by clicking the arrows at the side of the page, or by using the toolbar.
by clicking anywhere on the page.
by dragging the page around when zoomed in.
by clicking anywhere on the page when zoomed in.
web sites or send emails by clicking on hyperlinks.
Email this page to a friend
Search this issue
Index - jump to page or section
Archive - view past issues
Concrete In Australia : December 2013
50 Concrete in Australia Vol 39 No 4 DISCUSSION PAPER 13.1.2 .7 Part 5 AS5100) as necessary. Note that a larger value of θ means that more shear reinforcement will be required. In Colin’s example, anchorage becomes less critical as the distance from the support increases and as a result θ decreases. That is if cracking parallel to the compression struts were to be visible, these cracks would, initially at least, get less steep as their distance from the support increased. This is confirmed in Colin’s Figures 5, 6 and 7. However, these cracks are typically observed to become steeper as their distance from the support increases. This would indicate that more anchorage is typically available at the support than has been utilised in Colin’s example. In the AS/RTA April 2011 Interim Version of Part 5 of the Australian Bridge Code AS5100 – 2004, minimum values of θ have been stipulated as 30° for the design of shear in RC beams, and 20° for the angle between a strut and a tie in strut-tie analysis. Since the design of shear reinforcement in RC beams may be presented as a strut-tie analysis, this provides a degree of ambiguity. In the following I have considered both 20° and 30° as possible values of θmin . Initially, I have determined the anchorage force required as a proportion of the maximum flexural reinforcement force to permit the adoption of θmin . This will be influenced by the load distribution on the beam, but for simplicity let us confine ourselves to a uniformly distributed load “w u ”. Then, the mid-span flexural tension is Tmax =w u L2/8d And the maximum shear force of Vu = w u (L/2–dcotθ),is associated with an anchorage force Tanc such that: (1) = 0.1992 for (d/L) = 0.05 and θ = 20° = 0.1432 for (d/L) = 0.05 and θ = 30° That is for a UDL, θ at the support may be taken as 20° provided the anchorage capacity is > 0.1992 x the maximum flexural reo force at mid-span, and for a UDL, θ at the support may be taken as 30° provided the anchorage capacity is > 0.1432 x the maximum flexural reo force at mid-span. Colin’s example requires θ at the support = 40.6° with (d/L) = 0.102 which, from the above, would require Tanc / Tmax = 0.181, compared with 0.25 provided in the example. Therefore, my original expectation that the degree of anchorage would fully utilise the available capacity is not confirmed. This leaves the application of the load at the bottom of the beam as a possible explanation. Load application at the bottom of the beam may be seen to have two adverse effects. (a) the effective value of S is reduced by wu , and (b) the value of Vu is increased by wu d cot θ. However these amount to the same effect seen in different ways. Taking this effect into account, equation (1) becomes (2) = 0.238 for (d/L) = 0.102 and θ = 40.6° This is reasonably consistent with the value of 0.25, considering the totally independent derivation, and that I have ignored minor variations in “d” along the beam caused by variations in the depth of the compression block. Then, with θ at the support based on the designer’s choice, rather than determined by the anchorage available, the value of θ at intermediate locations between the support and mid-span may be determined from my Figure1, where (3) The variation of θ along the span is therefore influenced by the variation of S. In Table 1, Cases 1 and 3 represent a constant value for S throughout the span, and Cases 2 and 4 represent a linear decrease of 50% in the value of S, between the support and mid-span. The quarter point is taken as a = L/4, and The 3/8 point is taken as a = L/8. Case 1 (θ1 = the quarter point value of θ, and S is constant) Case 2 (θ1 = the quarter point value of θ, and S reduces linearly to 50% at mid-span) Case 3 (θ1 = the 3/8 point value of θ, and S is constant) Case 4 (θ1 = the 3/8 point value of θ, and S reduces linearly to 50% at mid-span) ,or 49-56 - Discussion.indd 50 49-56 - Discussion.indd 50 25/11/13 4:15 PM 25/11/13 4:15 PM