Development Length of Steel in Shear

[sybie99]

I have read an earlier thread on the topic, but the matter remains unclear.

Codes specify that the main tension rebar crossing a shear plane considered must be fully anchored (developed) either side of this plane. My understanding is that this is to keep the concrete together so that it can resist shear by means of friction. But if one provides shear links (stirrups) to increase shear capacity, why must the main bars be fully developed? If the concrete cannot resist the shear stress by friction the shear is resisted directly by the shear links which must shear off for the beam to fail.

The very reason one places shear links into the beam is because the shear stress exceeds capacity of the concrete section in friction. So for these shear links to "start working" the shear stress in the concrete resisted by friction must have been exceeded, so there is no need any longer to rely on the shear resistance is provided by the shear links.

 

[rapt]

Sybie,

No, it is to satisfy the logic of the Truss Analogy. Think of it as a normal truss with tension and compression cords and diagonal compression struts. The tension cord must develop its tension force at the point where it meets the diagonal strut.

 

[hokie66]

I think you are confusing what is now called "shear friction" with beam shear which is more correctly diagonal tension. In order to understand diagonal tension, consider a truss analogy. The stirrups in the truss carry tension, the concrete carries web compression, and the reinforcement on the tension side is the flange. Like a truss, at an end condition, the force in the flange is small in relation to the force at midspan, so the reinforcement anchorage conditions are less than full development.

 

[hokie66]

I posted before seeing rapt's post, but no matter, we agree, I think.

 

[Archie264]

'Enjoying watching you two great minds thinking alike.

To hopefully add to what you posted, it might help to consider that in an ideal world the stirrups would be placed at a 45 degree angle to be best positioned to take said diagonal shear. Since that is impractacle they are placed vertically at close enough intervals to accomodate the shear/diagonal tension.

 

[hokie66]

There was a time when much of the shear steel was placed on a diagonal. This involved bending some of the top bars to intercept the diagonal tension cracks, and even in some cases, leaning the stirrups. We no longer do that, as it is labor intensive.

 

[Archie264]

Very interesting, I didn't know that. I've seen longitudinal bar bends but I always assumed that was for the transition from positive to negative moment and for anchorage in general. I never considered that it could also take the shear. And didn't know that about the diagonal stirrups. Thanks.

 

[sybie99]

Thanks for the replies, I understand it better now.
My question would now be where does the code, in my case BS get the anchorage at support requirements from? BS needs 12x bar diameter past centre of support, but the support width of a precast beam for example needs only be enough to carry direct bearing stresses. Also the 12x bar diameter is hard to understand. If you use say ten 10mm bars with 12x bar diameter anchorage you are fine, but now if you use ten 12mm bars (almost double area) with 6 x bar diameter anchorage (so you can resist same tensile force) you are not satisfying code requirement. Doesnt seem right

 

[Lion06]

Is there any practical benefit to placing the bars perpendicular to the diagonal tension cracks in the shear zone? I can see this as helping to hold the cracks tighter, but not as efficient in resisting the actual shear force. Because the shear force is vertical it makes sense that the most efficient layout for the stirrups is vertical, at least in my mind.

 

[hokie66]

Can be done either way, but think of the truss analogy. The shear in a steel truss is resisted by a series of web members which are in tension and compression. The same holds for a concrete beam, but the tension members must be steel. If the steel webs are inclined as for a Pratt truss, they are in tension and thus are optimal in resisting the diagonal tension. We actually place the stirrups vertically, so the type truss approximates a Howe truss, in which the diagonal tension is inclined to the actual member, so the member at that position would have a larger area, but with a decreased length. Six of one, half dozen the other...

 

[rapt]

Sybie,

Codes provide simplified solutions that they think are logical but do not tell the full story (and often have limitations. For example, the BS8110 rule on this is far too simplified and assumes for small loads you will use small bars for a start as well as not necessarily providing the termination development really required. If you want to understand this better, read Eurocode (which has supposedly replaced BS8110 by now anyway)on it. Especially 9.2.1.3 and associated clauses.

 

[a2mfk]

happen to have a scan of this in my library, in the ballpark of this discussion