8.2 Shear and ending-oment Diagrams: Equation Form
8.2 Shear and ending-oment Diagrams: Equation Form Eample 1, page 1 of 6 1. Epress the shear and bending moment as functions of, the distance from the left end of the beam to an arbitrary point on the beam. Plot and vs.. 9 kip 6 kip 3 ft 5 ft 7 ft 1 Draw a free-body diagram and find the reactions. 9 kip 6 kip R 3 ft 5 ft 7 ft R + + ΣF y = 0: R 9 kip 6 kip + R = 0 Σ = 0: (9 kip)(3 ft) (6 kip)(3 ft + 5 ft) + R (3 ft + 5 ft + 7 ft) = 0 Solving gives R = 10 kip and R = 5 kip
8.2 Shear and ending-oment Diagrams: Equation Form Eample 1, page 2 of 6 2 Pass a section through the beam at a point between the left end and the 9-kip force. 9 kip 6 kip R = 10 kip 3 ft 5 ft 7 ft R = 5 kip 0 < < 3 ft 3 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. R = 10 kip (sign convention: positive counterclockwise, on right end of section) (sign convention: positive down, on right end of section) + + ΣF y = 0: 10 kip = 0 Σ = 0: (10 kip) + = 0 Solving gives = 10 kip and = 10 kip ft valid for 0 < < 3 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 1, page 3 of 6 4 Pass a section through the beam at a point between the 9-kip force and the 6-kip force. 9 kip 6 kip R = 10 kip 3 ft 5 ft 7 ft R = 5 kip 3 ft < < 8 ft 5 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. 9 kip + ΣF y = 0: 10 kip 9 kip = 0 + Σ = 0: (10 kip) + (9 kip)( 3 ft) + = 0 R = 10 kip 3 ft ( 3 ft) Solving gives = 1 kip (3) = ( + 27) kip ft (4) valid for 3 ft < < 8 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 1, page 4 of 6 6 Pass a section through the beam at a point between the 6-kip force and the right end of the beam. 9 kip 6 kip 3 ft 5 ft R = 10 kip 8 ft < < 15 ft 7 ft R = 5 kip 7 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. 9 kip 6 kip + + ΣF y = 0: 10 kip 9 kip 6 kip = 0 Σ = 0: (10 kip) + (9 kip)( 3 ft) + (6 kip)( 8 ft) + = 0 Solving gives 3 ft 5 ft ( 8 ft) R = 10 kip ( 3 ft) = 5 kip (5) = ( 5 + 75) kip ft (6) valid for 8 ft < < 15 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 1, page 5 of 6 8 Collect the results from Eqs. 1-6: 0 < < 3 ft = 10 kip = 10 kip ft 3 ft < < 8 ft = 1 kip = ( + 27) kip ft ns. 8 ft < < 15 ft = 5 kip = ( 5 + 75) kip ft
8.2 Shear and ending-oment Diagrams: Equation Form Eample 1, page 6 of 6 9 Plot and versus. 9 kip 6 kip R = 10 kip 3 ft 5 ft 7 ft R = 5 kip (kip) 10 1 (kip ft) 30 35 5
8.2 Shear and ending-oment Diagrams: Equation Form Eample 2, page 1 of 3 2. Epress the shear and bending moment as functions of, the distance from the left end of the beam to an arbitrary point on the beam. Plot and vs.. 2 N/m 20 N m 14 m 1 Draw a free-body diagram and find the reactions. 14 m = 7 m 2 Resultant = (2 N/m)(14 m) = 28 N 2 N/m 14 m 20 N m + + R ΣF y = 0: R 28 N = 0 Σ = 0: (28 N)(7 m) + 20 N m = 0 2 couple-moment reaction must always be included at a built-in end of a beam. Solving gives R = 28 N and = 176 N m
8.2 Shear and ending-oment Diagrams: Equation Form Eample 2, page 2 of 3 3 Pass a section through the beam at an arbitrary point (located by ) 2 N/m = 176 N m 14 m 20 N m R = 28 N 0 < < 14 m 4 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. Resultant = (2 N/m)() 2 + + ΣF y = 0: 28 N 2 = 0 Σ = 0: 176 N m 28 + ( )(2 N/m)() + = 0 2 Solving gives = 176 N m = ( 2 + 28) N ns. = ( 2 + 28 176) N m ns. valid for 0 < < 14 m. R = 28 N
8.2 Shear and ending-oment Diagrams: Equation Form Eample 2, page 3 of 3 5 Plot and versus. 2 N/m = 176 N m 14 m 20 N m R = 28 N (N) 28 (N m) 20 176
8.2 Shear and ending-oment Diagrams: Equation Form Eample 3, page 1 of 6 3. Epress the shear and bending moment as functions of, the distance from the left end of the beam to an arbitrary point on the beam. Plot and vs.. 10 lb/ft 7 ft 10 ft 3 ft 1 Draw a free-body diagram and find the reactions. 10 ft 2 = 5 ft Resultant = (10 lb/ft)(10 ft) = 100 lb 7 ft 10 ft 3 ft R + + ΣF y = 0: R 100 lb + R = 0 Σ = 0: (100 lb)(7 ft + 5 ft) + R (7 ft + 10 ft + 3 ft) = 0 R Solving gives R = 40 lb and R = 60 lb
8.2 Shear and ending-oment Diagrams: Equation Form Eample 3, page 2 of 6 2 Pass a section through the beam at a point between the left end of the beam and the beginning of the distributed load. 10 lb/ft R = 40 lb 7 ft 10 ft 3 ft R = 60 lb 0 < < 7 ft 3 Draw a free-body diagram and find the reactions. + ΣF y = 0: 40 lb = 0 R = 40 lb + Σ = 0: (40 lb) + = 0 Solving gives = 40 lb (1) = (40) lb ft (2) valid for 0 < < 7 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 3, page 3 of 6 4 Pass a section through the beam at a point between the beginning and end of the distributed load. 10 lb/ft R = 40 lb 7 ft 10 ft 3 ft R = 60 lb 7 ft < < 17 ft 5 Draw a free-body diagram of the portion of the beam to the left of the section and solve for and at the section. + ΣF y = 0: 40 lb (10 lb/ft)( 7 ft) = 0 Resultant = (10 lb/ft)( 7 ft) 7 ft 2 + Σ = 0: (40 lb) + [(10 lb/ft)( 7 ft)] ( 7 ft ) + = 0 2 Solving gives R = 40 lb 7 ft 7 ft = ( 10 + 110) lb (3) = ( 5 2 + 110 245) lb ft (4) valid for 7 ft < < 17 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 3, page 4 of 6 6 Pass a section through the beam at a point between the right end of the distributed load and the right end of the beam. 10 lb/ft R = 40 lb 7 ft 10 ft 3 ft R = 60 lb 17 ft < < 20 ft 7 Draw a free-body diagram of the portion of the beam to the left of the section and solve for and at the section. 10 ft = 5 ft Resultant = (10 lb/ft)(10 ft) = 100 lb 2 R = 40 lb 7 ft 10 ft 17 ft
8.2 Shear and ending-oment Diagrams: Equation Form Eample 3, page 5 of 6 8 + + ΣF y = 0: 40 lb 100 lb = 0 Σ = 0: (40 lb) + (100 lb)[( 17 ft) + 5 ft] + = 0 Solving gives = 60 lb (5) = ( 60 + 1200) lb ft (6) valid for 17 ft < < 20 ft. 9 Collect the results from Eqs. 1-6: 0 < < 7 ft = 40 lb = 40 lb ft 7 ft < < 17 ft = ( 10 + 110) lb = ( 52 + 110 245) lb ft ns. 17 ft < < 20 ft = 60 lb = ( 60 + 1200) lb ft
8.2 Shear and ending-oment Diagrams: Equation Form Eample 3, page 6 of 6 10 Plot and versus. 10 lb/ft (lb) R = 40 lb 40 7 ft 10 ft 3 ft R = 60 lb 40 (lb ft) 280 360 60 60 180
8.2 Shear and ending-oment Diagrams: Equation Form Eample 4, page 1 of 6 4. Epress the shear and bending moment as functions of, the distance from the left end of the beam to an arbitrary point on the beam. Plot and vs.. 27 kip ft 18 kip ft 3 ft 5 ft 7 ft 1 Draw a free-body diagram and find the reactions. 27 kip ft 18 kip ft R 3 ft 5 ft 7 ft R + + ΣF y = 0: R + R = 0 Σ = 0: R (3 ft + 5 ft + 7 ft) 27 kip ft 18 kip ft = 0 Solving gives R = 3 kip = 3 kip R = 3 kip
8.2 Shear and ending-oment Diagrams: Equation Form Eample 4, page 2 of 6 2 Pass a section through the beam at a point between the left end and the 27 kip ft moment couple. 27 kip ft 18 kip ft R = 3 kip 3 ft 5 ft 7 ft R = 3 kip 0 < < 3 ft 3 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. R = 3 kip + ΣF y = 0: 3 kip = 0 + Σ = 0: (3 kip) + = 0 Solving gives = 3 kip (1) = 3 kip ft (2) valid for 0 < < 3 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 4, page 3 of 6 4 Pass a section through the beam at a point between the 27 kip ft and 18 kip ft moment couples. R = 3 kip 3 ft < < 8 ft 27 kip ft 18 kip ft 3 ft 5 ft 7 ft R = 3 kip 5 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. ΣF y = 0: 3 kip = 0 R = 3 kip 27 kip ft 3 ft 3 ft + + Σ = 0: (3 kip) 27 kip ft + = 0 Solving gives = 3 kip (3) = ( 3 + 27) kip ft (4) valid for 3 ft < < 8 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 4, page 4 of 6 6 Pass a section through the beam at a point between the 18 kip ft moment couple and the right end of the beam. 27 kip ft 18 kip ft 3 ft 5 ft R = 3 kip 3 ft < < 8 ft 7 ft R = 3 kip 7 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. 3 ft R = 3 kip 27 kip ft 5 ft 18 kip ft + + ΣF y = 0: 3 kip = 0 Σ = 0: (3 kip) 27 kip ft 18 kip ft + = 0 Solving gives = 3 kip (5) = ( 3 + 45) kip ft (6) valid for 8 ft < < 15 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 4, page 5 of 6 8 Collect the results from Eqs. 1-6: 0 < < 3 ft = 3 kip = 3 kip ft 3 ft < < 8 ft = 3 kip = ( 3 + 27) kip ft ns. 8 ft < < 15 ft = 3 kip = ( 3 + 45) kip ft
8.2 Shear and ending-oment Diagrams: Equation Form Eample 4, page 6 of 6 9 Plot and versus. 27 kip ft 18 kip ft 3 ft R = 3 kip 5 ft 7 ft R = 3 kip (kip) 3 3 (kip ft) 18 21 3 9
8.2 Shear and ending-oment Diagrams: Equation Form Eample 5, page 1 of 6 5. Epress the shear and bending moment in the horizontal portion CD of the beam as functions of, the distance from the left end of the beam to an arbitrary point on the beam. Plot and versus. 4 kip 4 kip C D 2 ft 2 ft 4 ft 2 ft 2 ft 1 Draw a free-body diagram and find the reactions. 4 kip 4 kip + + ΣF y = 0: R 4 kip 4 kip + R = 0 Σ = 0: (4 kip)(2 ft) (4 kip)(10 ft) + R (12 ft) = 0 Solving gives R 2 ft 2 ft 4 ft 2 ft 2 ft R R = 4 kip R = 4 kip
8.2 Shear and ending-oment Diagrams: Equation Form Eample 5, page 2 of 6 2 Pass a section through the beam at a point between the left end and the attachment point for the first arm. 4 kip 4 kip 2 ft R = 4 kip 0 < < 4 ft 2 ft 4 ft 2 ft 2 ft R = 4 kip 3 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. Note carefully that the 4-kip force on the left arm does not act on this free body. + + ΣF y = 0: 4 kip = 0 Σ = 0: (4 kip) + = 0 Solving gives = 4 kip (1) = 4 kip ft (2) R = 4 kip valid for 0 < < 4 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 5, page 3 of 6 4 Pass a section through the beam at a point between the attachment points of the two arms. 4 kip 4 kip 2 ft R = 4 kip 2 ft 4 ft 2 ft 2 ft R = 4 kip 4 ft < < 8 ft 5 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. 4 kip + + ΣF y = 0: 4 kip 4 kip = 0 Σ = 0: (4 kip) + (4 kip)( 4 ft + 2 ft) + = 0 Solving gives = 0 (3) 2 ft R = 4 kip 2 ft 4 ft = 8 kip ft (4) valid for 4 ft < < 8 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 5, page 4 of 6 6 Pass a section through the beam at a point between the point of attachment of the right arm and the right end of the beam. 4 kip 4 kip 2 ft 2 ft 4 ft 2 ft 2 ft R = 4 kip R = 4 kip 7 8 ft < < 12 ft Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. 4 kip 8 Note that the 4-kip force on the right arm acts on the free body. 4 kip + + ΣF y = 0: 4 kip 4 kip 4 kip = 0 Σ = 0: (4 kip) + (4 kip)( 2 ft) + (4 kip)( 10 ft) + = 0 Solving gives = 4 kip (5) 2 ft R = 4 kip 2 ft 4 ft 2 ft 10 ft = ( 4 + 48) kip ft (6) valid for 8 ft < < 12 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 5, page 5 of 6 9 Collect the results from Eqs. 1-6: 0 < < 4 ft = 4 kip = 4 kip ft 4 ft < < 8 ft = 0 kip = 8 kip ft ns. 8 ft < < 12 ft = 4 kip = 4( 12) kip ft
8.2 Shear and ending-oment Diagrams: Equation Form Eample 5, page 6 of 6 10 Plot and versus. 4 kip 4 kip (kip) 2 ft R = 4 kip 2 ft 4 4 ft 2 ft 2 ft R = 4 kip 11 Note that the jumps in the diagrams occur at the attachment points of the arms, not at the points where the 4-kip eternal loads act. 4 16 16 (kip ft) 8 8
8.2 Shear and ending-oment Diagrams: Equation Form Eample 6, page 1 of 8 6. Epress the shear and bending moment as functions of, the distance from the left end of the beam to an arbitrary point on the beam. Plot and vs.. 4 kn 8 kn Hinge C 2 m 2 m 2 m 2 m 1 Draw a free-body diagram and find the reactions. Hinge 4 kn 8 kn C R 2 m 2 m 2 m R 2 m R C + + ΣF y = 0: R + R 4 kn 8 kn + R C = 0 (1) Σ = 0: R (2 m) (4 kn)(2 m + 2 m) (8 kn)(2 m + 2 m + 2 m) 2 Two equations but three unknowns. n additional equation is needed. + R C (2 m + 2 m + 2 m + 2 m) = 0 (2)
8.2 Shear and ending-oment Diagrams: Equation Form Eample 6, page 2 of 8 3 Pass a section through the beam at a point immediately to the right of the hinge. Hinge 4 kn 8 kn C 2 m 2 m 2 m 2 m 4 Draw a free-body diagram of the portion of the beam to the right of the section. 8 kn 5 ecause the section is net to a hinge, the moment is known to be zero there (that's what we mean by a "hinge"). = 0 2 m 2 m C R C 6 Write the equilibrium equation for the sum of moments about the hinge. + Σ hinge = 0: (8 kn)(2 m) + Rc(2 m + 2 m) = 0 (3) 7 Note that we don't use the equation ΣF y = 0, because this equation would introduce an additional unknown, the shear at the hinge.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 6, page 3 of 8 8 Solving Eqs. 1-3 gives R = 8 kn = 8 kn R = 16 kn 9 R C = 4 kn Pass a section through the beam at a point between the left end and the reaction at. 4 kn 8 kn Hinge R = 8 kn 2 m 2 m 2 m R = 16 kn 0 < < 2 m ΣF y = 0: 8 kn = 0 + 2 m C R C = 4 kn 10 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. R = 8 kn + Σ = 0: (8 kn) + Μ = 0 Solving gives = 8 kn (4) = ( 8) kn m (5) valid for 0 < < 2 m.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 6, page 4 of 8 11 Pass a section through the beam at a point between the the reaction at and the hinge. 4 kn 8 kn Hinge C 2 m 2 m 2 m R = 16 kn R = 8 kn 2 m < < 4 m 2 m R C = 4 kn 12 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. 2 ft + + ΣF y = 0: 8 kn + 16 kn = 0 Σ = 0: (8 kn) (16 kn)( 2 m) + = 0 R = 8 kn R = 16 kn Solving gives = 8 kn (6) = (8 32) kn m (7) valid for 2 m < < 4 m.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 6, page 5 of 8 13 Pass a section through the beam at a point between the hinge and the 8-kN force. 4 kn 8 kn Hinge C R = 8 kn 4 m < < 6 m 2 m 2 m 2 m R = 16 kn 2 m R C = 4 kn 14 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. 2 m Hinge 2 m 4 kn Σ F y = 0: 8 kn + 16 kn 4 kn = 0 Σ = 0: (8 kn) (16 kn)( 2 m) + (4 kn)( 4 m) + = 0 Solving gives = 4 kn (8) = (4 16) kn m (9) R = 8 kn R = 16 kn valid for 4 m < < 6 m.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 6, page 6 of 8 15 Pass a section through the beam at a point between the 8-kN force and end C of the beam. 4 kn 8 kn Hinge 2 m 2 m 2 m R = 8 kn R = 16 kn 2 m C R C = 4 kn 6 m < < 8 m 16 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. F y = 0: 8 kn + 16 kn 4 kn 8 kn = 0 R = 8 kn 2 m 2 m R = 16 kn Hinge 4 kn 2 m 8 kn = 0: (8 kn) (16 kn)( 2 m) + (4 kn)( 4 m) + (8 kn)( 6 m) + = 0 Solving gives = 4 kn (10) = ( 4 + 32) kn m (11) valid for 6 m < < 8 m.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 6, page 7 of 8 17 Collect the results from Eqs. 4-11: 0 < < 2 m = 8 kn = 8 kn m 2 m < < 4 m = 8 kn 4 m < < 6 m = 4 kn = (8 32) kn m ns. = (4 16) kn m 6 m < < 8 m = 4 kn = ( 4 + 32) kn m
8.2 Shear and ending-oment Diagrams: Equation Form Eample 6, page 8 of 8 18 Plot and versus. 4 kn 8 kn Hinge C 2 m 2 m 2 m R = 8 kn R = 16 kn 2m R C = 4 kn (kn) 8 4 8 4 4 4 8 (kn m) 8 16
8.2 Shear and ending-oment Diagrams: Equation Form Eample 7, page 1 of 7 7. Epress the shear and bending moment as functions of, the distance from the left end of the beam to an arbitrary point on the beam. Plot and vs.. 20 kip 4 kip/ft Hinge 5 ft 5 ft 8 ft 1 Draw a free-body diagram and find the reactions. 20 kip 5 ft 5 ft Hinge 2 couple moment must always be included at a built-in end of a beam. R + ΣF y = 0: R 20 kip 32 kip + R = 0 (1) 8 ft = 4 ft Resultant = (4 kip/ft)(8 ft) = 32 kip 2 8 ft R + Σ = 0: (20 kip)(5 ft) (32 kip)(5 ft + 5 ft + 4 ft) + R (5 ft + 5 ft + 8 ft) = 0 (2) 3 Two equations but three unknowns. n additional equation is needed.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 7, page 2 of 7 4 5 Pass a section through the beam at a point immediately to the right of the hinge. 20 kip Draw a free-body diagram of the portion of the beam to the right of the section. Hinge R 5 ft 5 ft 8 ft 8 ft 2 4 ft = 4 ft Resultant = (4 kip/ft)(8 ft) = 32 kip Resultant = 32 kip R 6 ecause the section is net to a hinge, the moment is known to be zero there (that's what we mean by a "hinge"). = 0 8 ft 7 Write the equilibrium equation for the sum of moments about the hinge. R + Σ hinge = 0: (32 kip)(4 ft) + R (8 ft) = 0 (3) 8 Note that we don't use the equation ΣF y = 0, because this equation would introduce an additional unknown, the shear at the hinge.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 7, page 3 of 7 9 Solving Eqs. 1-3 gives R = 36 kip R = 16 kip = 260 kip ft 10 Pass a section through the beam at a point between the left end and the 20-kip force. = 260 kip ft 20 kip Hinge 4 kip/ft 5 ft 5 ft 8 ft R = 36 kip R = 16 kip 0 < < 5 ft + ΣF y = 0: 36 kip = 0 11 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. = 260 kip ft R = 36 kip + Σ = 0: 260 kip ft (36 kip) + = 0 Solving gives = 36 kip (4) = (36 260) kip ft (5) valid for 0 < < 5 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 7, page 4 of 7 12 Pass a section through the beam at a point between the 20-kip force and the hinge. = 260 kip ft 20 kip Hinge 4 kip/ft 5 ft 5 ft 8 ft R = 36 kip R = 16 kip 5 ft < < 10 ft 13 Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. 20 kip + ΣF y = 0: 36 kip 20 kip = 0 = 260 kip ft + Σ = 0: 260 kip ft (36 kip) + 20 kip( 5 ft) + = 0 R = 36 kip 5 ft Solving gives = 16 kip (6) = (16 160) kip ft (7) valid for 5 ft < < 10 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 7, page 5 of 7 14 Pass a section through the beam at a point between the hinge and the right end of the beam. 20 kip 4 kip/ft = 260 kip ft Hinge 10 ft < < 18 ft 15 R = 36 kip 5 ft 5 ft Draw a free-body diagram of the portion of the beam to the left of the section and find and at the section. = 260 kip ft R = 36 kip 5 ft 20 kip Resultant = (4 kip/ft)( 10 ft) 10 ft 2 Hinge 5 ft + + 8 ft R = 16 kip ΣF y = 0: 36 kip 20 kip (4 kip/ft)( 10 ft) = 0 Σ = 0: 260 kip ft (36 kip) + 20 kip( 5 ft) + (4 kip/ft)( 10 ft)( 10 ft ) 2 + = 0 Solving gives = 4 + 56 kip (8) = ( 2 2 + 56 360) kip ft (9) valid for 10 ft < < 18 ft.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 7, page 6 of 7 16 Collect the results from Eqs. 4-9: 0 < < 5 ft = 36 kip = (36 260) kip ft 5 ft < < 10 ft = 16 kip = (16 160) kip ft ns. 10 ft < < 18 ft = ( 4 + 56) kip = ( 2 2 + 56 360) kip ft
8.2 Shear and ending-oment Diagrams: Equation Form Eample 7, page 7 of 7 17 Plot and versus. 20 kip 4 kip/ft = 260 kip ft Hinge R = 36 kip 5 ft 5 ft 8 ft R = 16 kip 36 36 (kn) 16 16 32 16 (kn m) 80 260
8.2 Shear and ending-oment Diagrams: Equation Form Eample 8, page 1 of 8 8. Epress the shear and bending moment as functions of, the distance from the left end of the beam to an arbitrary point on the beam. Plot and vs.. 2 kn/m 4 kn/m 6 m 6 m 1 Draw a free-body diagram and find the reactions. 2 kn/m 2 kn/m 6 m R 6 m R 2 Replace the trapezoidal distributed load by the sum of a rectangular and triangular load.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 8, page 2 of 8 3 Resultant of rectangular load 4 = (12 m)(2 kn/m) = 24 kn 12 m 6 m 6 m Resultant of triangular load 5 1 = (12 m)(2 kn/m) 2 = 12 kn 2 kn/m 12 m 1 3 6 2 kn/m (6 m + 6 m) = 4 m (acts through centroid of triangle) 6 m 4 m = 2 m R R + + ΣF y = 0: R 24 kn 12 kn + R = 0 Σ = 0: (12 kn)(2 m) + R (6 m) = 0 Solving gives R = 32 kn R = 4 kn
8.2 Shear and ending-oment Diagrams: Equation Form Eample 8, page 3 of 8 7 Pass a section through the beam at a point between the left end and the support at. 2 kn/m 4 kn/m 0 < < 6 m 6 m R = 32 kn 6 m R = 4 kn 8 Draw a free-body diagram of the portion of the beam to the left of the section and solve for and. 2 kn/m w = distributed load (kn/m) at location
8.2 Shear and ending-oment Diagrams: Equation Form Eample 8, page 4 of 8 9 efore we can solve for and, we have to epress w as a function of. This can be done by noting that w is a linear function of and then using the slope-intercept equation for a line. w (0, 2 kn/m) 10 (, w) Now the distributed load on the free-body of length can be replaced by the resultant of a rectangular and triangular load. (12 m, 4 kn/m) Slope Intercept w = m + b 4 kn/m 2 kn/m = + 2 kn/m 12 m 0 = + 2 (1) 6 11 Resultant of rectangular load 12 Resultant of triangular load 2 kn/m Free-body diagram w w 2 = = (2 kn/m) 2 3 = (w 2) 2
8.2 Shear and ending-oment Diagrams: Equation Form Eample 8, page 5 of 8 13 + + ΣF y = 0: (2 kn/m) (w 2) = 0 2 Σ = 0: (2 kn/m)()( ) + [ (w 2)( )] + = 0 2 2 3 Replacing w in these equations by w = (/6) + 2 from Eq. 1 and solving gives 2 = ( 2) kn (2) 12 3 = ( 2 ) kn m (3) 36 valid for 0 < < 6 m. 14 Pass a section through the beam at a point between the support at and the support at. 2 kn/m 4 kn/m 6 m < < 12 m 6 m R = 32 kn 6 m R = 4 kn
8.2 Shear and ending-oment Diagrams: Equation Form Eample 8, page 6 of 8 15 Free-body diagram 2 kn/m w() 6 m < < 12 m 6 m 6 m R = 32 kn 16 We can save some work if we note that this free-body diagram is identical to the previous one ecept that an additional vertical force of 32 kn is present. This increases the shear in Eq. 2 by 32 kn and the moment in Eq. 3 by (32 kn)( 6 m) so 2 = ( 2 + 32) kn (4) 12 3 = ( 2 + 32 192) kn m (5) 36 valid for 6 m < < 12 m.
8.2 Shear and ending-oment Diagrams: Equation Form Eample 8, page 7 of 8 17 Collect the results from Eqs. 4-11: 0 < < 6 m = ( 1 2 2) kn 12 = ( 1 3 2 ) kn m 36 6 m < < 12 m = ( 1 2 2 + 32) kn 12 ns. = ( 1 3 2 + 32 192) kn m 36
8.2 Shear and ending-oment Diagrams: Equation Form Eample 8, page 8 of 8 18 Plot and versus. 4 kn/m 2 kn/m (kn) 6 m 17 R = 32 kn 6 m R = 4 kn 4 15 2.03 (kn m) 42