Example Calculation of Shear Reinforcement in a Reinforced Concrete Lintel

1. Situation

A 6-1/4 in. thick by 16 3/4 in. deep lintel is to span an opening of 10 ft in reinforced concrete (ICF) wall. The factored axial (gravity) load applied to the lintel is 1900 plf. Detail shear reinforcement, if necessary. The specified concrete strength is 3000 psi and all reinforcement is to be Gr. 60 ksi. Assume that the top and bottom longitudinal reinforcement is centered 3 in. from the top and bottom of the lintel.

2. Approach

Our approach will be to determine the factored shear load and see if it can be carried by the concrete alone. If not, we will determine the required shear reinforcement.

Factored Shear Load

1. Determine the effective depth, d.
2. Determine the design shear (Vu @ x = d).

1. Determine Vc.
2. Do the Design Check ... Is Vu ≤ ½ φ (Vc)? If so, then done! ... if not, then ...

Determine Shear Reinforcement Details

1. Select convenient stirrup (shear reinforcement) spacing, s.
2. Select trial stirrup size and number of legs.
3. Determine shear strength provided by reinforcement, Vs.
4. Check ... Is Vu ≤ ½ φ (Vc + Vc)?
5. Adjust size, number of legs, spacing, etc. as appropriate.

Further Design Checks.

1. Check Vs against Code limits of 4 b d √f'_c and 8 b d √f'_c (see lesson on shear reinforcement, here).

3. Calculations

Load ...

The effective depth, d, is ... 16-3/4 in. - 3 in. = 13-3/4 in. = 1.15 ft.

The factored applied shear load at distance d from the face of the support (Vu @ x = d) is ...

Vu = 1900 plf (10/2 ft - 1.15 ft) = 1900 plf (3.85 ft) = 7323 lb.

Let's add in the weight of the lintel (clear to the face of the support) ...

... ω _s.w. = 6.25/12 x 16.75/12 x 150 = 109 plf.

Factored, is ... ω _us.w. = 1.2 (109 plf) = 131 plf.

Total factored shear load,

Vu = 7323 + 131 = ...

Vu = 7454 lb.

Concrete Strength ...

Vc = 2 b d √f'c = 2 (6.25 in.)(13.75 in.)√3000 psi = ...

Vc = 9414 lb.

Design Check ... can we live without shear reinforcement? ... Is Vu ≤ ½ φ (Vc)? ...

... φ = 0.75

Let's see ... is 7454 lb ≤ ½ (0.75) 9414 lb = 3530 lb? ... NO! ... we need shear reinforcement.

4. Shear Reinforcement ...

The maximum spacing of the shear reinforcement is ... d/2 = 13.75 in. / 2 = 6.875 in. But this is a weird number, so let's go with s = 6.0 in.

A common size stirrup size (shear reinforcement) is # 3; let's try a single leg stirrup, so,

A_v = 0.11 in.² ...

Before we get too carried away, let's check this against the Code minimum Av ...

A _{v min} = 0.75 √ f '_c b s / f_y = 0.75 √3000 psi (6.25 in.)(6 in.)/(60,000 psi) = 0.026 in.² ... good!

Now let's calc Vs ...

V_s = A_v f_y (d / s) = 0.11 in.² (60,000 psi)(13.75/6) = ...

Vs = 15,125 lb.

Wait! ... let's check against the max amount of Vs ...

V_{s max} = 4 b d √f'c (if we don't want to half) the spacing ... = 2 Vc = 2 (9414 lb) = 18,828 lb. Good!

But, interestingly, it's fairly close! If we had exceeded the 18,828 lb ... I would investigate using deformed wire stirrups.

So, let's proceed ...

Design Check ... is Vu ≤ φ (Vc + Vs)? ...

Well, let's see ... is 7454 lb ≤ 0.75 (9414 + 15,125) lb = 18,404 lb? ... yes! Sweet.

Okay, in some regards we're done! ... We could detail stirrups all the way across the lintel @ 6 in. o.c. with the ending stirrups not farther than s/2 = 3 in. from the face of the supports.

So, single leg # 3 stirrups @ 6 in. o.c. ...

If we want to get fancy, however, there could be some region of the beam that would not require stirrups at all. This would be where Vu at the section is less than ½ φ Vc. We could get this from a shear diagram for the beam, or, recognizing that the shear is theoretically zero at the mid-span ... and increasing by ... ω _u in both directions ...

ω _u = 1900 plf + 131 plf = 2031 plf.

The distance x from the mid-span to where the shear reaches ½ φ Vc is ...

... ½ φ Vc = 3530 lb = 2031 plf x ... x = 3530 / 2031 = 1.74 ft.

So, the middle 2 (1.74) = 3.5 ft of the beam need not shear reinforcement.

5. Conclusion

Detailing is an art of balancing complexity with economy. For some applications it may be appropriate to simply detail single leg # 3 stirrups at 6 in. o.c. across the entire lintel. In other applications, especially repeated ones, it may be appropriate to detail where the stirrups may be omitted. For a single lintel like the one examined, it would be the difference of a half dozen or so stirrups. But if there are multiples of the lintel exampled, then we might save a more significant quantity of steel.

6. References

Building Code Requirements for Structural Concrete, ACI 318, American Concrete Institute, P.O. Box 9094, Farmington hills, Michigan, 48333.

Beam Shear in Reinforced Concrete, Jeff Filler, Associated Content.

Shear Reinforcement in Reinforced Concrete Beams, Jeff Filler, Associated Content.