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Esfuerzos cortantes q

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AlberthOo Louis
AlberthOo Louis

This document contains calculations for a beam subjected to a 60 kN point load. It includes: 1) A free body diagram of the beam showing the sections and loads. 2) Calculation of the reactions at supports A and C, which are 35 kN and 25 kN respectively. 3) Calculation of the shear force and bending moment diagrams, showing the maximum bending moment is 175 kNm at the point of load application.

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ESFUERZOS CORTANTES Q Q (0 - 5.0) Q (5.0 - 7.0) Q (7.0 - 12.0) 350.00 -250.00 -250.00 MOMENTOS FLECTORES MOMENTO X P=600.00 TOTAL a=5.00 0.00 0.00 0.00 1.20 420.00 420.00 2.40 840.00 840.00 3.60 1,260.00 1,260.00 4.80 1,680.00 1,680.00 5.00 1,750.00 1,750.00 6.00 1,500.00 1,500.00 7.20 1,200.00 1,200.00 8.40 900.00 900.00 9.60 600.00 600.00 10.80 300.00 300.00 12.00 0.00 0.00
C A L C U L O D E V IG A 60kn I. D IA G R A M A D E C U E R P O LIB R E B II.C A L C U L O D E R E A C C IO N E S w + M B =0 = -P (5 )+ R C (12)= 0 = -6 0 (5 )+ R C (12)= 0 = -3 0 0 + R C (12) C A RC= 60kn 5 .00 w B 7 .00 A C R A = 35 300 12 =25 y=o = R A + R C -P = 0 = R A + 2 5 -60= R A = -2 5 + 60 R A = 35 III.C A L C U L O D E M O M E N T O Y CORTANTE R C = 25 w TRAMO 1 + F y= 0 = -v + 3 5= 0 v = 35 + M =0 = M -3 5 (x)= 0 M = 3 5 (x) V= M w A X R A=35 v=35 TRAMO 2 w 5 kn P=60 x x V= X 5 w R A=35 + F y= 0 = R A -P -V = 0 v = 3 5 -60-V v = -25 + M =0 = M -R A (x )+ P (x-5)= 0 = M -3 5 x + 6 0 (x-5)= 0 = M -3 5 x + 6 0 x-300 = M + 2 5 x -300 M = -2 5 x + 300 v = -25 X M m a x .= 1 75 M = 3 5 (x) M = 3 5 (x) 1 2 3 4 5 M = -2 5 x + 300 X 0 35 70 105 140 175 0 5 6 7 8 9 10 11 12 M = -2 5 (x )+ 300 175 150 125 100 175 75 25 0
C A L C U L O D E V IG A 6 0 kn I. D IA G R A M A D E C U E R P O LIB R E B II.C A LC U L O D E R E A C C IO N E S w + M B=0 = -P (5 )+ R C (12)= 0 = -6 0 (5 )+ R C (12)= 0 = -3 0 0 + R C (12) C A RC= 6 0 kn 5 .00 w B 7 .00 A C R A = 35 300 1 2 = 25 y= o = R A + R C -P = 0 = R A + 25-60= R A = -25+ 60 R A = 35 III.C A L C U LO D E M O M E N T O Y C O RTANTE R C = 25 w TRAM O 1 + F y= 0 = -v+ 35= 0 v= 35 + M =0 = M -3 5 (x)= 0 M = 3 5 (x) V= M w A X R A = 35 v= 35 TRAM O 2 w 5 kn P=60 x x V= X 5 w R A = 35 + F y= 0 = R A -P -V = 0 v= 3 5 -60-V v= -25 + M =0 = M -R A (x)+ P (x-5)= 0 = M -3 5 x+ 6 0 (x-5)= 0 = M -3 5 x+ 60x-300 = M + 2 5 x-300 M = -2 5 x+ 300 v= -25 X M m a x.= 175 M = 3 5 (x) M = 3 5 (x) 1 2 3 4 5 M = -2 5 x+ 300 X 0 35 70 105 140 175 0 5 6 7 8 9 10 11 12 M = -2 5 (x)+ 300 175 150 125 100 175 75 25 0
Esfuerzos cortantes q
Esfuerzos cortantes q
Esfuerzos cortantes q

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Esfuerzos cortantes q

  • 1. ESFUERZOS CORTANTES Q Q (0 - 5.0) Q (5.0 - 7.0) Q (7.0 - 12.0) 350.00 -250.00 -250.00 MOMENTOS FLECTORES MOMENTO X P=600.00 TOTAL a=5.00 0.00 0.00 0.00 1.20 420.00 420.00 2.40 840.00 840.00 3.60 1,260.00 1,260.00 4.80 1,680.00 1,680.00 5.00 1,750.00 1,750.00 6.00 1,500.00 1,500.00 7.20 1,200.00 1,200.00 8.40 900.00 900.00 9.60 600.00 600.00 10.80 300.00 300.00 12.00 0.00 0.00
  • 2. C A L C U L O D E V IG A 60kn I. D IA G R A M A D E C U E R P O LIB R E B II.C A L C U L O D E R E A C C IO N E S w + M B =0 = -P (5 )+ R C (12)= 0 = -6 0 (5 )+ R C (12)= 0 = -3 0 0 + R C (12) C A RC= 60kn 5 .00 w B 7 .00 A C R A = 35 300 12 =25 y=o = R A + R C -P = 0 = R A + 2 5 -60= R A = -2 5 + 60 R A = 35 III.C A L C U L O D E M O M E N T O Y CORTANTE R C = 25 w TRAMO 1 + F y= 0 = -v + 3 5= 0 v = 35 + M =0 = M -3 5 (x)= 0 M = 3 5 (x) V= M w A X R A=35 v=35 TRAMO 2 w 5 kn P=60 x x V= X 5 w R A=35 + F y= 0 = R A -P -V = 0 v = 3 5 -60-V v = -25 + M =0 = M -R A (x )+ P (x-5)= 0 = M -3 5 x + 6 0 (x-5)= 0 = M -3 5 x + 6 0 x-300 = M + 2 5 x -300 M = -2 5 x + 300 v = -25 X M m a x .= 1 75 M = 3 5 (x) M = 3 5 (x) 1 2 3 4 5 M = -2 5 x + 300 X 0 35 70 105 140 175 0 5 6 7 8 9 10 11 12 M = -2 5 (x )+ 300 175 150 125 100 175 75 25 0
  • 3. C A L C U L O D E V IG A 6 0 kn I. D IA G R A M A D E C U E R P O LIB R E B II.C A LC U L O D E R E A C C IO N E S w + M B=0 = -P (5 )+ R C (12)= 0 = -6 0 (5 )+ R C (12)= 0 = -3 0 0 + R C (12) C A RC= 6 0 kn 5 .00 w B 7 .00 A C R A = 35 300 1 2 = 25 y= o = R A + R C -P = 0 = R A + 25-60= R A = -25+ 60 R A = 35 III.C A L C U LO D E M O M E N T O Y C O RTANTE R C = 25 w TRAM O 1 + F y= 0 = -v+ 35= 0 v= 35 + M =0 = M -3 5 (x)= 0 M = 3 5 (x) V= M w A X R A = 35 v= 35 TRAM O 2 w 5 kn P=60 x x V= X 5 w R A = 35 + F y= 0 = R A -P -V = 0 v= 3 5 -60-V v= -25 + M =0 = M -R A (x)+ P (x-5)= 0 = M -3 5 x+ 6 0 (x-5)= 0 = M -3 5 x+ 60x-300 = M + 2 5 x-300 M = -2 5 x+ 300 v= -25 X M m a x.= 175 M = 3 5 (x) M = 3 5 (x) 1 2 3 4 5 M = -2 5 x+ 300 X 0 35 70 105 140 175 0 5 6 7 8 9 10 11 12 M = -2 5 (x)+ 300 175 150 125 100 175 75 25 0