RC Beam ACI Code Design (Concrete)

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Design Example: 1

A Cantilever beam 2.4 m long and 450 mm wide supports its own dead load plus a concentrated live load located 150 mm from the end of the beam and 112.5 mm away from the vertical axis of the beam. The concentrated load is 75 KN dead load and 100 KN live load. Design reinforcement for flexure, shear, and torsion using ACI code. Use f'c = 30 Mpa for concrete and fy = 460 Mpa for all steel.

Solution:

Schematic diagram of the problem

Compute the bending moment, shear force and torsion diagram

Assume h = 800 mm

Weight of normal concrete = ~ 24 KN/m³

Self-weight of the beam w = 0.45 x 0.8 x 24 = 8.64 KN/m

Factored UDL w_u = 1.2 x 8.64 = 10.368 KN/m

Factored Concentrated load P_u = 1.2 x 75 + 1.6 x 100 = 250 KN

Using Continuous beam analyzer tool (CBAns) below is the bending moment and shear force diagram

The torque acting on the beam is 250 * 0.1125 = 28.125 KN-m

M_u = 619.9 KN-m

V_u = 275.92 KN

T_u = 28.125 KN-m

Computing d and h for flexure

M_u/ɸK = bd²/10⁶

ɸK = ɸ(f’cω (1 - 0.59ω )

ω = ρfy/f’c

ɸ = 0.9 for flexure assuming tension controlled & ɸ = 0.75 for tension and shear

Try steel ratio ρ = 0.01

Re-Computing the bending moment, shear force and torsion diagram (for new self-weight of the beam)

Chosen h = 700 mm

Weight of normal concrete = ~ 24 KN/m³

Self-weight of the beam w = 0.45 x 0.7 x 24 = 7.56 KN/m

Factored UDL w_u = 1.2 x 8.64 = 9.072 KN/m

Factored Concentrated load P_u = 1.2 x 75 + 1.6 x 100 = 250 KN

Using Continuous beam analyzer tool (CBAns) below is the bending moment and shear force diagram

The torque acting on the beam is still the same (as we have only the concentrated load as the eccentric load on the beam) 250 * 0.1125 = 28.125 KN-m

M_u = 615.85 KN-m

V_u = 272.68 KN

T_u = 28.125 KN-m

Now Use the RC Beam ACI Code design tool calculate the steel area requirement

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