Dyma enghraifft waith ar gyfer dylunio WikiHouse Skylark llawr mewn perthynas â llwythi disgyrchiant yn ôl yr Eurocodes.

Sut mae Skylark yn delio â disgyrchiant?

O safbwynt llwyth disgyrchiant, gellir meddwl am siasi Skylark fel dilyniant o byrth sy'n gweithio'n annibynnol ar ei gilydd. Mae pob porth wedi'i wneud o drawstiau a cholofnau, sydd wedi'u cysylltu drwy ddefnyddio clymau bwa.

Enghraifft wedi'i gweithio

Dylunio trawst llawr Skylark 250 rhychwantu dros 5.7 m yn ôl yr Eurocodes. Mae'r adeilad at ddibenion preswyl. Mae'r trawst yn cael ei gefnogi gan golofnau 2.7 m. Mae ffeil taenlen hefyd ar gael yma.

Model strwythurol

O ystyried y rhychwant, dewisir trawst L. Mae'r trawst L yn rhychwantu ar draws 5738 mm (canol-golofn i ganol colofn), a gellir ei fodelu fel elfen trawst Euler-Bernoulli gyda ffynhonnau cylchdroadol mewn gohebiaeth o'r cymalau castellog (gweler Ffigur isod). Gellir ystyried y trawst yn syml.

Ar gyfer trawst L, mae'r gwerthoedd canlynol yn cael eu cymryd yn ôl y canllawiau peirianneg:

• \(W_{L/2}=1135296\,mm^3\)
• \(W_{j}=1135296\,mm^3\)
• \(I=28765\,cm^4\)
• \(k_r=2295\,kNm\)

Tybir bod trawstiau wedi'u gwneud o WISA plywood, sy'n eiddo i'w hadrodd yn y Ffigwr isod.

Diffiniad llwyth

The permanent load \(G_{beam}\) acting on the beam can be calculated according to the equation below:

\(G_{beam}= \overbrace{SW_{beam}}^{0.16 kN/m}+\overbrace{G_{imposed}}^{0.5 kN/m^2} \cdot \overbrace{i_{beams}}^{0.6m}=0.46 kN/m\)

ble mae \(SW_{trawst}\) yn hunan-bwysau'r trawstiau, \(G_{gosodedig}\) yw'r super imposed load a \(i_{beams}\) yw'r rhyng-echelin rhwng y trawstiau. Ystyrir llwyth mympwyol a osodir yn hafal i \(0.5\, kN/m^2\) er mwyn yr enghraifft waith hon.

According to Eurocode 1, the imposed load \(IL)\) for residential category is recommended to be taken equal to 2 kN/m\(^2\). Hence, the linear live load on the beam \(Q_{beam}\) is equal to:

\(Q_{beam}= \overbrace{IL}^{2 kN/m^2} \cdot \overbrace{i_{beams}}^{0.6m}=1.2 kN/m\)

Cyflwr terfyn serviceability (SLS)

Yn y cyflwr terfyn serviceability (SLS), gwirir bod: 1) y daflection elastig uchaf o dan y cyfuniad llwyth aml yn is na throthwy o L/250 (gyda L y rhychwant), a 2) ac mae'r daflection tymor hir o dan y cyfuniad llwyth parhaol led-barhaol yn is na throthwy o L/250.

The value of \(q_{fr}\) (frequent) and \(q_{qp}\) are calculated as:

\(q_{fr} =\overbrace{G_{beams}}^{0.46}+\overbrace{\psi_2}^{0.7}\overbrace{Q_{beams}}^{1.2}=1.3\,kN/m\)

\(q_{qp} =\overbrace{G_{beams}}^{0.46}+\overbrace{\psi_2}^{0.3}\overbrace{Q_{beams}}^{1.2}=0.82\,kN/m\)

The maximum vertical displacement \(v_{max}\) of the beam in the midspan can be calculated as:

\(v_{max}=\frac{5qL^4}{384EI}+\frac{qa^2}{2k_r}[L-a]\)

ble mae \(E\) yn modwlws elastig pren, \(I\) yw'r ail foment effeithiol o inertia, \(a\) yw'r pellter rhwng y cymal castellog a'r gefnogaeth, \(L\) yw'r rhychwant a \(k_r\) yw anystwythder cylchdroadol y cymal.

Adroddir y gwiriad ar y daflection elastig uchaf isod:

\(d_{el}=\frac{5\overbrace{q_{fr}}^{1.3}\overbrace{L}^{5738} \,^4}{384\underbrace{E}_{8170}\underbrace{I}_{28765\cdot10^{4}}}+\frac{\overbrace{q_{fr}}^{1.6}\overbrace{a}^{1700}\,^2}{2\underbrace{k_r}_{2295\cdot10^{6}}}[L-a]=11.1\,mm \le\frac{L}{250} \,\checkmark\)

A check on the long term deflection \(d_{lt}\) is reported below. Note that it is assumed that the beams are working in Service Class I, defined by Eurocode 5 as: characterized by a moisture content in the material corresponding to a temperature of 20 C and the relative humidity of the surrounding air only exceeding 65% for a few weeks per year.

\(d_{lt}=[\frac{5\overbrace{q_{qp}}^{1.12}\overbrace{L}^{5738} \,^4}{384\underbrace{E}_{8170}\underbrace{I}_{28765\cdot10^{4}}}+\frac{\overbrace{q_{qp}}^{1.12}\overbrace{a}^{1700}\,^2}{2\underbrace{k_r}_{2295\cdot10^{6}}}(L-a)](1+\overbrace{k_{def}}^{0.8})=12.6\,mm \le\frac{L}{250} \,\checkmark\)

Cyflwr terfyn eithaf (ULS)

Yn y cyflwr terfyn eithaf (ULS), mae'n cael ei wirio bod capasiti'r trawstiau yn fwy na'r galw. Yn benodol, cynhelir y gwiriad hwn ar gyfer y llwythi parhaol (quasi static combination) ac ar gyfer y llwythi tymor byr (cyfuniad prin). Adroddir isod y cyfuniad llwyth prin:

\(q_{ra} =\overbrace{\gamma_G}^{1.35}\overbrace{G_{beams}}^{0.46}+\overbrace{\gamma_Q}^{1.5}\overbrace{Q_{beams}}^{1.2}=2.42\,kN/m\)

Gwirio canol-rhychwant

The value of the maximum bending moment for permanent loads \(M_{Sd,qp,L/2}\) and value of the maximum bending moment for short term loads \(M_{Sd,ra,L/2}\) are reported below:

\(M_{Sd,qp,L/2}=\frac{\overbrace{q_{qp}}^{0.82}\overbrace{L}^{5738} \,^2}{8}=3.4\,kNm\)

\(M_{Sd,ra,L/2}=\frac{\overbrace{q_{ra}}^{2.42}\overbrace{L}^{5738} \,^2}{8}=10.0\,kNm\)

The moment capacity \(M_{Rd,L/2}\) in the mid-span of the beams is calculated as:

\(M_{Rd,L/2}=\overbrace{W_{L/2}}^{1135296\cdot 10^{-6}}\cdot \overbrace{f_{c,0}}^{16.7}=18.9\,kNm\)

Adroddir y gwiriadau yn yr hafaliadau isod. Sylwch fod y foment plygu gwrthsefyll ar lwythi parhaol yn cael ei nodi gyda \(M_{Rd,p,L/2}\), a nodir y foment plygu gwrthsefyll ar lwythi ar unwaith gyda \(M_{Rd,st,L/2}\).

\(M_{Rd,p,L/2}=\frac{\overbrace{k_{mod}}^{0.6}\overbrace{M_{Rd,L/2}}^{18.9} }{\underbrace{\gamma_M}_{1.25}}=9.0\,kNm\ge M_{Sd,qp,L/2} \, \checkmark\)

\(M_{Rd,st,L/2}=\frac{\overbrace{k_{mod}}^{1.1}\overbrace{M_{Rd,L/2}}^{18.9} }{\underbrace{\gamma_M}_{1.25}}=16.6\,kNm\ge M_{Sd,ra,L/2} \, \checkmark\)

Gwiriad ar y cyd Dovetail

The value of the maximum bending moment for permanent loads \(M_{Sd,qp}\) and value of the maximum bending moment for short term loads \(M_{Sd,ra}\) at the joint are reported below:

\(M_{Sd,qp,j}=\frac{4\overbrace{a}^{1700}(\overbrace{L-a}^{5738-1700})}{\underbrace{L^2}_{5738^2}}M_{Sd,qp,L2}=0.83M_{Sd,qp,L/2}=2.35\,kNm\)

\(M_{Sd,ra,j}=\frac{4\overbrace{a}^{1700}(\overbrace{L-a}^{5738-1700})}{\underbrace{L^2}_{5738^2}}M_{Sd,ra,L2}=0.83M_{Sd,ra,L/2}=8.3\,kNm\)

‍

The moment capacity \(M_{Rd,j}\) at the joint is calculated as:

\(M_{Rd,j}=\overbrace{W_{j}}^{1135296\cdot 10^{-6}}\cdot \overbrace{f_{c,0}}^{16.7}=18.9\,kNm\)

Adroddir y gwiriadau yn yr hafaliadau isod. Sylwch fod y foment blygu gwrthsefyll ar lwythi parhaol yn cael ei nodi gyda \(M_{Rd,p,j}\), a nodir y foment plygu gwrthsefyll ar lwythi ar unwaith gyda \(M_{Rd,st,j}\).

\(M_{Rd,p,j}=\frac{\overbrace{k_{mod}}^{0.6}\overbrace{M_{Rd,j}}^{18.9} }{\underbrace{\gamma_M}_{1.25}}=9.0\,kNm\ge M_{Sd,qp,j} \, \checkmark\)

\(M_{Rd,st}=\frac{\overbrace{k_{mod}}^{1.1}\overbrace{M_{Rd,j}}^{18.9} }{\underbrace{\gamma_M}_{1.25}}=16.6\,kNm\ge M_{Sd,ra,j} \, \checkmark\)

Gwirio colofn

The axial demand for permanent loads \(N_{Sd,qp}\) and the axial demand for short term loads \(N_{Sd,ra}\) is equal to:

\(N_{Sd,qp}=\frac{\overbrace{q_{qp}}^{0.82}\overbrace{L}^{5738}}{2}=2.4\,kN\)

\(N_{Sd,ra}=\frac{\overbrace{q_{ra}}^{2.42}\overbrace{L}^{5738}}{2}=6.9\,kN\)

The value of \(N_{rd}\) for a 2.7 m column is equal to 101.6 kN as per engineering guidelines. Therefore:

\(N_{Rd,p}=\frac{\overbrace{k_{mod}}^{0.6}\overbrace{N_{Rd}}^{101.6} }{\underbrace{\gamma_M}_{1.25}}=48.8\,kN\ge N_{Sd,qp} \, \checkmark\)

\(N_{Rd,st}=\frac{\overbrace{k_{mod}}^{1.1}\overbrace{N_{Rd}}^{101.6} }{\underbrace{\gamma_M}_{1.25}}=89.4\,kN\ge N_{Sd,ra} \, \checkmark\)