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Minimum reinforcement of flexural members

Due to architectural or functional requirement, large dimensions for some members are selected, these members will have a large dimension compared to their bending requirement. therefore, a light amount of reinforcement will be sufficient to resist the service loads and moments.
Figure 1

for members with very light reinforcement, the ultimate resisting moment will be less than their cracking moments. 

Mcr is the cracking moment.
fr is the modulus of rupture 
I is the moment of inertia
y is the distance from neutral axis to extreme compression fiber

a smaller area of steel means a smaller nominal flexural strength of the member. on the other hand, large dimension means a larger moment of inertia and this will increase the cracking moment magnitude. if this member cracks it will fail suddenly and it will jeopardize the whole structure.


Mn is the nominal flexural moment
As is area of steel
d is the effective depth of member
a is the depth of Whitney compression block

ACI [10.5.1] specifying the minimum area of steel required

for SI units

As,min=(0.25√fc'/fy)*bw*d and not less than 1.4bw*d/fy

For us units

As,min=(3√fc'/fy)*bw*d and not less than 200bw*d/fy

ACI[10.5.2] stating that bw shall be substituted with 2*bw  or the flange width whichever is smaller for a statically determined member where the tension reinforcing steel is in the flange

                                                               Figure 2

we can rewrite the previous equation using ρ=As/(b*d)

for SI units

ρmin=(0.25√fc'/fy) and not less than 1.4/fy

For us units

ρmin=(3√fc'/fy) and not less than 200/fy


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