sr,max=3.4c+0.425⋅k1⋅k2⋅ϕρp,effs sub r comma m a x end-sub equals 3.4 c plus the fraction with numerator 0.425 center dot k sub 1 center dot k sub 2 center dot phi and denominator rho sub p comma e f f end-sub end-fraction (high-bond bars), (pure tension).
fcd=αcc⋅fckγc=0.85⋅401.5=22.67 MPaf sub c d end-sub equals the fraction with numerator alpha sub c c end-sub center dot f sub c k end-sub and denominator gamma sub c end-fraction equals the fraction with numerator 0.85 center dot 40 and denominator 1.5 end-fraction equals 22.67 MPa
The story of Worked Examples to Eurocode 2: Volume 2 is one of a project left unfinished. While engineers may lament the absence of this specific collection of examples, the situation is far from a dead end. The intended topics—Foundations, Serviceability, Fire, and Retaining walls—are well-covered by a range of alternative, authoritative, and publicly available documents. By combining the published Volume 1 with the JRC report, Tony Threlfall's design guide, and The Concrete Centre's own "How to" series, a designer can assemble a comprehensive library of worked examples that surpasses even the original ambitious plan for Volume 2.
As=MEdfyd⋅zcap A sub s equals the fraction with numerator cap M sub cap E d end-sub and denominator f sub y d end-sub center dot z end-fraction worked examples to eurocode 2 volume 2
3. Worked Example 2: ULS Shear and Variable Strut Inclination Method
Fs=400⋅(2000.8⋅450)+80=222.2 kN+80 kN=302.2 kNcap F sub s equals 400 center dot open paren the fraction with numerator 200 and denominator 0.8 center dot 450 end-fraction close paren plus 80 equals 222.2 kN plus 80 kN equals 302.2 kN Step 2: Determine Required Steel Area ( Ascap A sub s Using high-yield steel (
Providing clear mathematical workflows that can be translated into design spreadsheets or verified against commercial FEA (Finite Element Analysis) software. Core Technical Areas Covered in Volume 2 sr,max=3
Run the same geometry in your structural analysis software (Tekla, ETABS, or Autodesk Robot). Typically, software gives you a "pass/fail" flag. The worked example shows you how the software calculated the factor of safety. You will often find software uses the simplified method for columns; Volume 2 will show you the general method for verification.
: Includes derived formulae and design charts to simplify routine calculations for column slenderness, reinforcement areas, and shear capacity. Worked Example to Eurocode 2 Vol. - Academia.edu
No resource is perfect. Volume 2 has two notable limitations: Worked Example 2: ULS Shear and Variable Strut
Concrete strut capacity: ( \nu = 0.6(1 - f_ck/250) = 0.6(1-0.14)=0.516 ) ( f_cd = 35/1.5 = 23.3 \text MPa ) ( \theta = 22^\circ ) initially: ( \cot \theta = 2.5 ) Check ( \frac\tau_t,Edf_cd \sin\theta \cos\theta + \frac\tau_v,Edf_cd \sin\theta \cos\theta \le 1 )? No – better use: [ \fracT_EdT_Rd,max + \fracV_EdV_Rd,max \le 1 ] But easier: ( \tau_total = \sqrt\tau_t,Ed^2 + \tau_v,Ed^2 = \sqrt2.31^2 + 0.74^2 = 2.43 \text MPa ) Allowable ( \tau_max = 0.5 \nu f_cd = 0.5 \times 0.516 \times 23.3 \approx 6.0 \text MPa ) → OK.
MEd = 1.35 x (10 x 6^2 / 8) + 1.5 x (5 x 6^2 / 8) = 63.9 kNm