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Beams Shear & Moment Diagrams

Beams Shear & Moment Diagrams. Prepared by Prof. V. S. Nikam. Beams. Members that are slender and support loads applied perpendicular to their longitudinal axis. Concentrated Load, P. Distributed Load, w(x). Longitudinal Axis. Span, L. F H. Pin. F V. F V. Fixed. F H. Roller. M.

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Beams Shear & Moment Diagrams

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  1. BeamsShear & Moment Diagrams Prepared by Prof. V. S. Nikam

  2. Beams • Members that are slender and support loads applied perpendicular to their longitudinal axis. Concentrated Load, P Distributed Load, w(x) Longitudinal Axis Span, L

  3. FH Pin FV FV Fixed FH Roller M Fv Roller Pin FH FV FV Types of Beams • Depends on the support configuration

  4. Statically Indeterminate Beams • Can you guess how we find the “extra” reactions? Continuous Beam Propped Cantilever Beam

  5. P a b L Internal Reactions in Beams • At any cut in a beam, there are 3 possible internal reactions required for equilibrium: • normal force, • shear force, • bending moment.

  6. M N V Internal Reactions in Beams • At any cut in a beam, there are 3 possible internal reactions required for equilibrium: • normal force, • shear force, • bending moment. Positive Directions Shown!!! Left Side of Cut Pb/L x

  7. V M N Internal Reactions in Beams • At any cut in a beam, there are 3 possible internal reactions required for equilibrium: • normal force, • shear force, • bending moment. Positive Directions Shown!!! Right Side of Cut Pa/L L - x

  8. Finding Internal Reactions • Pick left side of the cut: • Find the sum of all the vertical forces to the left of the cut, including V. Solve for shear, V. • Find the sum of all the horizontal forces to the left of the cut, including N. Solve for axial force, N. It’s usually, but not always, 0. • Sum the moments of all the forces to the left of the cut about the point of the cut. Include M. Solve for bending moment, M • Pick the right side of the cut: • Same as above, except to the right of the cut.

  9. Example: Find the internal reactions at points indicated. All axial force reactions are zero. Points are 2-m apart. P = 20 kN 1 2 3 4 5 8 9 10 6 7 8 kN 12 kN 12 m 20 m Point 6 is just left of P and Point 7 is just right of P.

  10. P = 20 kN 1 2 3 4 5 8 9 10 6 7 8 kN 12 kN 12 m 20 m 8 kN V (kN) x -12 kN 96 80 72 64 48 48 32 24 16 M (kN-m) x

  11. What is the slope of this line? What is the slope of this line? V & M Diagrams P = 20 kN 8 kN 12 kN 12 m 20 m 8 kN V (kN) x -12 kN 96 kN-m 96 ft-kips/12’ = 8 kips b -12 kips M (kN-m) a c x

  12. V & M Diagrams P = 20 kN 8 kN 12 kN 12 m 20 m 8 kN V (kN) x What is the area of the blue rectangle? -12 kN 96 kN-m What is the area of the green rectangle? 96 kN-m b -96 kN-m M (kN-m) a c x

  13. Draw Some Conclusions • The magnitude of the shear at a point equals the slope of the moment diagram at that point. • The area under the shear diagram between two points equals the change in moments between those two points. • At points where the shear is zero, the moment is a local maximum or minimum.

  14. The Relationship Between Load, Shear and Bending Moment

  15. Common Relationships

  16. Common Relationships M

  17. Example: Draw Shear & Moment diagrams for the following beam 12 kN 8 kN A C D B 1 m 3 m 1 m RA = 7 kN  RC = 13 kN 

  18. 12 kN 8 kN A C D B 1 m 3 m 1 m 8 7 8 7 V (kN) -15 -5 7 M (kN-m) 2.4 m -8

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