How can a wider beam have a higher required reinforcement rather than a narrow beam for the exact same load?
This information is not dependant on the type of reinforcement (normal or prestressed)
Beam A
b=400mm, h=800mm, length = 10m
Deadload = 0.4 m * 0.8 m * 25 kN/m3 = 8 kN/m
Permanent load = 22 kN/m
Total load = 8 kN/m + 22 kN/m = 30 kN/m
Beam B
b=500mm, h=800mm, length = 10m
Deadload = 0.5 m * 0.8 m * 25 kN/m3 = 10 kN/m
Permanent load = 20 kN/m
Total load = 10 kN/m + 20 kN/m = 30 kN/m
By setting the load to 20 kN/m instead of 22 kN/m for beam B the total load will be the same, giving the same moment, shear force and deflections
Settings in PRE-Stress:
Activate the “Automatic design of stirrups”-feature in Stirrup reinforcement.
Since both beams have the same height the different shear field divisions will be the same over the length.
As for the stirrups, set them to ø8 in Reinforcement details. This will let the program know what the default diameter should be in the automatic design.
Result
If we take a look at x=2.00m, shear force = 98.4kN for both beams, but for Beam A (b=400mm) the program applies ø8s240 and for Beam B (b=500mm) the program applies ø8s190.
After a successful design, go back to the stirups-tab and uncheck “Automatic design of stirrups” to view the applied stirrups:
How come the reinforcement amount is 20% higher for a wider beam with the same loads?
The reason is that it is the minimum reinforcement that is governing and the requirement of stirrups due to the load is lower.
And when examining the expressions we see that the width of the element is a factor.
Conclusion: The width of the element may affect the minimum reinforcement of a beam, so for low shear forces a wider beam is less optimal.