T029 (c) ibhxws webservices
Tragwerke, Schnittgroessen, Vorbemessung
Zweigelenk-Rahmen nach KLEINLOGEL
Rahmenform 29
Prinzipieller Ablauf der Nutzung des Dienstes:
1. Parameter eingeben bzw. waehlen
2. Dienst starten mit 'go' oder
2. Beispiel in Liste anklicken
3. Parameter aendern (weitere Nachweise)
4. Dienst starten mit 'go'
5. Ergebnistext markieren und kopieren
6. Ergebnistext in Ihre Dateien einfuegen
7. Mit ONLINE-PDF drucken oder lokal speichern
Structures, internal forces, predimensioning
Two-hinged frame according to KLEINLOGEL
Frame shape 29
Using Webservice
1. Input or choose values
2. Start webservice with 'go' or
2. Start example in black list
3. Change values (new calculation, new action)
4. Start webservice once more with 'go'
5. Copy result sets of service to your local editors or programs or
6. Use ONLINE-PDF for printing or local saving
Hintergrundinformationen zum Webdienstes:
Background informations webservice:
Rahmenform 29 Rechteckrahmen
Rahmenform 29 - frame shape 29
Festwerte - fixed values
k = (J2/J1)*(h/l)
N = 2*k + 3
Fall 29/1 - case 29/1
Gleichmäßige Wärmezunahme im ganzen Rahmen
MB = MC = -3*E*J2*alphat*t/(h*N)
My = (y/h)*MB
HA = HD = -MB/h
Fall 29/2 - case 29/2
Rechteck-Vollast auf dem Riegel (Symmetrischer Lastfall)
MB = MC = - q*l^2/(4*N)
maxM = q*l^2/8 + MB
Mx = q*x*x'/2 + MB
My = (y/h)*MB
HA = HD = - MB/h
VA = VD = q*l/2
Fall 29/3 - case 29/3
Rechteck-Streckenlast von links her auf dem Riegel
beta = b/l
VD = q*a^2/(2*l)
E*J1*uB = q*a^2*h^2*beta^2/(24*k)
MB = MC = - q*a^2*(1+2*beta)/(4*N)
HA = HD = - MB/h
VA = q*a-VD
Mx = (VA-q*x/2)*x + MB Bereich a
Mx = VD*x' + MC Bereich b
My = (y/h)*MB
Fall 29/4 - case 29/4
Symmetrische Rechteck-Streckenlasten vom Rand her
beta = b/l
MB = MC = - q*a^2*(2+beta)/(4*N)
maxM = (q*a^2)/2 + MB
VA = VD = q*a
HA = HD = - MB/h
Mx = q*x*(a-x/2) + MB
My = (y/h)*MB
Fall 29/5 - case 29/5
Symmetrische Rechteck-Streckenlast in mittiger Lage
alpha = a/l
VA = VD = q*a/2
MB = MC = - q*a*l*(3-alpha^2)/(8*N)
maxM = q*a*(l+2*b)/8 + MB
My = (y/h)*MB
Mx = VA*x + MB Bereich b
Mx = VA*x - q*(x-b)^2 + MB Bereich a
HA = HD = - MB/h
Fall 29/6 - case 29/6
Einzellast an beliebiger Stelle des Riegels
MB = MC = - 3*P*a*b/(2*l*N)
MP = P*a*b/l + MB
VA = P*b/l
VD = P*a/l
HA = HD = - MB/h
Mx = VA*x + MB Bereich a
Mx = VD*x' + MC Bereich b
My = (y/h)*MB
E*J1*uB = P*a*b*h^2*(b-a)/(12*l^2*k)
Fall 29/7 - case 29/7
Zwei gleiche Einzellasten symmetrisch auf dem Riegel
alpha = a/l
MB = MC = - 3*P*a*(1-alpha)/(N)
VA = VD = P
HA = HD = - MB/h
MP = P*a + MB
Mx = P*x + MB
My = (y/h)*MB
Fall 29/8 - case 29/8
Einzellast in der Mitte des Riegels
MB = MC = - 3*P*l/(8*N)
VA = VD = P/2
HA = HD = - MB/h
MP = P*l/4 + MB
Mx = P*x/2 + MB
My = (y/h)*MB
Fall 29/9 - case 29/9
Drei gleiche Einzellasten in den Viertelpunkten des Riegels
MB = MC = - 15*P*l/(16*N)
VA = VD = 3*P/2
HA = HD = - MB/h
M1 = 3*P*l/8 + MB
M2 = P*l/2 + MB
Mx = VA*x + MB
My = (y/h)*MB
Fall 29/10 - case 29/10
Zwei gleiche Einzellasten in den Drittelpunkten des Riegels
MB = MC = - 2*P*l/(3*N)
VA = VD = P
HA = HD = - MB/h
MP = P*l/3 + MB
Mx = P*x + MB
My = (y/h)*MB
Fall 29/11 - case 29/11
Vier gleiche Einzellasten in den Fünftelpunkten des Riegels
MB = MC = - 6*P*l/(5*N)
VA = VD = 2P
HA = HD = - MB/h
M1 = 2*P*l/5 + MB
M2 = 3*P*l/5 + MB
Mx = 2*P*x + MB
My = (y/h)*MB
Fall 29/12 - case 29/12
Riegel beliebig senkrecht belastet
VA = Sr/l
VD = Sl/l
MB = MC = - (L+R)/(2*N)
HA = HD = - MB/h
Mx = Mx0 + MB
My = (y/h)*MB
VA = VD = F/2 bei R=L (Sonderfall 29/12a symmetrische Last)
MB = MC = - L/N bei R=L (Sonderfall 29/12a symmetrische Last)
E*J1*uB = ((h^2)*(L-R))/(12*k)
==> Belastungsglieder - load terms: L, R, Sl, Sr, F, W, Mx0, My0
Fall 29/13 - case 29/13
Riegel beliebig antimetrisch belastet
VA = - VD = Sr/l
MB = MC = 0
HA = HD = 0
E*J1*uB = (h^2)*L/(6*k)
==> Belastungsglieder - load terms: L, R, Sl, Sr, F, W, Mx0, My0
Fall 29/14 - case 29/14
Waagerechte Einzellast in Riegelhöhe
VA = - VD = P*h/l
HD = - HA = P/2
MB = - MC = + P*h/2
Mx = P*h*(1/2-x/l)
My1 = - My2 = P*y/2
E*J1*uB = P*(h^3)*(N-2)/(12*k)
Fall 29/15 - case 29/15
Beide Stiele gleich belastet, von aussen her
MB = MC = - R*k/N
HA = HD = - (Sr+MB)/h
My = My0 + y*MB/h
==> Belastungsglieder - load terms: L, R, Sl, Sr, F, W, Mx0, My0
Fall 29/16 - case 29/16
Beide Stiele gleich belastet, von links her
MB = - MC = Sl
My = My0 + y*Sl/h
Mx = Sl*(x'-x)/l
VD = -VA = 2*Sl/l
HD = - HA = W
E*J1*uB = ((h^2)/(6*k))*[Sl*(2*k+1) + Rk]
==> Belastungsglieder - load terms: L, R, Sl, Sr, F, W, Mx0, My0
Fall 29/17 - case 29/17
Linker Stiel beliebig waagerecht belastet
MB = - R*k/(2*N) + Sl/2
MC = - R*k/(2*N) - Sl/2
MB - MC=Sl
VD = -VA = Sl/l
HD = -MC/h
HA = - (W-HD)
Mx = MB - x*Sl/l
My1 = My0 + y1*MB/h
My2 = y2*MC/h
E*J1*uB = ((h^2)/(12*k))*[Sl*(2*k+1) + Rk]
==> Belastungsglieder - load terms: L, R, Sl, Sr, F, W, Mx0, My0
Fall 29/18 - case 29/18
Einzellast an beliebiger Stelle des linken Riegels
alpha = a/l
beta = b/l
alpha + beta = 1
MB = (P*a/2)*(-((1+alpha)*beta*k)/N +1)
MC = (P*a/2)*(-((1+alpha)*beta*k)/N -1)
MP = alpha*(P*b + MB)
HD = - MC/h
HA = - (P - HD)
VA = - VD = - P*a/l
My1 = - HA*y1 Bereich a
My1 = P*a - HD*y1 Bereich b
Mx = MC + VD*x'
My2 = - HD*y2
E*J1*uB = (P*h^3*alpha/12)*(beta*(1+alpha) + (N-2)/k)
Fall 29/19 - case 29/19
Rechteck-Vollast an beiden Seiten (symmetrisch)
MB = MC = - (k*q*h^2)/(4*N)
My = (q*y*y')/2 + MB*y/h
HA = HD = - (q*h/2 + MB/h)
Fall 29/20 - case 29/20
Rechteck-Vollast am linken Stiel
MB = (q*h^2)/4*(-k/(2*N) +1)
MC = (q*h^2)/4*(-k/(2*N) -1)
HD = - MC/h
HA = - (q*h - HD)
VD = - VA = (q*h^2)
My1 = q*y1*y1'/2 + MB*y1/h
My2 = - HD*y2
Mx = MC + VD*x'
E*J1*uB = ((q*h^4)*(5*k+2)/(48*k)
Fall 29/21 - case 29/21
Rechteck-Streckenlast, linker Stiel, von unten
alpha = a/h
VA = - VD - (q*a^2)/(2*l)
MB = ((q*a^2)/4)*[(-(2-alpha^2)*k)/(2*N) +1]
MC = ((q*a^2)/4)*[(-(2-alpha^2)*k)/(2*N) -1]
HD = -MC/h
HA = - (q*a - HD)
My1 = (-HA - q*y1/2)*y1 Bereich a
My1 = 0.5*q*a^2 - HD*y1 Bereich b
Mx = MC + VD*x'
My2 = -HD*y2
Fall 29/22 - case 29/22
Rechteck-Streckenlast, beide Stiele, von unten, symm
alpha = a/h
beta = b/h
alpha + beta = 1
MB = MC = (-q*a^2*(2-alpha^2)*k)/(4*N)
HA = HD = -[q*a*(1+beta)/2 + MB/h]
My = [-HA - q*y/2]*y Bereich a
My = ((q*a^2)*y')/(2*h) + y*MB/h Bereich b
Fall 29/23 - case 29/23
Dreiecklast an beiden Stielen, symm
MB = MC = -7*p*h^2*k/(60*N)
HA = HD = -(p*h/3 + MB/h)
My = ((p*h^2)/6)*omegaD' + y*MB/h
omegaD' = y'/h - (y'/h)^3
Fall 29/24 - case 29/24
Dreiecklast am linken Stiel
MB = ((p*h^2)/12)*[-7*k/(10*N) + 1]
MC = ((p*h^2)/12)*[-7*k/(10*N) - 1]
HD = - MC/h
HA = - (p*h/2 - HD)
VA = - VD = -p*h^2/(6*l)
Mx = MC + VD*x'
My2 = -HD*y2
My1 = ((p*h^2)/6)*omegaD' + y1*MB/h
omegaD' = y1'/h - (y1'/h)^3
E*J1*uB = ((p*h^4)*(27*k+10))/(720*k)
Fall 29/25 - case 29/25
Momente an den Ecken, symm
MB1 = MC1 = + 3*M/N
MB2 = MC2 = - 2*M*k/N
HA = HD = - MB1/h
My = y*MB1/h
MB1 - MB2 = M
Fall 29/26 - case 29/26
Momente an den Ecken, asymm
MC2 = - MB2 = M
MB1 = MC1 = 0
HA = HD = 0
Mx = M*(x-x')/l = 2*M*((x/l) - 0.5)
VA = - VD = 2*M/l
E*J1*uB = - (M*(h^2)/(6*k)
Fall 29/27 - case 29/27
Moment am Eckpunkt B
MB1 = MC = (3*M)/(2*N)
VA = - VD = M/l
HA = HD = - MB1/h
MB2 = -M + MB1
Mx = x'*MB2/l + x*MC/l
My = (y/h)*MB1
E*J1*uB = - (M*(h^2)/(12*k)
Fall 29/28 - case 29/28
Konsollast am linken Stiel
alpha = a/h
MB = 0.5*P*c*[(((3*alpha^2) -1)*k)/N + 1]
MC = 0.5*P*c*[(((3*alpha^2) -1)*k)/N - 1]
M1 = -HA*a
M2 = P*c - HA*a
VD = P*c/l
VA = P - VD
HA = HD = -MC/h
My1 = - HA*y1 Bereich a
My1 = P*c - HA*y1 Bereich b
Mx = MC + VD*x'
My2 = -HD*y2
E*J1*uB = (P*c*h^2/12)*(3 -3*alpha^2 + 1/k)
Fall 29/29 - case 29/29
Konsollast an beiden Stielen, symm
alpha = a/h
MB = MC = (P*c*[(3*alpha^2) -1)]*k)/N
HA = HD = (P*c - MB)/h
VA = VD = P
M1 = -HA*a
M2 = P*c - HA*a
My = -HA*y Bereich a
My = P*c - HA*y Bereich b