Doubly reinforced and flanged beams

What to do when a beam is too small for tension steel alone, and how a slab cast with the beam helps it. Two worked examples to SANS 10100-1, by hand and in STATIX.

1. Learning outcomes

After this lesson you can:

recognise when K is more than K′ and what that means for the section;
design the compression steel A_s′ and the tension steel A_s of a doubly reinforced beam;
check that the compression steel yields, using d′/x and the strain;
find the effective flange width of T and L beams;
check whether the stress block stays in the flange, and design a flanged beam.

2. Before you start

Lesson 1: K, K′, the lever arm z and A_s.
The free STATIX RC beam calculator, open in a second tab.
A pocket calculator, paper and a pencil.
About 60 minutes.

3. Theory

Doubly reinforced beams

In lesson 1, K ≤ K′ meant tension steel alone was enough. When K > K′, the concrete in compression has reached its limit: with no moment redistribution the neutral axis may go down to x = 0.5d, and no further. The concrete can carry the moment K′ f_cu b d² and no more. Steel placed in the compression zone, A_s′, carries the rest. (SANS 10100-1, 4.3.3)

A beam section as the moment grows. 1: a small moment needs only bottom steel. 2: at the limit moment K′ f_cu b d² the neutral axis has reached 0.5d; the concrete can give no more. 3: for a larger moment the neutral axis stays at 0.5d, and top bars (C_s) with extra bottom steel carry the rest.

The design route. When K is more than K′ the concrete is already at its limit, so steel in the compression zone takes the extra moment. Then check that this steel really reaches its design stress.

The lever arm of the concrete is fixed at its value for K′:

z = d [0.5 + √(0.25 − K′/0.9)] = 0.777d (for K′ = 0.156) x = (d − z)/0.45 = 0.5d

The compression steel takes the moment above K′, with lever arm (d − d′) to the tension steel:

A_s′ = (K − K′) f_cu b d² / [0.87 f_y (d − d′)] A_s = K′ f_cu b d² / (0.87 f_y z) + A_s′

Does the compression steel yield? The formulas above assume the top bars work at 0.87 f_y. That needs a strain of at least 0.87 f_y/E_s = 391.5/200 000 = 0.00196 (f_y = 450 MPa, E_s = 200 GPa). The strain at the top bars is 0.0035 (x − d′)/x, so they yield when d′/x ≤ 1 − 0.00196/0.0035 = 0.44. If d′/x is larger, use the lower stress ε_sc E_s instead of 0.87 f_y.

Compression bars must be held by links so they cannot buckle outwards. Lesson 3 covers links and lesson 8 the detailing rules (SANS 10100-1, 4.11).

Compression bars are slender columns inside the beam. Without links round them they bow outwards under load and burst the cover off. Links at close spacing stop that (lesson 8 gives the spacing rule).

Flanged beams

A slab cast with its beams works as a flange in compression at mid-span. Only part of the slab counts. The effective width b_eff (SANS 10100-1, 4.3.1.4) is:

T-beam: b_eff = b_w + L_z/5 L-beam: b_eff = b_w + L_z/10 but not more than the actual flange width (the beam spacing)

L_z is the distance between points of zero moment. For a simply supported beam it is the span.

Why only part of the slab counts. The compression in the flange is highest over the web and fades away from it (shear lag). b_eff is the width of a rectangle at the peak stress that carries the same force. Not to scale.

If the stress block stays inside the flange (0.9x ≤ h_f), design the beam as a rectangle of width b_eff, exactly as in lesson 1. A quick test is the moment the flange alone can carry:

M_f = 0.45 f_cu b_eff h_f (d − h_f/2) If M ≤ M_f, the stress block is in the flange.

Over supports of continuous beams the flange is in tension, so there the beam is designed as a rectangle of width b_w. Minimum steel for a flanged beam depends on b_w/b_eff; this lesson uses 0.18% of b_w h for a web in tension with b_w/b_eff < 0.4. Check the value for your case in your copy of the code (SANS 10100-1, 4.11).