1. Beams Shear & Moment Diagrams E. Evans 2/9/06

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

3. Types of Beams Depends on the support configuration M FvFv FHFH Fixed FVFV FVFV FHFH Pin Roller Pin Roller FVFV FVFV FHFH

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

5. 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. L P ab

6. 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. Pb/L x Left Side of Cut V M N Positive Directions Shown!!!

7. 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. Pa/L L - x Right Side of Cut V M N Positive Directions Shown!!!

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-ft apart. 20 ft P = 20 kips 12 kips 8 kips 12 ft 1 7 10 6 239458 Point 6 is just left of P and Point 7 is just right of P.

10. 20 ft P = 20 kips 12 kips 8 kips 12 ft 1 7 10 6 239458 V (kips) M (ft-kips) 8 kips -12 kips 96 48 64 48 72 24 80 16 32 x x

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

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

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. Load 0ConstantLinear Shear ConstantLinearParabolic Moment LinearParabolicCubic Common Relationships

16. Load 00Constant Shear Constant Linear Moment Linear Parabolic Common Relationships M

17. Example: Draw Shear & Moment diagrams for the following beam 3 m1 m 12 kN8 kN A C B D R A = 7 kN  R C = 13 kN 

18. 3 m1 m 12 kN A C B D V (kN) M (kN-m) 7 -5 8 8 kN 7 -15 8 7 -8 2.4 m