single corbel.pdf

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Summary ofACI 318-02 Appendix A

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Single Corbel Design per ACI 318-02 Appendix A, SI UnitDeep 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 supportprecast beam reaction forces at 100 mm from the face of the column. The factored verticalload to be carried is 250 kN. A horizontal force of 50 kN is assumed to develop to account forcreep and shrinkage deformations.The concrete strength is 35 MPa (normal density), and the yield strength of reinforcement istaken 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 Since thisis less than the bearing stress limit, i.e. thebearing size is adequate.

Choose the Corbel Dimensions:Choose an overall corbel depth at column face of 450 mm. The ACI Code requires that thedepth 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 theselected 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 thecenter of bearing plate to allow for load eccentricities and erection tolerances. Thus, theposition 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 DAis 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 requiredcompressive force in strut DD, NDD, can be found by taking moments about Node A asfollows:

As the stress on the nodal zone at D is to be limited to

we haveSolving these two equations gives andThis fixes the geometry of the truss and means that member CD has a horizontal projection of10 + 125 + 79/2 = 175 mm while member BC has a horizontal projection of 300 - 79/2 = 260mm.


Determine the Required Truss Forces by Statics:The required forces in the other important members of the truss are given in the followingtable. Note that positive indicates tension, negative compression.

Member CD CB BD BA DA DD'Force (kN) -273 +159 -292 +245 +50 -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 forceshould be resisted by column longitudinal reinforcement. Therefore, continue the 4 No. 13 mmbars down the column just to have a sufficient development length.


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The area of reinforcement required for tie DA isChoose 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 thenodal zone stress limit, the tie reinforcement must engage an effective depth of concrete atleast 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 isless than the available length, i.e. 50 + 150 (25 + 10) = 165 mm, the anchorage length isadequate.

Check the Struts:The struts will be checked by computing the strut widths and checked whether they will fit inthe 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 toa, i.e. 79 mm.As shown in Figure 3, all the strut widths fit into the outline of the corbel region. Thus, thissolution is accepted. Figure 3 also shows a summary of the stress demand for each strut and itscorresponding stress limit (in brackets).


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 minimumspacing 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 fortie 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 isthe 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 closedstirrups 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 toAppendix A are shown in Figure 4.


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Q & A | Related LinksThis page was created and is maintained by Tjen TjhinUniversity of Illinois at Urbana-ChampaignLast update: June 01, 2002


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