Shear Force and Bending oment Diagrams for a Beam Steven ukazich San Jose State University
General procedure for the construction of internal force diagrams 1. Find all of the eternal forces and draw the eternal force diagram; 2. Choose a sign convention for each diagram; 3. If necessary, choose a reference coordinate system: 4. Use equilibrium analysis or differential and integral relationships to construct internal force functions; Cut structure at appropriate sections, The FBD on either side of the cut may be analyzed, Indicate unknown internal forces consistent with the chosen sign convention, Plot the internal force function for each segment, 5. Check each diagram for errors; Check discontinuities at location of applied forces in shear diagram, Check discontinuities at location of applied moment in moment diagram, Check differential and integral relationships between distributed load, shear, and bending moment.
Shear and Bending oment Diagram ample beam is supported by a pin support at point and etends over a roller support at point D. The beam is and subjected to a linearly varying load from to B, a point moment at point D a point load at point as shown. Draw the diagram and the bending moment diagram for the beam. Label all local maimum and minimum values and their locations and show your sign convention for each diagram. /ft 3-ft B C D 9 ft 2 ft 3 ft 4 ft
1. Find all of the eternal forces and draw the eternal force diagram; FBD of beam /ft 3-ft B C y D y D 9 ft 2 ft 3 ft 4 ft
Find support reactions using equations of equilibrium 1 2 /ft 9 ft = 9 k 3-ft B C y 6 ft 3 ft D y D 9 ft 2 ft 3 ft 4 ft! # = 0 D y = 10 k
Find support reactions using equations of equilibrium 1 2 /ft 9 ft = 9 k 3-ft B C y 6 ft 3 ft D y D 9 ft 2 ft 3 ft 4 ft! F 1 = 0 y =
Find support reactions using equations of equilibrium 1 2 /ft 9 ft = 9 k 3-ft B C y 6 ft 3 ft D y D! F 2 = 0 9 ft 2 ft 3 ft 4 ft = 0
ternal force diagram FBD of beam showing all known eternal forces /ft 3-ft B C 10 k D 9 ft 2 ft 3 ft 4 ft
2. Choose a sign convention for each diagram; 3. If necessary, choose a reference coordinate system /ft 3-ft B C 10 k D 9 ft 2 ft 3 ft 4 ft Reference coordinate Positive shear Positive bending moment
4. Use equilibrium analysis or differential and integral relationships to construct internal force functions; Cut structure at appropriate sections, The FBD on either side of the cut may be analyzed, Indicate unknown internal forces consistent with the chosen sign convention, Plot the internal force function for each segment, a /ft b c d 3-ft B C a b c d 10 k D 9 ft 2 ft 3 ft 4 ft
nalysis of Segment B FBD of sections on either side of cut at section a-a a a /ft 3-ft a a B C 10 k Notes The FBD on either side of the cut may be analyzed; Unknown internal forces are consistent with the chosen sign convention; For this segment the FBD to the left of section a-a is probably the best choice to analyze.
FBD of section to the left of cut a-a w = /ft 9 ft a = 2 9 w R = 1 2 R 2 9 a = 1 9 7 a 2 3 3 a Notes Find the intensity of the distributed load as a function of ; Replace the distributed load as an equivalent point load; Two unknowns (, ) and two equations of equilibrium available.
Solve for and functions for segment B 1 9 7 a! F 1 = 0 2 3 3 a 0 9 0 = # = 9 = < = 7 k = 2 1 9 7
Solve for and functions for segment B 1 9 7 a! > = 0 2 3 3 a 0 9 0 = # = 0 9 = < = 9 k-ft = 2 1 27?
Relationships between w,, and w = 2 9 = 2 1 9 7 w = 2 9 a = 2 1 27? a 0 9 ft d d = 1 9 2 = w d d = 2 1 27 37 =
Plot and functions for segment B /ft 3-ft B C D 10 k 9 ft 2 ft 3 ft 4 ft 0 3 2 = 4.24 ft -7 k C>2 = 4 2 = 5.66 k ft -9 k-ft
nalysis of Segment BC FBD of sections on either side of cut at section b-b /ft b b 3-ft B C b b 10 k D 9 ft 3 ft 4 ft
FBD of section to the left of cut b-b /ft b 9 k b 9 ft B b 6 ft 9 ft 3 ft B b
Solve for and functions for segment BC 9 k b! F 1 = 0 6 ft 3 ft B b 9 ft 9 11 9 = <F = 7 k 11 = GH = 7 k = 7
Solve for and functions for segment BC 9 k b! I = 0 6 ft 3 ft B b 9 ft 9 11 9 = < = 9 k ft 11 = GH = 2-ft = 54 7
Relationships between w,, and w = 0 /ft b = 7 = 54 7 9 ft B b d d = 0 = w d d = 7 =
Plot and functions for segment BC /ft 3-ft B C D 10 k 9 ft 2 ft 3 ft 4 ft 0 3 2 = 4.24 ft -7 k C>2 = 4 2 = 5.66 k ft -9 k-ft -2-ft
nalysis of Segment CD FBD of sections on either side of cut at section c-c /ft c 3-ft c B C c c 10 k D 9 ft 2 ft 4 ft
FBD of section to the left of cut c-c 9 k c 3-ft 6 ft 9 ft 3 ft B 2 ft C c
Solve for and functions for segment CD 9 k c 3-ft! F 1 = 0 6 ft 9 ft 3 ft B 2 ft C c 11 14 11 = GF = 7 k 14 = JH = 7 k = 7
Solve for and functions for segment CD 9 k c 3-ft! K = 0 6 ft 9 ft 3 ft B 2 ft C c 11 14 11 = GF = 9 k ft 14 = J = 1-ft = 86 7
Relationships between w,, and w = 0 /ft c 3-ft = 7 = 86 7 B 9 ft 2 ft C c d d = 0 = w d d = 7 =
Plot and functions for segment CD /ft 3-ft B C D 10 k 9 ft 2 ft 3 ft 4 ft 0 3 2 = 4.24 ft -7 k C>2 = 4 2 = 5.66 k ft 9 k-ft -7 k -9 k-ft -2-ft -1-ft
nalysis of Segment D FBD of sections on either side of cut at section d-d /ft d 3-ft d B C 10 k d d 9 ft 2 ft 3 ft D 18 -
FBD of section to the right of cut d-d Solve for and functions for segment CD d! F 1 = 0 d 18-14 18 14 = JF = 18 = = = 3
Solve for and functions for segment D d! N = 0 d 18-14 18 14 = JF = 1 ft 18 = = 0 = 54 3
Relationships between w,, and /ft d 3-ft w = 0 = 3 = 54 3 B C 10 k 9 ft 2 ft 3 ft D d d d = 0 = w d d = 3 =
Plot and functions for segment D /ft 3-ft B C D 10 k 9 ft 2 ft 3 ft 4 ft 0 3 2 = 4.24 ft -7 k C>2 = 4 2 = 5.66 k ft 9 k-ft -7 k 0-9 k-ft -2-ft -1-ft