Design of PSC Beam for Shear Strength – Worked Example

This article presents a worked example on evaluating the shear strength of a prestressed concrete (PSC) beam using EN 1992-1-1-2004. The beam shown below is an I beam which has been designed in an article published here for serviceability, it is then checked for ultimate bending capacity in another article published here, where the bending strength is achieved by using both prestressed tendons and non-stressed bonded reinforcement bars.  In this article we shall evaluate the shear capacity of the same beam.

The cross-section properties of the beam which were already evaluated when designing the beam for serviceability shall be reproduced below:

Geometric Properties of the Beam

Area of section (A) = = 1500000mm²

Distance of centroid from soffit (y_bot)  = 1510mm

Distance of centroid to top fiber(y_top) = 990mm

Second moment area (I) = 1.26 x 10¹²

Section modulus to top fiber (Z_t) = 1273585859

Section modulus to bottom fiber (Z_b)  = 835000000  

Material Properties

Below are the properties of the concrete, reinforcement, and tendons

Properties of concrete

Concrete grade (f_ck) = 40MPa

Material safety factor = 1.5

Design strength of concrete (f_cd) = 40/1.5 = 26.67MPa

f_ctm = 0.3f_ck^0.667 = 3.5MPa

f_{ctk, 0.05} = 0.7

fctm = 2.46MPa

f_ctd = f_{ctk, 0.05}/1.5 = 1.6MPa  

The properties of tendons and reinforcements provided during design of the beam for serviceability and flexure respectively shall be provided below:  

Tendon Properties

f_pk = 1860MPa

f_pk,0.1 = 0.85fpk = 1581MPa

f_pd = f_pk,0.1/1.15 = 1374MPa

Capacity of tendon = f_pd x As = 1374 x 112 x  = 154KN

Number of tendons required = 2424/154 = 15.7 = 16

Reinforcement properties

f_yk = 500MPa

f_yd = 500/1.15 = 434.8MPa

Diameter = 25mm

Area = (πd²)/4 = 490.9mm2

Total Area = 8 x 490.9 = 3927.2mm2

Force of rebar = f_yd x As = 434.8 x 490.9 x 10^-8= 213.5KN  

Loads

Permanent load = 37.5KN

Impose load = 80KN

Ultimate Load = 1.35 x 37.5 + 1.5 x 80 = 170.6KN

Ultimate moment on the beam = (170.6 x 20000²)/ (8 x 1000) = 8531KNm

Ultimate Shear Force on the beam = (170.6 x 20)/2 = 1706.25KN

Prestress force (P) = 2217.6KN

Prestress force at service (P_s) = 1848KN

Eccentricity of prestressed force at mid-span (e_mid) = 1335mm

Eccentricity of prestressed force at end span (e_end) = 0  

Design for Shear Strength

From fundamental knowledge of statics, the highest shear force in the simply supported beam occurs at its ends. So, we shall design the beam for shear at its ends since it is the most critical section, and then replicate the result throughout the entire span.

We shall check for the beam shear capacity without reinforcement when uncracked, and then when cracked under flexure; then we shall consider the lower of these two resistances as critical.

Afterwards, we shall check the critical resistance against the shear force. If the resistance is greater than the shear force, then the beam does not require shear reinforcement and only minimum area of shear reinforcement shall be provided. However, if the shear force is greater than the critical shear resistance, then we shall design for shear reinforcement.

Shear resistance without reinforcement of Uncracked Section in Bending

$ V_{Rdc\,\,}=\,\,\frac{I.b_w}{S}\,\,\sqrt{\left( f_{ctd} \right) ^2+\,\,\propto _1\sigma _{cp}f_{ctd}} $

σ_cp = N_Ed/A_c

σ_cp = 1848 x 10³/(1.5 x 10⁶)    ≤ 0.2fcd

σ_cp  = 1.23MPa

α₁ = 1

S = S_t + S_w

S_t = (300 x 200) x (2500 – 1510 – 200/2) = 5.3 x 10⁸

S_w = (300 x (2500 – 200 – 1510)²)/2 = 9.4 x 10⁷

S = S_t + S_w = 6.3 x 10⁸

$ V_{Rdc\,\,}=\,\,\frac{1.26 x 10^12 x 300}{6.3 x 10^8}\,\,\sqrt{\left( 1.6 \right) ^2+\,\,1 x 1.23 x 1.6} $ = 1306KN

It is obvious that the Shear force (1706.25KN) is greater than the shear capacity of uncracked section without reinforcement(1306KN)  

Shear Resistance of Cracked Section without Shear Reinforcement

v_{Rdc  x bwd    = {}C_{Rdc K(100ρLfck)^1/3 +} K₁σ_{cp} bwd ≥ (vmin +} K₁σ_cp)bwd

σ_cp = N_Ed/A_c σ_cp = 1848 x 10³/(1.5 x 10⁶)    ≤ 0.2fcd

σ_cp  = 1.23MPa

C_{Rdc =}  0.18/ϒ

C_{Rdc =}  0.18/1.5

_{CRdc =}  0.12

K =    (1 +√200/d)    ≤   2.0

d = e + yt = 1335 + 990 = 2325

K =    (1 +√200/2325) = 1.29

ρ_L = A_sl/bwd ≤   0.02

Since at end span there is no eccentricity and the tendons are anchored at the centroid of the beam, hence tendons  shall neglected when calculating Asl . Only reinforcement shall be considered in the calculation.

A_sl  = 3927.2mm²

ρ_l = (3927.2) / (300 × 2325) = 0.005

K₁  = 0.15

When the value of the parameters into the equation

v_{Rdc  = {0.12 x 0.15 (100 x 0.005 x 40)^{1/3  +} 0.5 x 1.23} x 300 x 2325}

V_Rdc = 941KN

Verifying whether the minimum permitted resistance using V_Rdc = _{{0.035K3/2fck^1/2}} shows that the requirement of minimum resistance is satisfied.

The shear resistance of uncracked section (1306KN) is greater than the shear resistance of cracked section (941KN), hence the resistance of cracked section shall be considered as critical. And since the shear force (1706.25KN) is greater than the critical shear resistance (941KN), then designing for shear reinforcement is required.

Design for Shear reinforcement

Check the upper limit of the compressive struct using:

V_Rdmax(45) = 0.18α_cwb_wd(1 – f_ck/250)f_ck

α_cw = 1 + σ_cp/f_cd

α_cw = 1 + 1.23/26.67 = 1.05

V_Rdmax(45) = 0.18 x 1.05 x 300 x 2325 (1 – 40/250)40

V_Rdmax(45) = 19860KN

Since the struct capacity (19860KN) is greater than the shear force (1706.25KN), then the beam section is sufficient. Now let’s also check whether the beam capacity at 22degree will suffice so that we go for the minimum permissible inclination

V_Rdmax(22) = 0.124b_wd(1 – f_ck/250)f_ck

V_Rdmax(22) = 0.124 x 1.05 x 300 x 2325 (1 – 40/250)40

V_Rdmax(22) = 3040KN

Since the struct capacity at 22degrees (3040KN) is greater than the shear force (1706.25KN), then we shall find the area of shear reinforcement at 22degrees.

 Calculate Asw/s at ϴ = 22

$ \frac{A_{sw}}{s}\,\,\,\,=\,\,\frac{V_{Ed}}{0.78xf_{yk}d\cot 22} $

$ \frac{A_{sw}}{s}\,\,\,\,=\,\,\frac{1706.25 x 10^3}{0.78x 2325 x 500\cot 22} $

$\frac{A_{sw}}{s}\,\,\,\,$ = 0.76

 

Check whether $\frac{A_{sw}}{s}\,\,\,\,$ satisfy the minimum requirement specified by the code.

$$ \frac{A_{sw\min}}{s}\,\,\,\,=\,\,\frac{0.08x\sqrt{f_{ck}}xb_w}{f_{yk}} $$

$$ \frac{A_{sw\min}}{s}\,\,\,\,=\,\,\frac{0.08x\sqrt{40}x300}{500} $$

= 0.3

Since A_smin/s (0.3) is less than As/s (0.76), shear reinforcement is greater than the required minimum area of reinforcement, minimum shear reinforcement requirement is satisfied

 

5) Calculate shear link reinforcement spacing requirement

Assume two-legged shear reinforcement of 10mm is to be used.

Area of 10mm shear reinforcement = 78.58mm2

Area of two-legged 10mm links = 2x 78.58   = 157mm2

157/s = 0.76

s = 157/0.76

s = 206

Provide Y10@200mm for shear reinforcement

Additional longitudinal reinforcement.

The requirement of additional longitudinal reinforcement required by EN 1992-1-2004 is met since the non-stressed rebars are provided all through the beam span and they also extend a full anchorage length.