APRĒĶINU PIEMĒRI

CALCULATION EXAMPLES

3 atrisināti piemēri ar pilnām atskaitēm — ievaddati, formulas, rezultāti

3 solved examples with full reports — input data, formulas, results

Liekts DZB šķērsgriezums

RC Bending Section Design

EN 1992-1-1 (Eurocode 2)

b × h	300 × 600 mm
L	6000 mm
c_nom	35 mm

M_Ed	350 kNm
V_Ed	180 kN
M_Ed,SLS	245 kNm

BetonsConcrete	C30/37
TēraudsSteel	B500B
Stiegr.Reinf.	4Ø25 (apakšābottom), 2Ø12 (augšātop)

CALCULATION REPORT

EN 1992-1-1 (Eurocode 2) - Reinforced Concrete Cross-Section Design

Bending resistance: Analytical calculation with rectangular concrete stress block

1. INPUT DATA

Geometry: b = 300 mm, h = 600 mm, L = 6000 mm, c_nom,bot = 35 mm, c_nom,top = 35 mm

Materials: Concrete f_ck = 30.0 MPa (f_cd = 20.00 MPa), Reinforcement f_yk = 500 MPa (f_yd = 435 MPa)

Reinforcement: Bottom 4Ø25.0 (A_s = 1963 mm²), Top 2Ø12.0 (A_s,top = 226 mm²)

Actions: M_Ed (ULS) = 350.00 kNm, M_Ed (SLS) = 245.00 kNm, V_Ed = 180.00 kN

2. EFFECTIVE DEPTH

d = h - c_nom - φ_w - φ_bot/2 = 600 - 35 - 8 - 25/2 = 544.5 mm

3. BENDING RESISTANCE

Neutral axis: x = (A_s × f_yd) / (0.8 × b × f_cd) = (1963 × 435) / (0.8 × 300 × 20.00) = 177.9 mm

Relative depth: ξ = x / d = 177.9 / 544.5 = 0.327 (ξ_bal = 0.45) ✓ OK

Internal lever arm: z = d - 0.4×x = 544.5 - 0.4×177.9 = 473.4 mm

Resistance: M_Rd = A_s × f_yd × z = 1963 × 435 × 473.4 / 10⁶ = 404.10 kNm

Check: M_Ed / M_Rd = 350.00 / 404.10 = 0.87 ✓ OK

4. SHEAR RESISTANCE

Concrete resistance without shear reinforcement (EN 1992-1-1, 6.2.2):

k = min(1 + √(200/d), 2.0) = 1.606

ρ_l = A_s/(b×d) = 0.01202

V_Rd,c = [0.18/γ_c × k × (100×ρ_l×f_ck)^1/3] × b × d = 104.01 kN

Shear reinforcement:

Type 1: 2Ø8 / 150 mm → A_sw/s = 101/150 = 0.670 mm²/mm

Σ A_sw/s = 0.670 mm²/mm

Optimal angle: cot θ = 2.00 (θ = 26.6°) | LV NA: 1.0 ≤ cot θ ≤ 2.0

Shear reinforcement resistance: V_Rd,s = (A_sw/s) × z × f_ywd × cot θ = 0.670 × 473.4 × 435 × 2.00 / 1000 = 275.87 kN

Concrete strut: V_Rd,max = 599.84 kN

Detailing requirements: A_sw,min = 39 mm² ✓, s_max = 408 mm ✓

Check: V_Rd = min(V_Rd,s, V_Rd,max) = min(275.9, 599.8) = 275.87 kN

Utilization: η = V_Ed / V_Rd = 180.0 / 275.9 = 0.65 ✓ OK

5. CRACK WIDTH (SLS)

Calculation per EN 1992-1-1, 7.3.4 (using M_Ed,SLS = 245.00 kNm):

Effective reinforcement ratio: ρ_p,eff = A_s / A_c,eff = 0.001963 / 0.041625 = 0.0472

Modular ratio: α_e = E_s / E_cm = 200000 / 32837 = 6.091

Reinforcement stress: σ_s = M_Ed,SLS/(A_s×z) = 254.6 MPa

Mean strain difference (EN 1992-1-1, 7.9):

ε_s-ε_cm = max(0.6×σ_s/E_s; [σ_s - k_t×f_ctm/ρ_p,eff×(1+α_e×ρ_p,eff)]/E_s)

= max(0.6×254.6/200000; [254.6 - 0.4×2.90/0.0472×(1+6.091×0.0472)]/200000)

= max(0.000764; 0.001115) = 0.001115

Maximum crack spacing (EN 1992-1-1, 7.11):

s_r,max = k₃×c + k₁×k₂×k₄×φ/ρ_p,eff

= 3.4×43 + 0.8×0.5×0.425×25/0.0472 = 236.3 mm

Crack width (EN 1992-1-1, 7.8):

w_k = (ε_s-ε_cm) × s_r,max = 0.001115 × 236.3 = 0.263 mm

Check per EN 1992-1-1, Table 7.1N (exposure class XC1): w_k ≤ w_max = 0.40 mm ✓ OK

6. DEFLECTION (SLS)

Simplified calculation (elastic method, using M_Ed,SLS = 245.00 kNm):

E_cm = 22×(f_ck/10)^0.3 (EN 1992-1-1, Table 3.1)

I_eff ≈ (I_cr + I_gross)/2 (simplified average of cracked and uncracked)

δ = k × M_Ed,SLS × L²/(E×I) ≈ 9.60 mm | L/d = 11.0

Check: δ ≤ L/250 = 24.0 mm ✓ OK

7. SUMMARY

Ultimate Limit State (ULS):

Bending	η_M = M_Ed/M_Rd = 350.0/404.1 = 0.87	✓ OK
Shear	η_V = V_Ed/V_Rd = 180.0/275.9 = 0.65	✓ OK

Serviceability Limit State (SLS):

Crack width	w_k = 0.263 mm ≤ w_max = 0.40 mm	✓ OK
Deflection	δ = 9.60 mm ≤ L/250 = 24.0 mm	✓ OK

✓ ALL CHECKS PASSED

8. CROSS-SECTION

DZB kolonnas aprēķins

RC Column Design

EN 1992-1-1 — Nominal Curvature

b × h	400 × 400 mm
L	3.5 m
c_nom	35 mm

N_Ed	1800 kN
M_Ed,top	120 kNm
M_Ed,bot	80 kNm

BetonsConcrete	C35/45
TēraudsSteel	B500B
Stiegr.Reinf.	8Ø20 (3+3 shēmalayout)

COMPRESSED ELEMENT DESIGN

EN 1992-1-1 (Eurocode 2) | Nominal Curvature Method

M-N diagram: concreteproperties v0.7 | EurocodeParabolicUltimate

1. INPUT DATA

Geometry:

Cross-section: rectangular, b = 400 mm, h = 400 mm
Length: L = 3.50 m
Cover: c_nom = 35 mm

Materials:

Concrete: C35/45 → f_ck = 35.0 MPa
Steel: B500B → f_yk = 500 MPa

Reinforcement:

Corner bars: 4×Ø20.0 mm
Side bars: 12×Ø20.0 mm
Total count: 16
Stirrup: Ø8.0 mm

Loads (ULS):

Axial load: N_Ed = 1800.0 kN (compression)
Moments Y-Y axis: M_y,top = 120.0 kNm, M_y,bot = 80.0 kNm
Moments Z-Z axis: M_z,top = 0.0 kNm, M_z,bot = 0.0 kNm

2. MATERIAL DESIGN VALUES

Concrete design compressive strength (EC2 3.1.6):

f_cd = α_cc × f_ck / γ_c = 0.85 × 35.0 / 1.50 = 23.33 MPa

Reinforcement design strength (EC2 3.2.7):

f_yd = f_yk / γ_s = 500 / 1.15 = 434.78 MPa

AXIAL UTILIZATION (EN 1992-1-1 §6.1)

N_Rd,max = (A_c − A_s)·f_cd + A_s·f_yd = 5802 kN; N_Ed/N_Rd,max = 0.31

The utilization above is the moment utilization at the given axial force (M_Ed/M_Rd). It correctly decides whether the point (N; M) lies inside the M-N envelope, but it does not show how close the section is to capacity overall. Programs that report a single "utilization" figure usually report proximity to the envelope, which is why their value tends to be considerably higher. Read both together.

DETAILING REQUIREMENTS (EN 1992-1-1 §9.5.3)

Minimum diameter: Ø_w ≥ max(6; Ø_main/4) = max(6; 5.0) = 6.0 mm | Ø8 ✓ OK

Y-Y AXIS CALCULATION

3.1 Effective (design) length:

Boundary conditions: Top: Pinned (F=1.0), Bottom: Pinned (F=1.0)

l_0,y = F_y × L = 1.00 × 3500 = 3500.0 mm

3.2 Slenderness:

Radius of gyration: i = h / √12 = 115.47 mm

λ_y = l₀ / i = 3500.0 / 115.47 = 30.31

3.3 Slenderness limit (EC2 5.8.3.1):

λ_lim = 20 · A · B · C / √n

A = 0.7 (creep effect)
B = 1.1 (reinforcement effect)
r_m = M₀₁/M₀₂ = 80.00/120.00 = 0.667 — end moments with the same sign (M₀₁ and M₀₂ give tension on the same side)
C = 1.7 − r_m = 1.7 − (0.667) = 1.033
n = N_Ed/(A_c·f_cd) = 0.4821

λ_lim,y = 22.92

3.4 Slenderness check:

λ_y = 30.31 ≥ λ_lim = 22.92 → Element IS slender

Second order effects must be considered!

3.5 Design moments:

First-order (input) moments:

M_top = 120.00 kNm, M_bot = 80.00 kNm
First order moment: M₀₁ = 95.75 kNm, M₀₂ = 135.75 kNm (ieskaitot imperfekcijas e_i)
Equivalent moment: M_0e = 119.75 kNm

Second-order moments (deflection effect):

Curvature distribution factor: c = 10 (auto)
A smaller c gives a larger second-order moment. EC2 5.8.8.2(3): c = 10 (≈ π²) for a constant cross-section; c = 8 — for a constant first-order moment (the more conservative bound). Automatically adopted c = 10 because the first-order moment varies along the member.
e₂ = (1/r)·L₀²/c = 0.000017 · 3500² / 10 = 20.38 mm
Second order moment: M₂ = 36.69 kNm (M₂ = N_Ed · e₂)

Minimum moment (EC2 6.1(4)):

Minimum eccentricity: e₀ = max(h/30, 20) = max(400/30, 20) = 20.0 mm
Minimum moment: M_min = e₀ · N_Ed = 20.0 · 1800.0 / 1000 = 36.00 kNm

Design moment (largest value):

M_Ed,calc = max(M₀₂; M_0e+M₂; M₀₁+0.5·M₂) = 156.44 kNm

M_min = 36.00 kNm

M_Ed,y = max(M_Ed,calc, M_min) = 156.44 kNm
(Governed by the input moment including second-order effects)

3.6 Resistance (accurate calculation from M-N diagram):

M_Rd,y calculated from the M-N interaction diagram at N_Ed = 1800.0 kN

Calculated using the concreteproperties library with the Eurocode 2 parabola-rectangle concrete model

M_Rd,y = 387.70 kNm

3.7 Utilization:

M_Ed,y / M_Rd,y = 156.44 / 387.70 = 0.40 ✓ OK

Z-Z AXIS CALCULATION

3.1 Effective (design) length:

Boundary conditions: Top: Pinned (F=1.0), Bottom: Pinned (F=1.0)

l_0,z = F_z × L = 1.00 × 3500 = 3500.0 mm

3.2 Slenderness:

Radius of gyration: i = h / √12 = 115.47 mm

λ_z = l₀ / i = 3500.0 / 115.47 = 30.31

3.3 Slenderness limit (EC2 5.8.3.1):

λ_lim = 20 · A · B · C / √n

A = 0.7 (creep effect)
B = 1.1 (reinforcement effect)
r_m = 1.0 — no end moments, or they are smaller than the imperfection moment — the first-order moment is predominantly from imperfections or transverse loading (EC2 5.8.3.1(1)) (M₀₂ = 0.00 kNm ≤ M_imp = 15.75 kNm)
C = 1.7 − r_m = 1.7 − (1.000) = 0.700
n = N_Ed/(A_c·f_cd) = 0.4821

λ_lim,z = 15.52

3.4 Slenderness check:

λ_z = 30.31 ≥ λ_lim = 15.52 → Element IS slender

Second order effects must be considered!

3.5 Design moments:

First-order (input) moments:

M_top = 0.00 kNm, M_bot = 0.00 kNm
First order moment: M₀₁ = 15.75 kNm, M₀₂ = 15.75 kNm (ieskaitot imperfekcijas e_i)
Equivalent moment: M_0e = 15.75 kNm

Second-order moments (deflection effect):

Curvature distribution factor: c = 8 (auto)
A smaller c gives a larger second-order moment. EC2 5.8.8.2(3): c = 10 (≈ π²) for a constant cross-section; c = 8 — for a constant first-order moment (the more conservative bound). Automatically adopted c = 8 because the first-order moment is constant (or arises only from imperfections).
e₂ = (1/r)·L₀²/c = 0.000017 · 3500² / 8 = 25.48 mm
Second order moment: M₂ = 45.86 kNm (M₂ = N_Ed · e₂)

Minimum moment (EC2 6.1(4)):

Minimum eccentricity: e₀ = max(b/30, 20) = max(400/30, 20) = 20.0 mm
Minimum moment: M_min = e₀ · N_Ed = 20.0 · 1800.0 / 1000 = 36.00 kNm

Design moment (largest value):

M_Ed,calc = max(M₀₂; M_0e+M₂; M₀₁+0.5·M₂) = 61.61 kNm

M_min = 36.00 kNm

M_Ed,z = max(M_Ed,calc, M_min) = 61.61 kNm
(Governed by the input moment including second-order effects)

3.6 Resistance (accurate calculation from M-N diagram):

M_Rd,z calculated from the M-N interaction diagram at N_Ed = 1800.0 kN

Calculated using the concreteproperties library with the Eurocode 2 parabola-rectangle concrete model

M_Rd,z = 387.23 kNm

3.7 Utilization:

M_Ed,z / M_Rd,z = 61.61 / 387.23 = 0.16 ✓ OK

BIAXIAL CHECK (EC2 5.8.9)

Biaxial bending formula:

(M_Ed,y/M_Rd,y)^a + (M_Ed,z/M_Rd,z)^a ≤ 1.0

Exponent 'a' (depends on axial load ratio):

n = N_Ed / N_Rd = 0.304
a = 1.17 (linear interpolation: N_Ed/N_Rd=0.1→a=1.0, 0.7→a=1.5, 1.0→a=2.0)

Check:

(156.44/387.70)^1.17 + (61.61/387.23)^1.17 = 0.462 ≤ 1.0 ✓ OK

REINFORCEMENT

Minimum reinforcement:

A_s,min = 0 mm² (EC2 9.5.2)

Required reinforcement:

A_s,req = 414 mm²

Provided reinforcement:

A_s,prov = 5027 mm² ✓ OK

Arrangement:

4 corner bars: Ø20.0 mm
12 side bars: Ø20.0 mm
Total: 16 bars

SUMMARY

Overall utilization (governing check): 0.46 (biaxial interaction)

✓ RESISTANCE SUFFICIENT

All checks passed.

N-M INTERACTION DIAGRAM (Y-Y AXIS)

Key points:

Maximum compression: N_max = 5802 kN (M = 0)
Maximum moment: M_max = 396.0 kNm
Pure bending (N≈0): M = 314.6 kNm

N-M INTERACTION DIAGRAM (Z-Z AXIS)

Key points:

Maximum compression: N_max = 5802 kN (M = 0)
Maximum moment: M_max = 395.6 kNm
Pure bending (N≈0): M = 314.5 kNm

3D N-M INTERACTION SURFACE

4 pāļu režģoga aprēķins (STM)

4-Pile Cap Design (STM)

EN 1992-1-1 cl. 6.5 — Strut-and-Tie

N_Ed	2400 kN
PāļiPiles	4 × Ø400
SolisSpacing	1500 × 1500 mm

h_cap	900 mm
KolonnaColumn	500 × 500 mm
c_nom	50 mm

BetonsConcrete	C30/37
TēraudsSteel	B500B
Stiegr.Reinf.	Ø20 (galv.main), Ø12 (sadal.distr.)

PILE CAP DESIGN — STRUT-AND-TIE METHOD

EN 1992-1-1 (EC2) section 6.5 + FIB Model Code

1. INPUT DATA

Configuration	4 piles
Concrete class	C30/37
Reinforcement class	B500B
Pile diameter D	400 mm
Column cross-section c₁ × c₂	500 × 500 mm
Cap depth h	900 mm
Cover c_nom,bot / c_nom,top	50 / 50 mm (sides = top: 50 mm)
Main reinforcement diameter φ_main	Ø20 mm
Distribution reinforcement diameter φ_dist	Ø12 mm

Material design parameters:

f_cd = α_cc × f_ck / γ_C = 1.0 × 30 / 1.5 = 20.00 MPa

f_yd = f_yk / γ_S = 500 / 1.15 = 434.8 MPa

ν' = 1 − f_ck / 250 = 1 − 30 / 250 = 0.880

Effective depth:

d = h − c_nom,bot − φ_main − φ_main/2 = 900 − 50 − 20 − 20/2 = 820 mm

2. LOADS

ULS limit state:

N_Ed (axial compression)	2400.0 kN
V_Ed,x (shear force X)	0.0 kN
V_Ed,y (shear force Y)	0.0 kN
M_Ed,x (moment about X axis)	0.0 kNm
M_Ed,y (moment about Y axis)	0.0 kNm

3. PILE REACTIONS

Formula:

R_i = N_Ed/n + M_x,tot × y_i / Σy_j² + M_y,tot × x_i / Σx_j²

Σx_j² = 2.2500 m², Σy_j² = 2.2500 m²

Example — Pile P1:

R₁ = 2400.0/4 + 0.0 × (-0.750) / 2.2500 + 0.0 × (-0.750) / 2.2500 = 600.0 kN

All pile reactions:

Pile	x (mm)	y (mm)	R_ULS (kN)	R_SLS (kN)
P1	-750	-750	600.0	425.0
P2	750	-750	600.0	425.0
P3	-750	750	600.0	425.0
P4	750	750	600.0	425.0

4. STRUT-AND-TIE MODEL

4.1. Struts

Strut angle and force:

θ_i = arctan(z / L_horiz,i), z = d − a₀/2

F_strut,i = R_i / sin(θ_i)

Example — Strut N0→P1:

L_horiz = 1061 mm

θ = arctan(678 / 1061) = 32.6°

F_strut = 600.0 / sin(32.6°) = 1114.6 kN

Strut	R_pile (kN)	θ (°)	F_strut (kN)
Strut N0→P1	600.0	32.6	1114.6
Strut N0→P2	600.0	32.6	1114.6
Strut N0→P3	600.0	32.6	1114.6
Strut N0→P4	600.0	32.6	1114.6

4.2. Ties

Tie force — horizontal component of the strut forces:

F_tie = R_i / tan(θ_i) (summed by directions)

The tie at the bottom of the pile cap forms a force couple: the vertical pile reaction R_i at distance L_horiz,i from the column axis produces a moment carried by a force couple (compression in the strut above + tension in the tie below) with lever arm z = d − a₀/2:

M_i = R_i × L_horiz,i; T_i × z = M_i ⟹ F_tie,i = R_i × L_horiz,i / z = R_i / tan(θ_i)

Example — Strut N0→P1:

M = R × L_horiz = 600.0 × 1061 / 1000 = 636.6 kNm

F_tie = M / z = 600.0 × 1061 / 678 = 938.9 kN

Tie	Direction	F_tie (kN)
Tie X (y=-750)	X	664.2
Tie X (y=750)	X	664.2
Tie Y (x=-750)	Y	664.2
Tie Y (x=750)	Y	664.2

Total bottom tie force (force-couple result):

Σ F_tie,X = 1328.4 kN | Σ F_tie,Y = 1328.4 kN

5. NODE CHECK (EN 1992-1-1 §6.5.4)

Permissible node stress (EN 1992-1-1 §6.5.4):

CCC (triaxial compression): σ_Rd,max = K₄ × ν' × f_cd = 3.0 × 0.880 × 20.00 = 52.80 MPa

EN 1992-1-1 §6.5.4(6): If triaxial compression is confirmed, the concrete resistance may be increased. In the CCC node 4 struts run to different sides + the vertical column — the concrete is compressed from at least 3 directions, providing a confinement effect. K₄ = 3.0 (≈ 3× compared with K₁ = 1.0 for a surface node).

CCT: σ_2Rd,max = K₂ × ν' × f_cd = 0.85 × 0.880 × 20.00 = 14.96 MPa

CTT: σ_3Rd,max = K₃ × ν' × f_cd = 0.75 × 0.880 × 20.00 = 13.20 MPa

Example — Column (top) (CCC):

A_node = 250000 mm²

F = 2400.0 kN = 2400000 N

σ_Ed = F / A = 2400000 / 250000 = 9.60 MPa

σ_Ed / σ_iRd,max = 9.60 / 52.80 = 0.18 ✓ OK

Node	Type	σ_Ed (MPa)	σ_iRd,max (MPa)	Utilization	Status
Column (top)	CCC	9.60	52.80	0.18	✓ OK
Pile P1	CCT	4.77	14.96	0.32	✓ OK
Pile P2	CCT	4.77	14.96	0.32	✓ OK
Pile P3	CCT	4.77	14.96	0.32	✓ OK
Pile P4	CCT	4.77	14.96	0.32	✓ OK

6. TIE REINFORCEMENT

Required reinforcement area:

A_s,req = F_tie / f_yd

Example — Tie X (y=-750):

A_s,req = F_tie × 10³ / f_yd = 664.2 × 10³ / 434.8 = 1528 mm²

A_s,min = max(0.26 × f_ctm/f_yk × b × d; 0.0013 × b × d) = 692 mm²

A_s,design = max(A_s,req; A_s,min) = max(1528; 692) = 1528 mm²

n = ⌈1528 / 314.2⌉ = 5 pcs Ø20

A_s,prov = 5 × 314.2 = 1571 mm² ✓ OK

Tie	F_tie (kN)	A_s,req (mm²)	Bars	A_s,prov (mm²)	Status
Tie X (y=-750)	664.2	1528	5Ø20	1571	✓ OK
Tie X (y=750)	664.2	1528	5Ø20	1571	✓ OK
Tie Y (x=-750)	664.2	1528	5Ø20	1571	✓ OK
Tie Y (x=750)	664.2	1528	5Ø20	1571	✓ OK

7. STRUT CHECK (EN 1992-1-1 §6.5.2)

7.1. Strut cross-section area:

Strut width at the node with compression zone depth a₀ optimisation (CCIP-057, Section 4.1; Santos et al.):

CCC node (column):

w_CCC = a₀ × sin(θ) + c_share × cos(θ)

Where a₀ — optimised compression zone depth below the column

CCT node (pile):

w_CCT = D_pile × sin(θ) + u × cos(θ) (EN 1992-1-1 Fig.6.26)

Where u = 2 × (h − d) — tie depth (2× distance to the bar centroid)

Lever arm: z = d − a₀/2, θ = arctan(z / L_horiz)

A_strut = w × b; σ_strut = F_strut / A_strut

Optimisation: a₀ is increased until the capacity of both ends is balanced (CCIP-057 s.4.2: "trial and error to find the depth of the top node at which the stress equals the design strength"). Additional sources: Santos et al. — SciELO; Schlaich, Schäfer, Jennewein (1987) — PCI Journal.

7.2. Strut resistance (EN 1992-1-1 §6.5.2):

Without shear reinforcement (concrete strut): σ_0Rd,max = K₀ × ν' × f_cd = 0.6 × 0.880 × 20.00 = 10.56 MPa

Without shear reinforcement the strut resistance is governed by the cracked compression zone reduction: K₀ = 0.6 (EN 1992-1-1 §6.5.2, Eq. 6.56) — for a strut with transverse tension (bottle-neck effect without confinement).

With shear reinforcement (CCT node stress): σ_2Rd,max = K₂ × ν' × f_cd = 0.85 × 0.880 × 20.00 = 14.96 MPa

Example — Strut N0→P1:

a₀ = 285 mm (optimised compression zone depth)

z = d − a₀/2 = 820 − 142 = 678 mm

θ = arctan(z/L) = 32.6°

CCC end (column part 250×250 mm):

X-Z plane: L_x = 750 mm, θ_x = arctan(z/L_x) = 42.1°

w_x = a₀×sin(θ_x) + c_x×cos(θ_x) = 285×0.670 + 250×0.742 = 377 mm

Y-Z plane: L_y = 750 mm, θ_y = arctan(z/L_y) = 42.1°

w_y = a₀×sin(θ_y) + c_y×cos(θ_y) = 285×0.670 + 250×0.742 = 377 mm

A_CCC = w_x × w_y = 377 × 377 = 141797 mm²

CCT end (pile D=400 mm):

w_bot = D×sin(θ) + u×cos(θ) = 400×0.000 + 160×0.000 = 350 mm

b_bot = D = 400 mm

A_CCT = w × b = 350 × 400 = 140065 mm²

Critical end: bottom, A_strut = min(A_CCC; A_CCT) = 140065 mm²

F_strut = 1115 kN = 1114603 N

σ_strut = F / A = 1114603 / 140065 = 7.96 MPa

σ_strut / σ_0Rd,max = 7.96 / 10.56 = 0.75 ✓ OK

Strut	θ (°)	σ_strut (MPa)	σ_0Rd,max (MPa)	Utilization	Status
Strut N0→P1	32.6	7.96	10.56	0.75	✓ OK
Strut N0→P2	32.6	7.96	10.56	0.75	✓ OK
Strut N0→P3	32.6	7.96	10.56	0.75	✓ OK
Strut N0→P4	32.6	7.96	10.56	0.75	✓ OK

✓ The concrete provides the strut resistance without shear reinforcement.

9. MINIMUM BAR SPACING CHECK (EC2 8.2)

Minimum clear spacing (EC2 8.2(2)):

s_min = max(φ; d_g + 5; 20) = max(20; 32 + 5; 20) = 37 mm

Tie	s_clear (mm)	s_min (mm)	Status
Tie X (y=-750)	90	37	✓ OK
Tie X (y=750)	90	37	✓ OK
Tie Y (x=-750)	90	37	✓ OK
Tie Y (x=750)	90	37	✓ OK

10. PUNCHING CHECK (EC2 6.4)

Punching resistance (without shear reinforcement):

k = min(1 + √(200/d); 2.0) = min(1 + √(200/820); 2.0) = 1.494

v_Rd,c = C_Rd,c × k × (100 × ρ_l × f_ck)^1/3 = 0.366 MPa

Example — Column:

V_Ed = 2400.0 kN

u₁ = 12304 mm

v_Ed = V_Ed × 10³ / (u₁ × d) = 2400000 / (12304 × 820) = 0.238 MPa

v_Ed / v_Rd,c = 0.238 / 0.366 = 0.65 ✓ OK

Location	v_Ed (MPa)	v_Rd,c (MPa)	Utilization	Status
Column	0.238	0.366	0.65	✓ OK
Pile P1	0.063	0.366	0.17	✓ OK
Pile P2	0.063	0.366	0.17	✓ OK
Pile P3	0.063	0.366	0.17	✓ OK
Pile P4	0.063	0.366	0.17	✓ OK

11. ANCHORAGE CHECK

Bond strength (EC2 8.4.2):

f_bd = 2.25 × η₁ × η₂ × f_ctd = 3.04 MPa

Basic anchorage length:

l_b,rqd = (φ/4) × (f_yd / f_bd) = (20/4) × (434.8 / 3.04) = 715 mm

l_bd = max(l_b,rqd; l_b,min) = 715 mm

l_bd,hook = 0.7 × l_bd = 0.7 × 715 = 500 mm

Available length:

l_b,avail = e_edge − c_nom,side = 600 − 50 = 750 mm

c_nom,side = c_nom,top = 50 mm

l_b,avail = 750 ≥ l_bd = 715 ✓ Straight anchorage sufficient

12. GRAPHICS

Plan view:

STM scheme (side view):

SUMMARY

✓ All verification checks satisfied.

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