of analysis for bending, there was no ultimate strength methodology—limit strength as it became known later—to calculate shear strength or development strength. Both are essential parts of beam design. At that time the provisions of the code and the technical literature were only good for pure bending, that is, for analysis of a section subjected to bending. It did not take into consid-eration the presence of other forces such as shear and compression that are inevitably combined with the external moment.
In a simply supported beam loaded with a uniformly distributed load there is only one cross section that receives a pure moment, while mathematically speaking, there are infinite cross sections subjected to both bending and shear.
What methods were then used to evaluate the limit-state effects of these forces? For shear, a parabolic stress distribution matched with a uniform stress distribution extending from a working stress evaluated the neutral axis to the
46 REINFORCED-CONCRETE STRUCTURES
centerline of the tensile reinforcement. Still, these stresses where applied to a nonexisting part of the cross section split by a crack. That is, stresses were applied over a void and results evaluated on a hypothesis that contradicted the laws of physics. Evaluation of resistance to bond failure of bars embedded in concrete was equally not adapted to ultimate strength or strength require-ments.
A homogeneous material—reinforced concrete is anything but homoge-neous—was assumed subjected to differential tensile forces at the location of the rebar. This means a continuum of infinitely small elements was assumed capable of forming a continuous chain with the tendency to shear off along a plain parallel to the axis of the beam. Again, splitting of concrete by nu-merous diagonal cracks was not considered. These cracks actually reacted against slippage under the high bearing pressure created by lugs of rebars that would either deform or pull out.
In 1963 the author presented the unique findings of research about the relationship of those hitherto ignored forces that affected the limit-state re-sponse and strength of reinforced concrete to the Wiesbaden Symposium on Shear of the European Concrete Committee. The paper contained the first mathematical model and rational method that showed by actual design ex-amples how to combine shear and moment in one ultimate strength operation, that is, to determine the amount and disposition of rebars for shear and flex-ural resistance. It was also the first time that it was noted that the stirrups take part in the resisting moment.
The author also derived mathematically the amount of moment sharing between stirrups and main tensile reinforcement. He named it shift in the moment diagram. The concept was quoted by others without crediting the original source and was eventually incorporated into ACI 318.
Application to Design of Structural Members
Consider a continuous beam with a uniformly distributed load. The following steps will determine the total shear reinforcement:
1. Divide the length over which shear reinforcement is required into equal increments; then calculate each cross section separately for the given static conditions.
2. Provide each zone with the reinforcement required for the worst case in that zone, that is, the biggest area of steel at the smallest pitch ob-tained from the two results of the calculations made on the two bound-aries of the zone—1-1 and 2-2 of zone II in Figure 3.7.
Thus a typical system of shear reinforcement will be as in Figure 3.7, where heavy lines represent stronger shear reinforcement and thinner lines lighter
www.4downloader.ir
3.3 DEVELOPMENT LENGTH 47
Figure 3.7 Typical system of shear reinforcement. Reproduced from the author’s paper presented at the European Concrete Committee Symposium on Shear.
reinforcement. The area of each pair of stirrups is large in zone II because this coincides with the point of zero bending moment. No shear resistance is provided by the bending moment and the intensity of the shear force is still high.
The pitch of the shear reinforcement is quite large, though, because the inclination of the crack is 45⬚. In zone III the area of stirrups is the smallest.
This is due to the decrease of the shear forces and the increased shear resis-tance of the compression zone caused by the increasing bending moment.
48 REINFORCED-CONCRETE STRUCTURES
The inclination of the cracks varies with the type of load, which makes it difficult to give actual figures for the spacing of stirrups. In beams reinforced with stirrups large secondary cracks tend to form at the final stages of loading, which also weakens the structure. It follows that it is more practical to com-bine stirrups with bent-up bars.
We will now present how the ACI addresses the development length re-quirement. Given the ACI equation (12-1) with Ktr⫽ 0,
3 ƒy ␣␥
ld⫽ 冉40兹ƒ⬘c [(c⫹K ) /d ]tr b冊db (12-1) we have
ƒ ƒ
ld 3 y ␣␥ 3 y ␣␥
⫽ ⫽
db 40兹ƒ⬘c [(c⫹K ) /d ]tr b 40 兹ƒ⬘c1.5 /db
We now develop a #7 bottom bar forƒ⬘c ⫽ 3.0 ksi and ƒy⫽ 40 ksi:
ld ⫽ 3⫻ 40000 (1.0)(1.0)(1.0)(1.0)⫽ 32 in. 42-in. top bar,–78 in.
db 40⫻ 54.77 1.5 / (7 / 8)
That is, the development length ld⫽ 32 in. for a -in. diameter bottom bar.–78
To develop a #8 bottom bar forƒ⬘c⫽ 3.0 ksi and ƒy⫽ 40 ksi, ld 3⫻ 40000 (1.0)(1.0)(1.0)(1.0)
⫽ ⫽ 36.5⬇ 36 in. 47-in. top bar, #8
db 40⫻ 54.77 1.5
To develop a #6 bottom bar forƒ⬘c⫽ 3.0 ksi and ƒy⫽ 40 ksi, ld (1.0)Bott(1.0)Epox(0.8)(1.0)Ltw
⫽ 54.77
db (1.5⫹ 0) / 0.75
⫽ 27.38⬇ 28 in. for #6 bottom bar
That is, 36-in. top bar, #6.
Redo the above for ƒy⫽ 60 ksi (ƒ⬘c ⫽ 3.0 ksi):
#7 Bottom bar 48 in. 62-in. Top bar
#8 Bottom bar 55 in. 71-in. Top bar
#6 Bottom bar 41 in. 54-in. Top bar