// Numbas version: finer_feedback_settings {"name": "Momentkapacitet balk - h\u00e5llfasthetsdom\u00e4nen", "extensions": [], "custom_part_types": [{"source": {"pk": 1, "author": {"name": "Christian Lawson-Perfect", "pk": 7}, "edit_page": "/part_type/1/edit"}, "name": "Yes/no", "short_name": "yes-no", "description": "
The student is shown two radio choices: \"Yes\" and \"No\". One of them is correct.
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", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "Klarar träbalken krav motsvarande R {rKrav}?
\nAntag att balken är brandutsatt på 3 sidor (under och på sidorna) samt att ingen vippning kan ske, dvs $k_{crit} = 1$. Använd metoden reducerat tvärsnitt.
", "advice": "Momentpåkänningen för en fritt upplagd balk beräknas som
\n$M_{Ed,fi}=\\frac{qL^2}{8}$
\nDär $q$ är den utbredda lasten vilken beräknas utifrån den dimensionerande lasten i brandfallet.
\n$E_{d,fi}=g_k+\\psi_1 q_k$
\nFör {occupancy} är $q_k$ = {qk} kN/m² och ur tabell kan vi läsa att $\\psi_1$ = {psi_1}. Då egentyngden, $g_k$, ges i uppgiften till {gk} kN/m² beräknas $E_{d,fi}$ till
\n$E_{d,fi}=\\var{precround(gk,1)}+\\var{psi_1}\\cdot \\var{qk}=\\var{precround(Ed_fi,1)}$ $kN/m²$
\nEtt centrumavstånd motsvarande {cc} meter gör att $E_{d,fi}$ kan antas som
\n$\\var{precround(Ed_fi,1)}\\cdot \\var{precround(cc,1)} = \\var{precround(Ed_fi_tot,1)}$ $kN/m$
\nDen första ekvationen kan därför justeras till
\n$M_{Ed,fi}=\\frac{E_{d,fi}L^2}{8}=\\frac{\\var{precround(Ed_fi_tot,1)}\\cdot\\var{length}^2}{8}=\\var{precround(MEd_fi,1)}$ $kNm$
\nI första steget beräknas inbränningshastigheten genom att beräkna den minsta bredd, $b_{min}$, som krävs för att rundningen av hörnen inte ska tas med
\n$b_{min}=\\huge\\{ \\normalsize{2d_{char,0}+80 \\atop 8.15d_{char,0}}\\text{ }{d_{char,0}\\geq13\\text{ }mm \\atop d_{char,0}<13\\text{ }mm}\\Rightarrow b_{min}=2 t_R \\beta _0+80=\\var{bmin}$ $mm$
\n$b_{min}=\\huge\\{ \\normalsize{2d_{char,0}+80 \\atop 8.15d_{char,0}}\\text{ }{d_{char,0}\\geq13\\text{ }mm \\atop d_{char,0}<13\\text{ }mm}\\Rightarrow b_{min}=8.15d_{char,0}=\\var{bmin}$ $mm$
\nmed $t_R = \\var{rkrav}\\text{ }min$ och $\\beta_0 = \\var{beta0}\\text{ }mm/min$ blir $b_{min} = \\var{bMin}\\text{ }mm$ vilket är $\\var{betaTrue}$ än $b=\\var{b}\\text{ }mm$ varför $\\beta_{design}=\\underline{\\underline{\\var{betaDesign}\\text{ }mm/min}}$.
\nMed tresidig inbränning beräknas $b_{eff}$ som:
\n$b_{eff}=b-2(\\beta_{design}t_R+k_0d_0)=\\var{b}-2\\left(\\var{betaDesign}\\cdot\\var{rKrav}+\\var{kZero}\\cdot\\var{dZero}\\right)=\\var{precround(bEff,1)}$ $mm$
\noch $h_{eff}$ som
\n$h_{eff}=h-(\\beta_{design}t_R+k_0d_0)=\\var{h}-\\left(\\var{betaDesign}\\cdot\\var{rKrav}+\\var{kZero}\\cdot\\var{dZero}\\right)=\\var{precround(hEff,1)}$ $mm$
\nUtifrån det beräknas böjmotståndet som:
\n$W_{eff}=\\frac{b_{eff}h_{eff}^2}{6}=\\frac{\\var{bEff}\\cdot\\var{hEff}^2}{6}=\\underline{\\underline{\\var{precround(wEff,1)}\\text{ }mm^3}}$
\nI brandfallet beräknas dimensionerande materialvärden med:
\n$f_{md,fi}=k_{mod,fi}\\frac{k_{fi}f_{mk}}{\\gamma_{M,fi}}=1\\frac{\\var{kFi}\\cdot\\var{fm}}{1}=\\underline{\\underline{\\var{precround(fmFi,1)}\\text{ }MPa}}$
\nMed $k_{crit}$ = 1, beräknas $M_{Rd,fi}$ som:
\n$M_{Rd,fi}=k_{crit}f_{md,fi}W_{eff}=1\\cdot\\var{precround(fmFi,1)}\\cdot\\var{precround(wEff,1)}=\\underline{\\underline{\\var{precround(mToT,1)}\\text{ }kNm}}$
\nKontrollera om hållfastheten, $M_{Rd,fi}$, är större än belastningen, $M_{Ed,fi}$, dvs.
\n$M_{Ed,fi}\\leq M_{Rd,fi}$
\n$\\var{precround(MEd_Fi,1)}\\leq\\var{precround(Mtot,1)}$
\nOm ovanstående är sant så kan konstruktionen anses klara krav motsvarande R {rKrav}.
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