Review Module – Reinforced Concrete Design (Shear)

SHEAR IN BEAMS Strength Reduction Factor, ∅= 0.75

SITUATION. A 6 m simply supported beam is carrying the following 1. Determine the critical design shear stress (MPa) in beam BE?
loads: 2. Calculate the shear strength, Vc (kN), provided by the concrete in
Superimposed dead load = 50kN/m beam BE.
Live Load = 30kN/m
3. Determine the spacing (mm) for the two legs of 10 mm Ø.
Factored axial compressive load = 250kN
4. Using NSCP, determine the maximum spacing (mm) of stirrups.
The beam is 300 mm wide and has a total depth of 700 mm and is
SITUATION.
reinforced at the bottom side with 3 – 32 mmø.
Concrete cover to the centroid of main reinforcements = 70mm A Continuous beam is reinforced as shown.
Material properties: Given:
f'c = 27.6 MPa fy = 276 MPa As=8-20mm diameter bars b=350mm
Unit weight of concrete = 23.54 kN/m3 As’=4-20mm diameter bars h1=100mm
Ultimate load combination = 1.2D + 1.6L ds=10mm diameter h2=400mm
Strength Reduction Factor, Φ= 0.75 a=45mm

1. Compute the ultimate shear force Vu (KN) at critical section.

2. Compute the modified moment Mm (KN-m).
3. Compute the nominal concrete shear capacity (KN) of the beam.
4. Compute the maximum permissible spacing (mm) of 12 mm Ø
stirrups.
5. Compute the actual spacing (mm) of 12 mm Ø stirrups.

SITUATION.

Concrete fc’=34MPa
Longitudinal steel, fyl=413MPa
Transverse ties, fyv=275MPa
Clear concrete cover=40mm
1. Calculate the shear strength (kN) provided by the concrete.
2. The 10mm diameter ties are spaced at 100mm on centers. What is
the shear strength provided by the shear reinforcement?
3. What is the maximum allowable tie spacing (mm)?

NSCP 2015 PROVISIONS

SHEAR STRENGTH OF CONCRETE

422.5.5 Vc for Non-Prestressed Members without Axial force
422.5.5.1 For non-prestressed members without axial force, Vc shall be
calculated by
𝑉𝐶 = 0.17𝜆√𝑓′𝑐𝑏𝑤 𝑑
Unless a more detailed calculation is made in accordance with Table
422.5.5.1
Table 422.5.5.1
Detailed Method for Calculating Vc
𝑉𝐶
𝑉𝑢 𝑑
Least of (0.16𝜆√𝑓 ′ 𝑐 + 17𝜌𝑤 )𝑏 𝑑 (a)
Given: 𝑀𝑢 𝑤
Beam Sections, b x h for AD, BE and CF=250mmx400mm (a), (b)
(0.16𝜆√𝑓 ′ 𝑐 + 17𝜌𝑤 )𝑏𝑤 𝑑 (b)
Girder Sections, b x h for ABC and DEF=350mm x 600mm and (c);
′
0.29𝜆√𝑓 𝑐𝑏𝑤 𝑑 (c)
Dimensions:
Mu occurs simultaneously with Vu at the section considered
S=3.0m
L=6.0m
422.5.6 Vc for Non-Prestressed Members with Axial Compression
Concrete Cover to the centroid of reinforcements=70mm
422.5.6.1 For non-prestressed members with axial compression, Vc
Concrete Compressive Strength, fc’=20.7MPa
shall be calculated by:
Reinforcing steel yield strength, fy=275MPa
𝑁𝑢
Loads: 𝑉𝐶 = 0.17(1 + )𝜆√𝑓 ′ 𝑐𝑏𝑤 𝑑
14𝐴𝑔
Dead Load, D=5.5kPa (beam weight already included)
Live Load, L=4.8kPa (all spans) unless a more detailed calculation is made in accordance with Table
Required Strength, U = 1.2D + 1.6L 422.5.6.1, where Nu is positive for compression
Table 422.5.6.1 Detailed Method for Calculating Vc SHEAR IN COLUMNS
𝑉𝐶
(0.16𝜆√𝑓 𝑐 ′ SITUATION. A column section shown is reinforced with 8 – 32 mm Ø
𝑉𝑢 𝑑 bars, with a clear concrete cover of 40 mm for the 12 mm Ø ties. The
+ 17𝜌𝑤 )𝑏 𝑑 (a)
4ℎ − 𝑑 𝑤 design axial load due to the reversal effect of DL, LL and WL changes
𝑀𝑢 − 𝑁𝑢 as follows
Least of 8
Equation not applicable if
(a), (b) Along the positive x – direction: Along the negative x – direction
4ℎ − 𝑑 (b)
and (c); 𝑀𝑢 − 𝑁𝑢 ≤0 Mu = -420 kN.m Mu = +420 kN.m
8
Vu = 370 kN Vu = 370 kN
0.29𝑁𝑢 Nu = 1320 kN Nu = 450 kN
0.29𝜆√𝑓 ′ 𝑐𝑏𝑤 𝑑√1 + (c)
𝐴𝑔 Reduction factor=0.75
Mu occurs simultaneously with Vu at the section considered 1. Determine the concrete shear strength (KN) for the positive x –
direction using simplified calculation.
422.5.7 Vc for Non-Prestressed Members with Significant Axial 2. Determine the concrete shear strength (KN) for the negative x –
direction using simplified calculation.
Tension
3. Determine the required spacing (mm) of shear reinforcement.
422.5.7.1 For non-prestressed members with significant axial tension, Apply provisions on spacing limits of reinforcement when
Vc shall be calculated by: applicable
𝑁𝑢 Use fc’ = 28 MPa and fy = 415 MPa
𝑉𝐶 = 0.17(1 + )𝜆√𝑓 ′ 𝑐𝑏𝑤 𝑑
3.5𝐴𝑔
where Nu is negative for tension, and Vc shall not be less than zero

SHEAR STRENGTH PROVIDED BY STEEL

422.5.10.5.3 Vs for shear reinforcement in Section 422.5.10.5.1 shall
be calculated by
𝐴𝑣 𝑓𝑦𝑡 𝑑
𝑉𝑆 =
𝑠
where s is the spiral pitch or the longitudinal spacing of the shear
reinforcement and Av is given in Section 422.5.10.5.5 or 422.5.10.5.6

Table 409.7.6.2.2 Maximum Spacing of Shear Reinforcement

Maximum s, mm SITUATION. A circular column is reinforced by 10-25mm diameter bars
Non- with a clear cover of 40mm for 12mm diameter spiral.
Vs Prestressed Given: Diameter, D = 500mm
prestressed
beam Strength Reduction Factor = 0.75
beam
𝑑 3ℎ Nu = 540kN
Lesser of: f’c = 28MPa and fy = 275MPa
≤ 0.33√𝑓 ′ 𝑐𝑏𝑤 𝑑 2 4 1. Determine the nominal shear capacity of the spiral column if the
600
𝑑 3ℎ 12mm diameter spirals are spaced 70mm on centers.
> 0.33√𝑓 ′ 𝑐𝑏𝑤 𝑑 Lesser of: 2. Determine the required spacing of the 12mm diameter spirals for a
4 8 factored shear force Vu=360kN.
300
3. Determine the nominal shear stress in the column if Vu=360kN.
Table 409.6.3.3 Required 𝑨𝒗,𝒎𝒊𝒏 SITUATION. The figure shows the column section of the building with
Beam Type 𝑨𝒗,𝒎𝒊𝒏 /𝑺 transverse confining reinforcement.
Non-prestressed 𝑏𝑤 Clear cover of 12mmØ ties = 40mm
and prestressed 0.062√𝑓′𝑐 a
𝑓𝑦𝑡 fc’ = 28 MPa
with 𝐴𝑝𝑠 𝑓𝑠𝑒 < Greater fy = 415 MPa
of: 𝑏𝑤
fyh = 278 MPa
0.40(𝐴𝑝𝑠 + 0.35 b
𝑓𝑝𝑢 + 𝐴𝑠 + 𝑓𝑦 ) 𝑓𝑦𝑡 Reduction factor = 0.75
𝑏𝑤
0.062√𝑓′𝑐 c
Prestressed with Greater 𝑓𝑦𝑡
𝐴𝑝𝑠 𝑓𝑠𝑒 ≥ of: 𝑏𝑤
Lesser 0.35 d
0.40(𝐴𝑝𝑠 + of: 𝑓𝑦𝑡
𝑓𝑝𝑢 + 𝐴𝑠 + 𝑓𝑦 ) 𝐴𝑝𝑠 𝑓𝑝𝑢 𝑑
√ e
80𝑓𝑦𝑡 𝑑 𝑏𝑤
1. Determine the required spacing (mm) if lateral reinforcement (mm)
for factored shear load Vuy = 450 KN if the allowable concrete
shearing stress is 0.88 MPa.
2. What is the maximum spacing (mm) of the 12mmØ transverse
reinforcement?
3. Determine the required spacing (mm) of confining hoop
reinforcement in accordance with the code for seismic design.
4. What is the maximum spacing (mm) of the 12mmØ transverse
reinforcement in accordance with the code for seismic design.

NSCP 2015 PROVISIONS

418.7.5 Transverse Reinforcement
418.7.5.1 Transverse reinforcement required in Sections 418.7.5.2
through 418.7.5.4 shall be provided over a length lo from each joint face
and on both sides of any section where flexural yielding is likely to occur
as a result of lateral displacements beyond the elastic range of behavior.
Length lo shall be at least the largest of (a) through (c):
a. The depth of the column at the joint face or at the section where
flexural yielding is likely to occur;
b. One-sixth of the clear span of the column;
c. 450 mm

418.7.5.3 Spacing of transverse reinforcement shall not exceed the

smallest of (a) through (c):
a. One-fourth of the minimum column dimension.
b. Six times the diameter of the smallest longitudinal bar;
c. So as calculated by
350 − ℎ𝑥
𝑠𝑜 = 100 + ( )
3
The value of So from Eq. 418.7.5.3 shall not exceed 150 mm, and need
not be taken less than 100 mm.

418.7.5.4 Amount of transverse reinforcement shall be in accordance

with Table 418.7.5.4 The concrete strength factor, kf, and confinement
effectiveness factor, kn, are calculated according to Eq. 418.7.5.4a and
418.7.5.4b.
𝐟′𝐜
𝐤𝐟 = + 𝟎. 𝟔 ≥ 𝟏. 𝟎
𝟏𝟕𝟓
(418.7.5.4a)
𝐧𝐥
𝐤𝐧 =
𝐧𝐥 − 𝟐
(418.7.5.4b)
where 𝐧𝐥 , is the number of longitudinal bars or bar bundles around the
perimeter of a column core with rectilinear hoops that are laterally
supported by the corner of hoops or by seismic hooks.