Shear and Moment Diagrams Today’s Objective: Students will be able to: 1.Derive shear and bending moment diagrams for a loaded beam using a) piecewise.

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Presentation transcript:

Shear and Moment Diagrams Today’s Objective: Students will be able to: 1.Derive shear and bending moment diagrams for a loaded beam using a) piecewise analysis b) differential/integral relations

These diagrams plot the internal forces with respect to x along the beam. APPLICATIONS They help engineers analyze where the weak points will be in a member

General Technique Because the shear and bending moment are discontinuous near a concentrated load, they need to be analyzed in segments between discontinuities

Detailed Technique 1) Determine all reaction forces 2) Label x starting at left edge 3) Section the beam at points of discontinuity of load 4) FBD each section showing V and M in their positive sense 5) Find V(x), M(x) 6) Plot the two curves

SIGN CONVENTION FOR SHEAR, BENDING MOMENT Sign convention for: Shear: + rotates section clockwise Moment: + imparts a U shape on section Normal: + creates tension on section (we won't be diagraming nrmal)

Example Find Shear and Bending Moment diagram for the beam Support A is thrust bearing (Ax, Ay) Support C is journal bearing (Cy) PLAN 1) Find reactions at A and C 2) FBD a left section ending at x where (0<x<2) 3) Derive V(x), M(x) 4) FBD a left section ending at x where (2<x<4) 5) Derive V(x), M(x) in this region 6) Plot

Example, (cont) 1) Reactions on beam  2) FBD of left section in AB  –note sign convention 3) Solve: V = 2.5 kN M = 2.5x kN-m 4) FBD of left section ending in BC: 5) Solve: V = -2.5 kN -2.5x+5(x-2)+M = 0 M = x

Example, continued Now, plot the curves in their valid regions: Note disconinuities due to mathematical ideals

Example2 Find Shear and Bending Moment diagram for the beam PLAN 1) Find reactions 2) FBD a left section ending at x, where (0<x<9) 3) Derive V(x), M(x) 4) FBD a left section ending somewhere in BC (2<x<4) 5) Derive V(x), M(x) 6) Plot

Example2, (cont) 1) Reactions on beam  2) FBD of left section  –note sign convention 3) Solve:

Example 2, continued Plot the curves: Notice Max M occurs when V = 0? could V be the slope of M?

A calculus based approach Study the curves in the previous slide Note that 1) V(x) is the area under the loading curve plus any concentrated forces 2) M(x) is the area under V(x) This relationship is proven in your text when loads get complicated, calculus gets you the diagrams quicker

derivation assumes positive distrib load Examine a diff beam section

Example3 Reactions at B