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Design per ACI 318-02 Appendix A, US Customary Unit Deep Beam (1) | Deep Beam (2) | Dapped-Beam End | Single Corbel | Double Corbel
A double corbel projecting from a 14 in.
14 in. column is to be designed to support precast beam reaction forces at 6 in. from the face of the column. The factored vertical load to be carried is 61.8 kips. Due to restraint of beam creep and shrinkage deformations, a factored horizontal force of 14.3 kips is assumed to develop. The upper column carries a factored axial load of 275 kips. Summary of ACI 318-02 Appendix A View Print-Friendly Version of this Page View PDF Version of this Page
The concrete strength is 4 ksi (normal density), and the yield strength of reinforcement is taken as 60 ksi. Determine the Bearing Plate Dimensions: Choose a 12 in. 6 in. 1/2 in. bearing pad. The bearing plate area is 12(6) = 72 in.2 and the bearing stress is 61.8(1000)/72 = 858 psi. Since this is less than the bearing stress limit, i.e. the bearing size is adequate.
Choose the Corbel Dimensions: Choose a depth at the column face of 18 in. The ACI Code requires that the depth at the outside of the bearing area is at least one-half of the column face depth. Therefore, select a depth of 10 in. at the free end of the corbel. The selected dimensions for the corbel are summarized in Figure 1.
Figure 1 (Click here to view a larger image) Establish the Strut-and-Tie Model: To allow for load eccentricities and erection tolerances, consider the vertical load to be placed 1 in. toward the edge of the corbel from center of bearing plate. Thus, the position of vertical load is 1 + 6 = 7 in. from the face of column. The geometry of the assumed truss is given in Figure 2. The center of the tie is assumed to be located 2 in. from the top of the corbel. Thus, d = 18 – 2 = 16 in. The horizontal strut BB’ is assumed to lie on the horizontal line passing through the sloping end of the corbel. The location of strut CB centerline can be found by calculating the required compressive force in strut CB, NCB, and the strut stress limit to obtain the strut width a. The strut CB force is and the limit stress on the nodal zone B (also strut CB) is Thus, we have
This fixes the geometry of the truss and means that member AB has a horizontal projection of 0.46 + 7 + 5.58/2 = 10.25 in.
Figure 2 (Click here to view a larger image) Determine the Required Truss Forces by Statics: The required forces in all the members of the truss are given in the following table. Note that positive indicates tension, negative compression. Member Force (kips)
AA' +54.1
AB -73.5
BB' -39.8
CB -199.3
BD -137.5
Design the Tie:
The area of reinforcement required for tie AA’ is
and the
minimum area of reinforcement is Choose 6 No. 4 bars,
Check the Struts: The struts will be checked by computing the strut widths and checked whether they will fit in the space available. The stress of the diagonal strut AB is limited to Hence, the required width for strut AB is
Select a width of 3.00 in. for strut AB.
The stress of the vertical struts BD and CB and horizontal strut BB’ is limited to Hence, the required widths for strut BD and BB’ are and respectively. Choose 4 in. width for strut BD and 2 in. width for strut BB’. The required width for strut CB is equal to a, i.e. 5.58 in.
As shown in Figure 3, all the strut widths fit into the outline of the corbel region. Thus, this solution is accepted. Note that Figure 3 also shows a summary of the stress in each strut and its corresponding stress limit (in brackets).
Figure 3 (Click here to view a larger image) Calculate the Minimum Reinforcement Required for Crack Control: According Appendix A, the minimum reinforcement provided must satisfy be able to take
to
as 0.75 for the diagonal struts, and the minimum spacing for the vertical
reinforcement is the smallest of 12 in. or d/2. In addition, the code requires closed stirrups or ties parallel to the reinforcement required for tie AA’ to be uniformly distributed with 2/3 of the effective depth adjacent to tie AA’, i.e. 2/3 (16) = 10.67 in. Use 11 in. The area of these ties must exceed , where is the area of reinforcement resisting the tensile force . Hence the minimum area required is
Try 3 No. 3 closed stirrups with average spacing of 11/3 = 3.67 in.
Since this amount of reinforcement satisfies both requirements, provide 3 No. 3 closed stirrups distributed over a depth of 11 in. from tie AA’ with a concrete cover of 1 in. Design the Nodal Zones and Check the Anchorages: The width a of nodal zone A was chosen to satisfy the stress limits on the nodal zone. To anchor tie AA’, the horizontal loop is used. The detail is shown in Figure 4. To satisfy the nodal
To anchor tie AA’, the horizontal loop is used. The detail is shown in Figure 4. To satisfy the nodal zone stress limit, the tie reinforcement must engage an effective depth of concrete at least equal to
This limit is easily satisfied since the nodal zone available is 4 in. The required anchorage length for tie AA’ is
Since this is
less than the available length, i.e. 3 + 6 – (1 + 3/8) = 7.63 in., the anchorage length is adequate. Summary of the Design: The reinforcement details for the corbel designed using the strut-and-tie model according to Appendix A are shown in Figure 4.
Figure 4 (Click here to view a larger image) Deep Beam (1) | Deep Beam (2) | Dapped-Beam End | Single Corbel | Double Corbel Site Map | Contact Us | Copyright Info Home | In the News | Strut-and-Tie Method | CAST Design Tool NSF CAREER Award Project | Other Educational Resources Q & A | Related Links This page w as created and is maintained by Tjen Tjhin University of Illinois at Urbana-Champaign Last update: June 01, 2002
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