Single Corbel.pdf

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Single Corbel

Single Corbel

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Design per ACI 318-02 Appendix A, SI Unit

Deep Beam (1) | Deep Beam (2) | Dapped-Beam End | Single Corbel | Double Corbel

A single corbel projecting from a 350 mm 350 mm column is to be designed to support

precast beam reaction forces at 100 mm from the face of the column. The factored vertical load to be carried is 250 kN. A horizontal force of 50 kN is assumed to develop to account for creep and shrinkage deformations. Summary of ACI 318-02 Appendix A View Print-Friendly Version of this Page View PDF Version of this Page

The concrete strength is 35 MPa (normal density), and the yield strength of reinforcement is taken as 420 MPa. Determine the Bearing Plate Dimensions: Choose a 300 mm

150 mm 13 mm bearing pad. The bearing plate area is

and the bearing stress is is less than the bearing stress limit, i.e.

Since this the

bearing size is adequate. Choose the Corbel Dimensions: Choose an overall corbel depth at column face of 450 mm. The ACI Code requires that the depth at the outside of the bearing area is at least one-half of the depth at the column face. Therefore, select a depth of 225 mm at the free end of the corbel. Figure 1 summarizes the selected dimensions for the corbel.

Figure 1 (Click here to view a larger image) Determine the Strut-and-Tie Model: The vertical load is assumed to be located 25 mm toward the edge of the corbel from the center of bearing plate to allow for load eccentricities and erection tolerances. Thus, the position of vertical load is 25 + 100 = 125 mm from the face of column. The geometry of the assumed truss is given in Figure 2. The center of tie CB is assumed to be

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located 50 mm from the top of the corbel. Thus, d = 450 – 50 = 400 mm. The horizontal tie DA is assumed to lie on the horizontal line passing through the sloping end of the corbel. The location of strut DD’ centerline is found by calculating the strut width a. The required compressive force in strut DD’, NDD’, can be found by taking moments about Node A as follows:

As the stress on the nodal zone at D is to be limited to we have Solving these two equations gives

and

This fixes the geometry of the truss and means that member CD has a horizontal projection of 10 + 125 + 79/2 = 175 mm while member BC has a horizontal projection of 300 - 79/2 = 260 mm.

Figure 2 (Click here to view a larger image) Determine the Required Truss Forces by Statics: The required forces in the other important members of the truss are given in the following table. Note that positive indicates tension, negative compression. Member Force (kN)

CD -273

CB +159

BD -292

BA +245

DA +50

DD' -495

Design the Ties: The area of reinforcement required for tie CB is

Choose 4 No. 13 mm bars, As shown in the above table, tie BA has a larger tension than tie CB. However, this tie force should be resisted by column longitudinal reinforcement. Therefore, continue the 4 No. 13 mm bars down the column just to have a sufficient development length.

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The area of reinforcement required for tie DA is Choose 2 No. 10 mm additional column ties at location DA, Design the Nodal Zones and Check the Anchorages: The width a of nodal zone D was chosen to satisfy the stress limits on the nodal zone. To anchor tie CB, 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 100 mm.

The required anchorage length for tie AA’ is Since this is less than the available length, i.e. 50 + 150 – (25 + 10) = 165 mm, the anchorage length is adequate. Check the Struts: The struts will be checked by computing the strut widths and checked whether they will fit in the space available. The stresses in the diagonal struts CD and BD is limited to Hence, the required widths for struts CD and BD are and respectively. Choose 50 mm width for both struts CD and BD. The required width for strut DD’ is equal to a, i.e. 79 mm. As shown in Figure 3, all the strut widths fit into the outline of the corbel region. Thus, this solution is accepted. Figure 3 also shows a summary of the stress demand for 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 to be able to take as 0.75 for the diagonal struts, and the minimum spacing for the vertical reinforcement is the smallest of 300 mm or d/2.

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In addition, the code requires closed stirrups or ties parallel to the reinforcement required for tie CB to be uniformly distributed with 2/3 of the effective depth adjacent to tie CB, i.e. 2/3 (400) = 267 mm; use 275 mm. 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. 10 mm closed stirrups with average spacing of 275/3 = 92 mm.

Since this amount of reinforcement satisfies both requirements, provide 3 No. 10 closed stirrups distributed over a depth of 275 mm from tie CB with a concrete cover of 25 mm. 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

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Single Corbel.pdf

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