Expert Tip: Concrete shear design based on AASHTO LRFD 3

AASHTO , midas Civil , concrete shear design .
Blog Article on Concrete Shear Design - AASHTO LRFD 3

2 min read

Author: Seungwoo Lee, Ph.D., P.E., S.E.

Publish Date: 2 Jun, 2023

Optimum crack angle θ

From the previous example, we can catch that there are some possible crack angle ranges for the given εx and vu/f’c. Now our question is which values of θ and β are the optimums? The previous example shows that, without considering longitudinal reinforcements, mostly (not always) the lowest crack angle results in the least number of stirrups. However, with considering longitudinal reinforcements, the optimum crack angle increases. The methodology to find out the optimum crack angle is proposed by Rahal and Collins (Background to the general method of shear design in the 1994 CSA-A23.3 standard, Canadian Journal of Civil Engineering, February 2011).


The total stirrup quantity per section is:


stirrup quantity per section

From the longitudinal reinforcement equation,


longitudinal reinforcement equation

Increasing 50% of stirrup quantity by considering relative workability for bending and placing,


Increasing 50% of stirrup quantity by considering relative workability for bending and placing

graph stirrup quantity

For εx=0.5/1000, vu/f’c=0.15 case, crack angle θ=32° corresponds to the minimum steel quantity case. AASHTO LRFD gives θ=32.1°, β=2.36, and close to what we got 2.29.


Detailed calculation

In the previous example, we calculated the maximum stirrup spacing with assumed longitudinal reinforcements. Now, with the help of computers, we can calculate the optimum stirrup spacing AND longitudinal reinforcements together. Detailed calculation steps are as follows.


Detailed calculation steps

Step 1) Assume ε1=0.003179

Step 2) Assume θ=37°

Step 3) Calculate f1 from fcr


Calculate f1 from fcr

Step 4) Calculate f1 from vci


Step4) Calculate f1 from vci


Assume fv=57.17 ksi, Av=0.22 in2, s=7.446 in.


Assume fv=57.17 ksi, Av=0.22 in2, s=7.446 in.

Step 5) Calculate β


Step5) Calculate β


Step 6) Calculate ε2


Step6) Calculate ε2

Step6) Calculate ε2 further

Step 7) Calculate εx

Assume As = 4.281 in2

Step7) Calculation


Step 8) Calculate ε1


Step8) Calculate ε1

Repeat Steps 1) to 8)

Step 9) Calculate fv


calculate fv


Repeat Steps 4) to 9)

Step 10)


step 10)


Repeat Steps 7) to 10)

Step 11)

step 11

The same calculation can be performed for different crack angles as shown below.


different crack angles graph

From the figure, we can see that the optimum crack angle is around 36.75 degrees. With εx=1.052/1000, vu/f’c=0.06604, we got approximately θ=37°, β=2.35.

AASHTO LRFD gives θ=36.4°, β=2.23 for εx=1/1000, vu/f’c=0.075 case and these are very close to what we got with iteration.


Simplified methods

AASHTO gives somewhat conservative results.





We can perform concrete shear design using modified compression field theory.

The modified compression field theory requires iterative calculations, but we have no problems with computers.

From the AASHTO LRFD appendix, we can find optimum crack angles and corresponding β values easily and these values are close to what we got from iterations.

AASHTO LRFD's main chapter shows a very simplified formula to calculate θ and β, and these are somewhat conservative values.

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