Shear ultimate limit state
EUROCODE-2

Shear ultimate limit state

DATA

Dimensions (?)

b(cm) h(cm) c(mm)

b: width section. Valid values from 10 to 150
h: depth section. Valid values from 10 to 150
c: Concrete cover. Valid values from 10 to h/3

Concrete strength (?)

fck (MPa) γc αcc

f_ck: the characteristic compressive cylinder strength of concrete. Valid values from 20 to 90
γc: Partial factor for concrete for ultimate limit state. National choice 2.4.2.4(1). Valid values from 1 to 2
α_cc: National choice 3.1.6 (1)P: Coefficient of fatigue strength of the concrete. Recommended value is 1.0. Valid values from 0.5 to 1

Reinforcement strength

fyk(MPa) (?) γs

f_yk: the characteristic strength of reinforcement. Valid values from 400 to 600
γs: Partial factor for reinforcement for ultimate limit state. National choice 2.4.2.4(1). Valid values from 1 to 1.8

shear reinforcement

REINFORCEMENT
Member with shear reinforcement

Transverse reinforc. (?)

A_sw(cm²) s(cm) α(^osex)

Asw: Cross-sectional area of shear reinforcement. Valid values from Ast,min >0 to Asw,max (Φ32 bars in width section separated 5cm).
s: The spacing of the stirrups. Valid values from 5 to 2h.
α: Angle between the reinforcements and the member′s axis. Valid values from 45 to 90.

You can enter the area directly or through the following data:

type	Num	Φ
1
2

Intro

Tensile reinforcement

A_s1 (cm²) (?) Φmax

As1: Cross-sectional area of tensile reinforcement. Valid values from 0 to As1,max (layer Φ40 with bar spacing s=5 cm).
Φmax: max. nominal diameter of a reinforcing bar. Valid values from 5 to 40

You can enter the area directly or through the following data:

type	Num	Φ
1
2

Intro

Compression reinforc.

A_s2 (cm²) (?)

As2: Cross-sectional area of compression reinforcement. Valid values from 0 to As2,max (0.5·fcd·Ac/fyd).

You can enter the area directly or through the following data:

type	Num	Φ
1
2

Intro

INTERNAL FORCES

Shear force V_Ed(Kn) (?)

Design shear force in the section considered. Valid values from 0 to 0.5·fcd·Ac

Axial force N_Ed(Kn) (?)

Design axial force in the section considered. Valid values from 0 to fcd·Ac

Inner lever arm (?)

z = · d

Inner lever arm corresponding to the bending moment. For concrete without axial force, use z = 0,9·d. Valid values from 0,9·d to 0,6·d

VALUES FOR USE IN A COUNTRY
Use values recommended

Value v_min (?)

·k^3/2·f_ck^1/2

Value design shear (see clause 6.2.2(1)
The recommended value is 0.035·k^3/2·f_ck^1/2(Valid values from (0.01 to 0.09)·k^3/2·f_ck^1/2)

C_Rd,c (?) k1

/γc

Values design shear (see clause 6.2.2(1)

C_Rd,c: The recommended value is 0.18/γc (Valid values from (0.10 to 0.25)/γc)
k1: The recommended value is 0.15 (Valid values from 0.1 to 0.2).

min(cotθ) (?) max(cotθ)

≤ cotθ ≤

θ is the angle between the concrete compression strut and the member axis.

min(cotθ): The recommended value is 1. Valid values from 0.2 to 1
max(cotθ): The recommended value is 2.5. Valid values from 1 to 4

Value v₁ (?)

·0.6·(1-f_ck/250)

Value design shear (see clause 6.2.3(3)
The recommended value is 0.6·(1-f_ck/250). You can multiply this value by a factor between 0.1 and 3

Value a_cw (?)

·Recom_value

Value design shear (see clause 6.2.3(3). The recommended value is

1 for non-prestressed structures
(1+σ_cp/f_cd), for 0<σ_cp≤0.25 f_cd
1.25, for 0.25·f_cd<σ_cp≤0.5·f_cd
2.5(1-σ_cp/f_cd), for 0.5·f_cd<σ_cp≤f_cd

You can multiply this value by a factor between 0.1 and 3

RESULT

Design shear force	Design shear strength		Requirement
V_Ed (KN)	V_Rd,max (KN)	V_Rd (KN)	V_Ed < V_Rd and V_Rd,max
200	305.01	239.77	OK

DETAILS OF CALCULATION

Notation and methodology according to clause 6.2 of EC2

V_Rd,max (Resistance crushing of compression struts)= 305.01 kN
V_Rd,max = α_cw · b_w · z · v₁ · f_cd·(cotθ + cotα) / (1+cot²θ)
V_Rd,max = 1·300·328·0.54·16.67·(2.5+0) / (1+2.5²) = 305007 N
where:

f_cd = α_cc · f_ck / γc = 1 · 25 / 1.5 = 16.67 MPa
α_cw = 1 · 1 = 1
Case: non-prestressed structure or non-axial force
z = 0.9 · d = 32.76 cm
d (effective depth) = h – c - Φ_max/2 = 40 – 3 – 1.2/2 = 36.4 cm
v₁ = 1·0.6·(1-25/250) = 0.54
cotθ = 2.5 (1 ≤ cotθ ≤ 2.5)
(it get the maximum value of V_Rd,s that satisfies V_Ed ≤ V_Rd,max)

V_Rd (Tensile strength in the web) = 239.77 KN
V_Rd = max (V_Rd,c ; V_Rd,s), in which:

V_Rd,c (Resistance without shear reinforcement)= 53.56 kN
V_Rd,c = [C_Rd,ck(100·ρ₁·f_ck)^1/3 + k₁·σ_cp] b_w·d
V_Rd,c = [0.12·1.741·(100·0.0052·25)^1/3 + 0.15·0]·300·364 = 53562 N
with a minimum of
V_{Rd,c (min)} = (v_min + k₁·σ_cp) b_w·d
V_{Rd,c (min)} = (0.402 + 0.15·0)·300·364 = 43909 N
where:
- C_Rd,c = 0.18/1.5 = 0.12
- k = min [1+(200/d)^1/2 , 2] = 1.741
  1+(200/d)^1/2 = 1+(200/364)^1/2 = 1.741
- ρ1 = min (A_s1/(b∙d) , 0.02) = 0.0052
  A_s1/(b∙d) = 5.65/(100∙36.4) = 0.0052
- σ_cp = min [N_Ed/(b·h) ; 0.2·f_cd] = min (0 ; 3.33) = 0 MPa
- v_min = 0.035·k^3/2·f_ck^1/2 = 0.035·1.741^3/2·25^1/2 = 0.402 MPa

V_Rd,s (Resistance with shear reinforcement)= 239.77 kN
V_Rd,s = A_sw/s · z · f_ywd · (cotθ+cotα)·sinα
V_Rd,s = 101/15 · 32.76 · 434.78 · (2.5+0)·1 = 239765 N
with f_ywd = f_yk / γa = 500 /1.15 = 434.78 N/mm2

Shear ultimate limit state EUROCODE-2

Shear ultimate limit state
EUROCODE-2