The design for longitudinal shear can readily be carried out using Equations 5.3(2) and 5.3(5). Note that both transverse shear reinforcement and torsional reinforcement (see Section 6.4) resist the corresponding design actions by tension in the legs of the stirrups or closed ties. Thus, stirrups or closed ties are effective in providing resistance to longi-tudinal shear, which is by dowel action of the vertical legs. In design, follow these steps:
1. Check that Equation 5.3(1) prevails.
2. If not, the required longitudinal reinforcement may be obtained from Equation 5.3(5) as
A s
b
f k f g b
sf f
sy.f
co ct p f
= ( * /τ φ− ′) /µ− / Equation 5.3(6)
3. Ensure that the existing transverse shear and/or torsional reinforcement equals or exceeds that given in Equation 5.3(6). Otherwise, provide the balance of the shear reinforcement as required in Equation 5.3(6).
Note that Clause 8.4.4 of the Standard limits the spacing to
s≤ 3 5. tf Equation 5.3(7)
where tf = thickness of the topping or flange anchored by the shear reinforcement.
Further, Clause 8.4.5 of the Standard stipulates that the average thickness of the structural components subject to longitudinal shear shall be not less than 50 mm, with no local thickness less than 30 mm.
5.3.5 Design example
Problem
For the T-beam in the preceding design example (Section 5.2.8), check and design for longitudinal shear, if applicable. Take ′fc=20 MPa and use N12 or N16 bars for ties as necessary. Assume monolithic construction.
Solution
From Section 5.2.8, self-weight = 8.722 kN/m and V* = 547.7 kN.
The critical shear plane, although not specified, may be taken as at the rectangular stress block depth level located (in this case) in the web – thus bf = 300 mm.
Equation 5.3(4): ß = 1.
142 part 1 reinforced concrete
With reference to Figure 5.2(4), imposing ΣFx = 0 gives
4960 × 500 = 0.85 × [1200 × 100 × 20 + 300(γ kud –100) × 20]
from which
γ kud = 186.27 mm
300 mm 1200 mm
Z
γ kud +
Centre of gravity
100 mm
86.27 mm
figure 5.3(4) determination of z for the example t-beam
Then, in Figure 5.3(4), the distance between the centre of gravity of the compressive resultant and the extreme compressive fibre
z= × × + × × +
× + ×
1200 100 100 2 86 27 300 100 86 27 2 1200 100 86 27 3
/ ( . ) ( . )
.
/
00 =66 52. mm
Equation 5.3(3) : Equation 5.3(2) :
mm z d z= − =800 66 52 733 48− . = .
/ MPa
τ*=547 7 10 733 48 300. × 3 ( . × )=2 49. The quantities required for Equation 5.3(5) are:
• Asf/s for the existing transverse reinforcement or N12 ties @ 180 mm = 220/180 = 1.222 mm2/mm
• gp = (100 × 1200 + 86.27 × 300) × 10–6× 24.85 × 103 = 3625.1 N/m = 3.63 N/mm From Equation 2.1(3),
• fct′ =0 36 20 1 61. = . MPa From
• Table 5.3(1), μ = 0.9; kco = 0.5 Thus Equation 5.3(5):
τu=0 9 1 222 500 300 3 63 300. ( . × / + . / )+0 5 1 61 2 65. × . = . MPa
chapter 5 Ultimate strength design for shear 143
Check Equation 5.3(1):
φτu=0 7 2 65 1 86. × . = . < =τ* 2 49. MPa, which is inadequate.
Thus, additional shear reinforcement is required.
Equation 5.3(6) : Af s
s = × − ×
−
300 500
2 49
0 7. 0 5 1 61 0 9 3 6
. . . 3 300
1 828
/ .
=
mm /m2 m
Try N12 ties: s=2 110× = mm, which is rather cl 1 828 120 35
. . ose
Thus, try N16 ties and
s=2 200× = 1 828 218 8
. . mm
Say, use N16 ties @ 220 mm, which is less than 3.5tf = 3.5 × 100 = 350 mm as required in Equation 5.3(7).
In summary, the final reinforcement arrangement for both transverse and longitu-dinal shear is N16 closed ties @ 220 mm, and the total number required over the 6-m span: (28 + 1) = 29.
144 part 1 reinforced concrete
(b) 200
350 100 40
40 500
1000 2000
(a) 90 kN
8 N28 8 N36
70 70 70
70
figure 5.4(1) details of a simple beam with overhang: (a) loading configuration and (b) cross-sectional details
note: all dimensions are in mm.
5.4 problems
1. Figure 5.4(1) shows a simple beam with overhang with its loading configuration and cross-sectional details. The concrete strength is ′fc=25 MPa.
Design and detail the transverse shear reinforcement as necessary for the critical shear section of the beam. Use R10 ties. The maximum shear reinforcement require-ment may be adopted throughout the beam.
chapter 5 Ultimate strength design for shear 145
200 80
4 N28 4 N32
400 100 q = 75 kN
g = 15 kN/m including self-weight
500 mm
A B
8 m 18 m
C
(a) 500 40 40
70
(b)
figure 5.4(2) details of a simple beam with a cantilever overhang: (a) loading configuration and (b) cross-sectional details
note: all dimensions are in mm unless otherwise specified.
2. A simple beam with a cantilever overhang and its loading configuration and cross-sectional details are shown in Figure 5.4(2). The concrete strength fc′ = 25 MPa. Note that the load combination formula 1.2g + 1.5q applies, in which g and q are dead and live loads, respectively.
Design and detail the transverse shear reinforcement as necessary for the critical shear section of the beam which may occur at the left or right of the support at B.
Use R10 ties and the maximum shear reinforcement requirement may be adopted throughout the beam.
146 part 1 reinforced concrete
56 mm 554 mm 300 mm
2500 mm 2500 mm
48 mm
(b) (a)
10 N32
42 mm 700 kN
5 N32
figure 5.4(3) details of a simple beam: (a) loading configuration and (b) cross-sectional details note: all dimensions are in mm.
3. For the beam shown in Figure 5.4(3), design the transverse and longitudinal shear reinforcement. Take fc′ = 25 MPa and use N16 bars for closed ties. Assume that the construction is monolithic.
4. For the cantilever beam shown in Figure 3.11(19) (for Problem 24 in Section 3.11), if q = 150 kN and there is no change to any other condition as given in Problem 24, design and detail, as necessary, the transverse shear reinforcement. Use N12 closed ties only.
5. Details of a simply-supported beam are given in Figures 5.4(4)a and b.
Note that the dead load, g, includes the self-weight.
Design the transverse shear reinforcement in terms of N10 ties @ s, the required spacing at the critical section which in this case may be taken at the appropriate support (not do from it). Use the same spacing throughout. Take fc′ = 32 MPa and you may assume θv = 45°.
chapter 5 Ultimate strength design for shear 147
(a) (b)
6065 610
3000
325
6 N32
3000 3200
Pu Pu = 300 kN
figure 5.4(5) details of a simple beam under two concentrated loads note: all dimensions are in mm.
q = 750 kN 60º
4 m 4 m
g = 20 kN/m
790
80
2 N28
8 N28
300
762 62
5652
(b) (a)
figure 5.4(4) details of a simply supported beam: (a) loading configuration and (b) cross-sectional details
note: all dimensions are in mm unless otherwise specified.
6. A simply supported beam is shown in Figure 5.4(5)a with details of the longitudinal reinforcement given in Figure 5.4(5)b. Taking ′ =fc 30 MPa, design the shear rein-forcement (using vertical N10 stirrups only). Sketch the final design.
148 part 1 reinforced concrete
a a
A
3000 mm
80 kN B
70 70
70
200
350 100 500
6 N36 3 N24
Section a•a figure 5.4(6) details of a cantilever beam
note: all cross-sectional dimensions are in mm.
8. The beam-and-slab system in Figure 3.11(18) (for Problem 23 in Section 3.11) is to be cast in two steps: first the webs up to level x-x, then the slab.
In addition to a dead load (including self-weight) g = 35 kN/m and a live load q = 80 kN/m, each of the T-beams has to carry a concentrated live load of 150 kN at the middle of a simply supported span of 8 m.
As the structural engineer, you are required to design the slab–web interface against longitudinal shear failure. You may assume μ = 0.9 and kco = 0.4; use N12 bars for ties if necessary. Take ′fc= 20 MPa, and the effective depth may be taken as d = 870 mm for this design.
7. A cantilever beam and its loading configuration are detailed in Figure 5.4(6).
Considering transverse shear only, design and detail the shear reinforcement as necessary. Use R10 ties and adopt the maximum shear reinforcement requirement throughout the span. Take ′fc=32 MPa.
chapter 5 Ultimate strength design for shear 149
(b) (a)
g P = 405 kN P = 405 kN
Cover = 40 mm Spacing = 40 mm 360 mm
10 N40 2000 mm
1000 mm 100 mm
100 mm 100 mm
5 m 5 m 5 m
100 mm
100 mm
80 mm 80 mm
60 mm
figure 5.4(7) a simply supported bridge beam and its (idealised) cross-section note: all cross-sectional dimensions are in mm.
9. Figures 5.4(7)a and b, respectively, give details of a simply supported bridge beam and its (idealised) cross-section. In addition to the dead load g = 25 kN/m (includ-ing self-weight), the beam is to carry the two concentrated live loads as shown. The characteristic strength of the concrete is ′ =fc 32 MPa.
(a) Design the transverse reinforcement to resist the vertical shear. Use (vertical) N12 closed ties only; show the design details in a sketch.
(b) Assuming monolithic construction, carry out a check for the longitudinal shear.
Is the design adequate?
150 part 1 reinforced concrete
200 80
(b) (a)
B q = 75 kN
10 m 10 m
A
g = 15 kN/m including self-weight
500 mm
70
40 40
4 N28
400 100 500
2 N16
figure 5.4(8) details of a simply-supported beam: (a) loading configuration and (b) cross-sectional details
note: all cross-sectional dimensions are in mm unless otherwise specified.
10. A simply-supported beam and its loading configuration and cross-sectional details are shown in Figure 5.4(8). The concrete strength is ′fc=25 MPa.
Design and detail the transverse shear reinforcement as necessary for the critical shear section of the beam. Use R10 ties and adopt the maximum shear reinforce-ment requirereinforce-ment throughout the beam.
chapter 5 Ultimate strength design for shear 151
(b) 300 50
2 N24 818 790
8 N28
82 56
54 (a)
5 m 45º
5 m
q = 1000 kN
g = 1000 kN/m
figure 5.4(9) details of a cantilever beam: (a) loading configuration and (b) cross-sectional details note: all cross-sectional dimensions are in mm unless otherwise specified.
11. Details of a cantilever beam are given in Figures 5.4(9)a and b. Note that the dead load, g, includes the self-weight.
Design the transverse shear reinforcement in terms of N12 ties @ s, the required spacing at the critical section which in this case may be taken at the root or support of the cantilever (not do from it). Use the same spacing throughout. Take ′ =fc 40 MPa.
152