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

Consider the 2 meter deep beam described in Figure 1 below. Use the strut-and-tie model to determine the required amount of reinforcement.

Additional details: and

Figure 1
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Step 1: Evaluate the Total Factored Load,

Step 2: Check Bearing Capacity at Loading and Support Locations

Bearing strength at points of loading = = 0.75(0.85)(25)(1.0)(450)(500)/1000
= 3586 kN > 1600 kN OK

Bearing strength at supports = = 0.75(0.85)(25)(0.80)(450)(500)/1000
= 2868 kN > 1600 kN OK

Step 3: Select the Strut-and-Tie Model to Use in Design

Figure 2
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Step 4: Isolate Disturbed Region and Estimate Member Forces and Dimensions

The entire deep beam is a disturbed region, but it is only necessary to consider the left third of the structure to complete the design. The horizontal position of nodes A and B are easy to define, but the vertical position of these nodes must somehow be estimated or determined. What we do know is that the design strength of strut BC must be greater than or equal to the factored load in strut BC. That is:

Strut BC: , where = 1.0 (prismatic strut)

Similarly, the design strength of tie AD must be greater than the factored load in tie AD. In addition, this tie must be anchored over a large enough area such that the factored load is less than

Tie AD: and
Tie AD: , where = 0.8 (on tie anchored in Node A)

By setting the design strength equal to the required capacity, jd will be a maximum and w_t = 1.25 w_c. The flexural lever arm will be jd = 2000 - w_c/2 - w_t/2 = 2000 – 1.125w_c.

Figure 3
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By taking summation of moments about point A:
By substituting for F_BC,
w_c = 231 mm, and therefore w_t = 288 mm.

If these values are used for the dimensions of the struts and ties, the stress in strut F_BC will be at its limit, and the force in tie F_AD will be anchored in just sufficient area. It is often wise to increase these values a little to leave some margin. w_c will be selected to be 240 mm, and w_t will be selected to be 300 mm.

jd = 2000 – 240/2 – 300/2 = 1730 mm and F_BC = F_AD = 1600(2000)/1730 = 1850 kN

Check capacity of strut BC:  = 0.75(0.85)(1.0)(25)(500)(240)/1000 
= 1912 kN  OK

Step 5: Select Reinforcement

Tie AD:

Consider 1 layer of 6 #36(11) bars = 6036 mm² @ 150 mm from bottom
Consider 2 layers of 5 #29(9) bars = 6450 mm² @ 80 mm and 220 mm from bottom
Consider 3 layers of 6 #22(7) bars = 6966 mm² @ 60, 150, and 240 mm from bottom

Check capacity of tie AD: = 0.75(6450)(420)/1000 = 2032 kN > 1850 kN  OK

Step 6: Calculate Force in Diagonal Compressive Strut F_AB and Check Capacity

and

Therefore, the force in the diagonal compressive strut,
Width at top of strut = =
Width at bottom of strut = =

Assuming that sufficient crack control reinforcement is used, then

Check Capacity of strut AB:  = 0.75(0.85)(0.75)(25)(500)(476)/1000 
= 2885 kN > 2444 kN OK

Figure 4
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Step 7: Minimum Distributed Reinforcement and Reinforcement for Bottle-Shaped Struts

Horizontal Web Reinforcement:

Use one #13(4) on each face at s_h = 300 mm over entire length,
A_h/(b s _h) = 2(129)/500/300 = 0.0017 > 0.0015 OK

Vertical Web Reinforcement:

Use one #16(5) on each face at s_v = 300 mm over entire length,
A_v/(b s_v) = 2(199)/500/300 = 0.00265 > 0.0025 OK

Check of Reinforcement to Resist Bursting Forces in Bottle-Shaped Struts: 
= 0.00312 > 0.003  OK

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