Base plaatontwerp voorbeeld met behulp van EN 1993-1-8:2005, IN 1993-1-1:2005, IN 1992-1-1:2004, and EN 1992-4:2018.
Probleemverklaring:
Determine whether the designed column-to-base plate connection is sufficient for a Vy=5-kN en Vz=5-kN schuifbelastingen.
Gegeven gegevens:
Kolom:
Kolomgedeelte: SHS 180x180x8
Kolomgebied: 5440 mm2
Kolommateriaal: S235
Bodemplaat:
Baseplaat afmetingen: 350 mm x 350 mm
Basisplaatdikte: 12 mm
Basisplaatmateriaal: S235
Vocht:
Vochtdikte: 6 mm
Grout material: 30 MPa
Beton:
Concrete dimensies: 350 mm x 350 mm
Betonnen dikte: 350 mm
Betonnen materiaal: C25/30
Gebarsten of ongescheurd: Gebarsten
Ankers:
Ankerdiameter: 12 mm
Effectieve inbeddingslengte: 150 mm
Ingebouwde plaatdiameter: 60 mm
Ingebedde plaatdikte: 10 mm
Ankermateriaal: 8.8
Andere informatie:
- Niet-countersunk ankers.
- Anker met gesneden draden.
- K7 factor for anchor steel shear failure: 1.0
- Degree of Restraint of Fastener: No restraint
Lassen:
Lastype: Fillet Weld
Weld leg size: 8mm
Vulmetaalclassificatie: E35
Ankergegevens (van Skyciv Calculator):
Definities:
Load Path:
De SkyCiv-software voor het ontwerpen van grondplaten follows IN 1992-4:2018 for anchor rod design. Shear loads applied to the column are transferred to the base plate through the welds and then to the supporting concrete through the anchor rods. Friction and shear lugs are not considered in this example, as these mechanisms are not supported in the current software.
Ankergroepen:
The software includes an intuitive feature that identifies which anchors are part of an anchor group for evaluating concrete shear breakout en concrete shear pryout mislukkingen.
Een ankergroep is defined as two or more anchors with overlapping projected resistance areas. In dit geval, the anchors act together, and their combined resistance is checked against the applied load on the group.
A single anchor is defined as an anchor whose projected resistance area does not overlap with any other. In dit geval, the anchor acts alone, and the applied shear force on that anchor is checked directly against its individual resistance.
This distinction allows the software to capture both group behavior and individual anchor performance when assessing shear-related failure modes.
Stapsgewijze berekeningen:
Controleren #1: Lascapaciteit berekenen
We assume that the Vz shear load is resisted by the top and bottom welds, Terwijl de Vy shear load is resisted exclusively by the left and right welds.
To determine the weld capacity of the top and bottom welds, we first calculate their total weld lengths.
\(
L_{w,top\&bodem} = 2 \links(b_{col} – 2t_{col} – 2R_{col}\Rechtsaf)
= 2 \keer links(180 \,\tekst{mm} – 2 \keer 8 \,\tekst{mm} – 2 \keer 4 \,\tekst{mm}\Rechtsaf)
= 312 \,\tekst{mm}
\)
De volgende, we berekenen de stresses in the welds.
Note that the applied Vz shear acts parallel to the weld axis, with no other forces present. This means the perpendicular stresses can be taken as zero, and only the shear stress in the parallel direction needs to be calculated.
\(
\sigma_{\dader} = frac{N}{(L_{w,top\&bodem})\,a\sqrt{2}}
= frac{0 \,\tekst{kN}}{(312 \,\tekst{mm}) \keer 5.657 \,\tekst{mm} \keer sqrt{2}}
= 0
\)
\(
\jouw_{\dader} = frac{0}{(L_{w,top\&bodem})\,a\sqrt{2}}
= frac{0 \,\tekst{kN}}{(312 \,\tekst{mm}) \keer 5.657 \,\tekst{mm} \keer sqrt{2}}
= 0
\)
\(
\jouw_{\parallel} = frac{V_{z}}{(L_{w,top\&bodem})\,een}
= frac{5 \,\tekst{kN}}{(312 \,\tekst{mm}) \keer 5.657 \,\tekst{mm}}
= 2.8329 \,\tekst{MPa}
\)
Gebruik makend van IN 1993-1-8:2005, Eq. 4.1, the design weld stress is obtained using the directional method.
\(
F_{w,Ed1} = sqrt{ (\sigma_{\dader})^ 2 + 3 \links( (\jouw_{\dader})^ 2 + (\jouw_{\parallel})^2 Juist) }
= sqrt{ (0)^ 2 + 3 \keer links( (0)^ 2 + (2.8329 \,\tekst{MPa})^2 Juist) }
= 4.9067 \,\tekst{MPa}
\)
Daarnaast, the design normal stress for the base metal check, per IN 1993-1-8:2005, Eq. 4.1, is taken as zero, sinds no normal stress is present.
\(
F_{w,Ed2} = \sigma_{\dader} = 0
\)
Nu, let us assess the left and right welds. As with the top and bottom welds, we first calculate the Totale laslengte.
\(
L_{w,left\&Rechtsaf} = 2 \links(d_{col} – 2t_{col} – 2R_{col}\Rechtsaf)
= 2 \keer links(180 \,\tekst{mm} – 2 \keer 8 \,\tekst{mm} – 2 \keer 4 \,\tekst{mm}\Rechtsaf)
= 312 \,\tekst{mm}
\)
We then calculate the components of the weld stresses.
\(
\sigma_{\dader} = frac{N}{(L_{w,left\&Rechtsaf})\,a\sqrt{2}}
= frac{0 \,\tekst{kN}}{(312 \,\tekst{mm}) \keer 5.657 \,\tekst{mm} \keer sqrt{2}}
= 0
\)
\(
\jouw_{\dader} = frac{0}{(L_{w,left\&Rechtsaf})\,a\sqrt{2}}
= frac{0 \,\tekst{kN}}{(312 \,\tekst{mm}) \keer 5.657 \,\tekst{mm} \keer sqrt{2}}
= 0
\)
\(
\jouw_{\parallel} = frac{V_y}{(L_{w,left\&Rechtsaf})\,een}
= frac{5 \,\tekst{kN}}{(312 \,\tekst{mm}) \keer 5.657 \,\tekst{mm}}
= 2.8329 \,\tekst{MPa}
\)
Gebruik makend van IN 1993-1-8:2005, Eq. 4.1, we determine both the design weld stress and the design normal stress for the base metal check.
\(
F_{w,Ed1} = sqrt{ \links( \sigma_{\dader} \Rechtsaf)^ 2 + 3 \links( \links( \jouw_{\dader} \Rechtsaf)^ 2 + \links( \jouw_{\parallel} \Rechtsaf)^2 Juist) }
\)
\(
F_{w,Ed1} = sqrt{ \links( 0 \Rechtsaf)^ 2 + 3 \keer links( \links( 0 \Rechtsaf)^ 2 + \links( 2.8329 \,\tekst{MPa} \Rechtsaf)^2 Juist) }
\)
\(
F_{w,Ed1} = 4.9067 \,\tekst{MPa}
\)
The next step is to identify the governing weld stress between the top/bottom welds and the left/right welds. Because the weld lengths are equal and the applied loads have the same magnitude, the resulting weld stresses are equal.
\(
F_{w,Ed1} = \max(F_{w,Ed1}, \, F_{w,Ed1})
= \max(4.9067 \,\tekst{MPa}, \, 4.9067 \,\tekst{MPa})
= 4.9067 \,\tekst{MPa}
\)
The base metal stress remains zero.
\(
F_{w,Ed2} = \max(F_{w,Ed2}, \, F_{w,Ed2}) = \max(0, \, 0) = 0
\)
Nu, we calculate the weld capacity. Eerste, the resistance of the hoeklassen is computed. Vervolgens, the resistance of the base metal is determined. Using EN 1993-1-8:2005, Eq. 4.1, the capacities are calculated as follows:
\(
F_{w,Rd1} = frac{f_u}{\beta_w \left(\gamma_{M2,weld}\Rechtsaf)}
= frac{360 \,\tekst{MPa}}{0.8 \keer (1.25)}
= 360 \,\tekst{MPa}
\)
\(
F_{w,Rd2} = frac{0.9 f_u}{\gamma_{M2,weld}}
= frac{0.9 \keer 360 \,\tekst{MPa}}{1.25}
= 259.2 \,\tekst{MPa}
\)
Uiteindelijk, we compare the weld stresses with the weld capacities, and the base metal stresses with the base metal capacities.
Sinds 4.9067 MPa < 360 MPa en 0 MPa < 259.2 MPa, the capacity of the welded connection is voldoende.
Controleren #2: Calculate concrete breakout capacity due to Vy shear
Following the provisions of IN 1992-4:2018, the edge perpendicular to the applied load is assessed for shear breakout failure. Only the anchors nearest to this edge are considered engaged, while the remaining anchors are assumed not to resist shear.
These edge anchors must have a concrete edge distance greater than the larger of 10·hef and 60·d, waar hebben is the embedment length and d is the anchor diameter. If this condition is not met, the thickness of the base plate must be less than 0.25·hef.
If the requirements in IN 1992-4:2018, Clausule 7.2.2.5(1), are not satisfied, the SkyCiv software cannot proceed with the design checks, and the user is advised to refer to other relevant standards.
From the SkyCiv software results, the edge anchors act as enkele ankers, since their projected areas do not overlap. For this calculation, Anchor 1 will be considered.
To calculate the portion of the Vy shear load carried by Anchor 1, the total Vy shear is distributed among the anchors nearest to the edge. This gives the perpendicular force on Anchor 1.
\(
V_{\dader} = frac{V_y}{N_{een,s}}
= frac{5 \,\tekst{kN}}{2}
= 2.5 \,\tekst{kN}
\)
Voor de parallel force, it is assumed that all anchors resist the load equally. Daarom, the parallel component of the load is calculated as:
\(
V_{\parallel} = frac{V_z}{N_{anc}}
= frac{5 \,\tekst{kN}}{4}
= 1.25 \,\tekst{kN}
\)
De total shear load on Anchor 1 is therefore:
\(
V_{Ed} = sqrt{ \links( V_{\dader} \Rechtsaf)^ 2 + \links( V_{\parallel} \Rechtsaf)^ 2 }
\)
\(
V_{Ed} = sqrt{ \links( 2.5 \,\tekst{kN} \Rechtsaf)^ 2 + \links( 1.25 \,\tekst{kN} \Rechtsaf)^ 2 } = 2.7951 \,\tekst{kN}
\)
The first part of the capacity calculation is to determine the alpha and beta factors. We gebruiken IN 1992-4:2018, Clausule 7.2.2.5, to set the lf dimension, en Vergelijkingen 7.42 en 7.43 to determine the factors.
\(
l_f = \min(h_{ef}, \, 12d_{anc})
= \min(150 \,\tekst{mm}, \, 12 \keer 12 \,\tekst{mm})
= 144 \,\tekst{mm}
\)
\(
\alpha = 0.1 \links(\frac{l_f}{c_{1,s1}}\Rechtsaf)^{0.5}
= 0.1 \keer links(\frac{144 \,\tekst{mm}}{50 \,\tekst{mm}}\Rechtsaf)^{0.5}
= 0.16971
\)
\(
\beta = 0.1 \links(\frac{d_{anc}}{c_{1,s1}}\Rechtsaf)^{0.2}
= 0.1 \keer links(\frac{12 \,\tekst{mm}}{50 \,\tekst{mm}}\Rechtsaf)^{0.2}
= 0.07517
\)
The next step is to calculate the initial value of the characteristic resistance of the fastener. Gebruik makend van IN 1992-4:2018, Vergelijking 7.41, the value is:
\(
V^{0}_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,c} = k_9 \left( \frac{d_{anc}}{\tekst{mm}} \Rechtsaf)^{\alfa}
\links( \frac{l_f}{\tekst{mm}} \Rechtsaf)^{\bèta}
\sqrt{ \frac{f_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt}}{\tekst{MPa}} }
\links( \frac{c_{1,s1}}{\tekst{mm}} \Rechtsaf)^{1.5} N
\)
\(
V^{0}_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,c} = 1.7 \keer links( \frac{12 \,\tekst{mm}}{1 \,\tekst{mm}} \Rechtsaf)^{0.16971}
\keer links( \frac{144 \,\tekst{mm}}{1 \,\tekst{mm}} \Rechtsaf)^{0.07517}
\keer sqrt{ \frac{20 \,\tekst{MPa}}{1 \,\tekst{MPa}} }
\keer links( \frac{50 \,\tekst{mm}}{1 \,\tekst{mm}} \Rechtsaf)^{1.5}
\keer 0.001 \,\tekst{kN}
\)
\(
V^{0}_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,c} = 5.954 \,\tekst{kN}
\)
Vervolgens, We berekenen de Referentie geprojecteerd gebied of a single anchor, als vervolg op IN 1992-4:2018, Vergelijking 7.44.
\(
EEN_{c,V }^{0} = 4.5 \links( c_{1,s1} \Rechtsaf)^ 2
= 4.5 \keer links( 50 \,\tekst{mm} \Rechtsaf)^ 2
= 11250 \,\tekst{mm}^ 2
\)
Daarna, We berekenen de Werkelijk geprojecteerd gebied of Anchor 1.
\(
B_{c,V } = \min(c_{links,s1}, \, 1.5c_{1,s1}) + \min(c_{Rechtsaf,s1}, \, 1.5c_{1,s1})
\)
\(
B_{c,V } = \min(300 \,\tekst{mm}, \, 1.5 \keer 50 \,\tekst{mm}) + \min(50 \,\tekst{mm}, \, 1.5 \keer 50 \,\tekst{mm}) = 125 \,\tekst{mm}
\)
\(
H_{c,V } = \min(1.5c_{1,s1}, \, t_{concerentie}) = \min(1.5 \keer 50 \,\tekst{mm}, \, 200 \,\tekst{mm}) = 75 \,\tekst{mm}
\)
\(
EEN_{c,V } = H_{c,V } B_{c,V } = 75 \,\tekst{mm} \keer 125 \,\tekst{mm} = 9375 \,\tekst{mm}^ 2
\)
We also need to calculate the parameters for shear breakout. We gebruiken IN 1992-4:2018, Vergelijking 7.4, to get the factor that accounts for the disturbance of stress distribution, Vergelijking 7.46 for the factor that accounts for the member thickness, en Vergelijking 7.48 for the factor that accounts for the influence of a shear load inclined to the edge. These are calculated as follows:
\(
\Psi_{s,V } = min links( 0.7 + 0.3 \links( \frac{c_{2,s1}}{1.5c_{1,s1}} \Rechtsaf), \, 1.0 \Rechtsaf)
= min links( 0.7 + 0.3 \keer links( \frac{50 \,\tekst{mm}}{1.5 \keer 50 \,\tekst{mm}} \Rechtsaf), \, 1 \Rechtsaf)
= 0.9
\)
\(
\Psi_{h,V } = max links( \links( \frac{1.5c_{1,s1}}{t_{concerentie}} \Rechtsaf)^{0.5}, \, 1 \Rechtsaf)
= max links( \links( \frac{1.5 \keer 50 \,\tekst{mm}}{200 \,\tekst{mm}} \Rechtsaf)^{0.5}, \, 1 \Rechtsaf)
= 1
\)
\(
\alfa_{V } = \tan^{-1} \links( \frac{V_{\parallel}}{V_{\dader}} \Rechtsaf)
= \tan^{-1} \links( \frac{1.25 \,\tekst{kN}}{2.5 \,\tekst{kN}} \Rechtsaf)
= 0.46365 \,\tekst{rad}
\)
\(
\Psi_{\alfa,V } = max links(
\sqrt{ \frac{1}{(\omdat(\alfa_{V }))^ 2 + \links( 0.5 \, (\zonder(\alfa_{V })) \Rechtsaf)^ 2 } }, \, 1 \Rechtsaf)
\)
\(
\Psi_{\alfa,V } = max links(
\sqrt{ \frac{1}{(\omdat(0.46365 \,\tekst{rad}))^ 2 + \links( 0.5 \times \sin(0.46365 \,\tekst{rad}) \Rechtsaf)^ 2 } }, \, 1 \Rechtsaf)
\)
\(
\Psi_{\alfa,V } = 1.0847
\)
One important note when determining the alpha factor is to ensure the perpendicular shear and parallel shear are identified correctly.
Uiteindelijk, We berekenen de breakout resistance of the single anchor using IN 1992-4:2018, Vergelijking 7.1.
\(
V_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,c} = V^0_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,c} \links(\frac{EEN_{c,V }}{A^0_{c,V }}\Rechtsaf)
\Psi_{s,V } \Psi_{h,V } \Psi_{eg,V } \Psi_{\alfa,V } \Psi_{= reductiefactor voor gesneden draad,V }
\)
\(
V_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,c} = 5.954 \,\tekst{kN} \keer links(\frac{9375 \,\tekst{mm}^ 2}{11250 \,\tekst{mm}^ 2}\Rechtsaf)
\keer 0.9 \keer 1 \keer 1 \keer 1.0847 \keer 1
= 4.8435 \,\tekst{kN}
\)
Applying the partial factor, the design resistance is 3.23 kN.
\(
V_{Rd,c} = frac{V_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,c}}{\gamma_{Mc}}
= frac{4.8435 \,\tekst{kN}}{1.5}
= 3.229 \,\tekst{kN}
\)
Sinds 2.7951 kN < 3.229 kN, the shear breakout capacity for Vy shear is voldoende.
Controleren #3: Calculate concrete breakout capacity due to Vz shear
The same approach is used to determine the capacity on the edge perpendicular to the Vz shear.
Because of the symmetric design, the anchors resisting Vz shear are also identified as enkele ankers. Let’s consider Anchor 1 again for the calculations.
Om de te berekenen perpendicular load on Anchor 1, we divide the Vz shear by the total number of anchors nearest to the edge only. Om de te berekenen parallel load on Anchor 1, we divide the Vy shear by the total number of anchors.
\(
V_{\dader} = frac{V_{z}}{N_{een,s}}
= frac{5 \,\tekst{kN}}{2}
= 2.5 \,\tekst{kN}
\)
\(
V_{\parallel} = frac{V_{j}}{N_{anc}}
= frac{5 \,\tekst{kN}}{4}
= 1.25 \,\tekst{kN}
\)
\(
V_{Ed} = sqrt{ \links( V_{\dader} \Rechtsaf)^ 2 + \links( V_{\parallel} \Rechtsaf)^ 2 }
\)
\(
V_{Ed} = sqrt{ \links( 2.5 \,\tekst{kN} \Rechtsaf)^ 2 + \links( 1.25 \,\tekst{kN} \Rechtsaf)^ 2 }
= 2.7951 \,\tekst{kN}
\)
Using a similar approach to Check #2, the resulting breakout resistance for the edge perpendicular to Vz shear is:
\(
V_{Rd,c} = frac{V_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,c}}{\gamma_{Mc}}
= frac{4.8435 \,\tekst{kN}}{1.5}
= 3.229 \,\tekst{kN}
\)
Sinds 2.7951 kN < 3.229 kN, the shear breakout capacity for Vz shear is voldoende.
Controleren #4: Calculate concrete pryout capacity
The calculation for the shear pryout resistance involves determining the nominal capacity of the anchors against tension breakout. The reference for tension breakout capacity is IN 1992-4:2018, Clausule 7.2.1.4. A detailed discussion of tension breakout is already covered in the SkyCiv Design Example with Tension Load and will not be repeated in this design example.
From the SkyCiv software calculations, the nominal capacity of the section for tension breakout is 44.61 kN.
We gebruiken dan IN 1992-4:2018, Equation 7.39a, to obtain the design characteristic resistance. Gebruik makend van k8 = 2, the capacity is 59.48 kN.
\(
V_{Rd,cp} = frac{k_8 N_{cbg}}{\gamma_C}
= frac{2 \keer 44.608 \,\tekst{kN}}{1.5}
= 59.478 \,\tekst{kN}
\)
In the shear pryout check, all anchors are effective in resisting the full shear load. From the image generated by the SkyCiv software, all failure cone projections overlap with each other, making the anchors act as an ankergroep.
Daarom, the required resistance of the anchor group is the total resultant shear load of 7.07 kN.
\(
V_{res} = sqrt{(V_y)^ 2 + (V_z)^ 2}
= sqrt{(5 \,\tekst{kN})^ 2 + (5 \,\tekst{kN})^ 2}
= 7.0711 \,\tekst{kN}
\)
\(
V_{Ed} = links(\frac{V_{res}}{N_{anc}}\Rechtsaf) N_{een,G1}
= links(\frac{7.0711 \,\tekst{kN}}{4}\Rechtsaf) \keer 4
= 7.0711 \,\tekst{kN}
\)
Sinds 7.0711 kN < 59.478 kN, the shear pryout capacity is voldoende.
Controleren #5: Calculate anchor rod shear capacity
The calculation of the anchor rod shear capacity depends on whether the shear load is applied with a moment arm. To determine this, we refer to IN 1992-4:2018, Clausule 6.2.2.3, where the thickness and material of the grout, the number of fasteners in the design, the spacing of the fasteners, and other factors are checked.
De Skyciv Base Plate Design Software performs all the necessary checks to determine whether the shear load acts with or without a lever arm. For this design example, it is determined that the shear load is niet applied with a lever arm. Daarom, we gebruiken IN 1992-4:2018, Clausule 7.2.2.3.1, for the capacity equations.
We begin by calculating the characteristic resistance of the steel fastener using IN 1992-4:2018, Vergelijking 7.34.
\(
V^0_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,s} = k_6 A_s f_{u,anc}
= 0.5 \keer 113.1 \,\tekst{mm}^2 \times 800 \,\tekst{MPa}
= 45.239 \,\tekst{kN}
\)
De volgende, we apply the factor for the ductility of the single anchor or the anchor group, taking k7 = 1.
\(
V_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,s} = k_7 V^{0}_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,s}
= 1 \keer 45.239 \,\tekst{kN}
= 45.239 \,\tekst{kN}
\)
We then obtain the partial factor for steel shear failure gebruik makend van IN 1992-4:2018, Tafel 4.1. For an anchor with 8.8 materiaal, the resulting partial factor is:
\(
\gamma_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,schuintrekken}
= max links( 1.0 \links( \frac{F_{u,anc}}{F_{j,anc}} \Rechtsaf), \, 1.25 \Rechtsaf)
= max links( 1 \[object Window]{800 \,\tekst{MPa}}{640 \,\tekst{MPa}}, \, 1.25 \Rechtsaf)
= 1.25
\)
Applying this factor to the characteristic resistance, the design resistance is 36.19 kN.
\(
V_{Rd,s} = frac{V_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,s}}{\gamma_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt,schuintrekken}}
= frac{45.239 \,\tekst{kN}}{1.25}
= 36.191 \,\tekst{kN}
\)
De required shear resistance per anchor rod is the resultant shear load divided by the total number of anchor rods, which calculates to 1.77 kN.
\(
V_{Ed} = frac{\sqrt{ (V_y)^ 2 + (V_z)^ 2 }}{N_{anc}}
\)
\(
V_{Ed} = frac{\sqrt{ (5 \,\tekst{kN})^ 2 + (5 \,\tekst{kN})^ 2 }}{4}
= 1.7678 \,\tekst{kN}
\)
Sinds 1.7678 kN < 36.191 kN, the anchor rod steel shear capacity is voldoende.
Ontwerp Samenvatting
De Skyciv Base Plate Design Software Kan automatisch een stapsgewijze berekeningsrapport genereren voor dit ontwerpvoorbeeld. Het biedt ook een samenvatting van de uitgevoerde controles en hun resulterende verhoudingen, De informatie in één oogopslag gemakkelijk te begrijpen maken. Hieronder is een sample samenvattende tabel, die is opgenomen in het rapport.
Skyciv Sample Report
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