Shear capacity without shear reinforcement

1. General

In FEM-Design it is possible to calculate the shells (plates and walls) with or without shear reinforcement. When calculating without the shear reinforcement, the shear force that is taken into consideration and the amount of longitudinal reinforcement that is used in the calculations may sometimes raise questions. Here is an example on how it is calculated in FEM-Design.

2. Shear forces and shear resultant

In the example below, we can look at one corner for the analysis. There are two shear forces – Tx’y’ = -456,28 kN/m (figure 1) and Tx’z’ = -105,59 kN/m (figure 2). We see the shear force as a value per meter. 


Figure 1 - Shear force Tx'z'


Figure 2 - Shear force Ty'z'


In the analysis, these two forces are combined into one resultant shear force. We use the regular Pythagoras formula v_Ed = sqrt((-456,28)² + (-105,59)²) = 468,34 kN.
We can see the same result in the design results – RC shell Design forces v_Ed (figure 3). This is the full shear force (without any reduction).


Figure 3 - Shear design force

3. Applied shear capacity

In FEM-Design version 21 and earlier, we had the shell shear capacity only with the longitudinal bars (shear capacity without shear reinforcement) and no reduction for the shear force. In the newest version 22, we made an improvement in the shear design, and it contains extra features now.
More info about this can be found in the WIKI (https://wiki.fem-design.strusoft.com/xwiki/bin/view/New%20features/New%20features%20in%20FEM-Design%2022/RC%20design/#HShearreinforcementdesign )

Mainly we have separated the shear capacities into three groups:
 • V_Rd,max = the concrete compression strut strength (EN1992-1-1 6.2.3.)
 • V_Rd,c = the concrete shear capacity without shear reinforcement (but with longitudinal reinforcement, according to EN1992-1-1 6.2.2.). This used to be the only calculation method for the shells and this is described in the example below. Now it is used together with the other capacities.
 • V_Rd,s = capacity with shear reinforcement (EN1992-1-1 6.2.3.). This is a new method and is described in our WIKI page.

4. Shear capacity without shear reinforcement

The formula for applied shear capacity is according to the Eurocode 1992-1-1 equation 6.2.a:
 V_Rd,c = [C_Rd,c * k * (100 * ρ₁ * f_ck)^(1/3) + k₁ * σ_cp] * b_w * d
which should be bigger than the V_Rd,c,min

V_Rd,c,min is according to the equation 6.2.b:
 V_Rd,c,min = (v_min + k₁ * σ_cp) * b_w * d

For the calculation we need the following data:
 • Plate thickness: 150mm
 • Concrete cover x’: 20mm
 • Concrete cover y’: 40mm
 • Longitudinal reinforcement on tensioned side (top): x’ direction 16/180 = 1117 mm²/m (figure 4) and y’ direction 16/160 = 1257 mm²/m (figure 5). 


Figure 4 - Reinforcement in x' direction


Figure 5 - Reinforcement in y' direction


 • Direction of the resultant shear force: α = arctan(105,59 / 456,28) = 13,03°
 • We consider a 1m wide plate
 • Concrete strength: f_ck = 45MPa
 • The direction of the reinforcement coincides with the local system of the plate and thus: ξ= 0°, η=90°
 • k₁=0,15
 • C_Rd,c = 0,18/γ_c = 0,18 / 1,5 = 0,12
 • d = (d_x + d_y) / 2 = ((150 - 20 - 8) + (150 - 40 - 8)) / 2 = 112 mm
 • k = 1 + sqrt(200 / d) = 1 + sqrt(200 / 112) = 2,34 which must be less or equal than 2, so k = 2
 • A_α = A_x * cos²(α - ξ) + A_y * cos²(α - η) = 1124 mm²
 • ρ₁ = A_α / (b_w * d) = 1124 / (1000 * 112) = 0,01004
 • σ_cp = 0, since there is no applied normal force 

Now we can calculate the V_Rd,c:
 V_Rd,c = [0,12 * 2 * (100 * 0,01004 * 45)^(1/3) + 0,15*0] * 1000 * 112 = 95,73 kN
 v_min = 0,0035 * 2^(3/2) * 45^(1/2) = 0,664078
 V_Rd,c,min = (0,664078 + 0,15 * 0) * 1000 * 112 = 74,38 kN

Since V_Rd,c is bigger than V_Rd,c,min then we can use the value 95,73 kN.
We can see the same value in FEM-Design result Shear Capacity - vRd,c (figure 6):


Figure 6 - Shear capacity


We can also see in the detail results the calculation (figure 7):


Figure 7 - Detailed results


We can see from the detailed result, that the capacity is compared against a reduced shear force 300,72 kN not the full 468,34 kN. In this example, we can see that the capacity without shear reinforcement is not enough and shear reinforcement is needed even for the reduced force.

The reduction is made based on the settings from the new extended design tab and can be read more about in the WIKI page.