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Beams Shear & Moment Diagrams E. Evans 2/9/06 Beams • Members that are slender and support loads applied perpendicular to their longitudinal axis. Distributed Load, w(x) Concentrated Load, P Longitudinal Axis Span, L Types of Beams • Depends on the support configuration FH Pin FH FV FV Fixed Roller M Fv Roller FV Pin FV FH Statically Indeterminate Beams Continuous Beam Propped Cantilever Beam • Can you guess how we find the “extra” reactions? 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. 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. M Left Side of Cut N V Pb/L x Positive Directions Shown!!! 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. M V Positive Directions Shown!!! Right Side of Cut N Pa/L L-x 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. Example: Find the internal reactions at points indicated. All axial force reactions are zero. Points are 2-ft apart. P = 20 kips 1 2 3 4 5 8 9 10 6 7 8 kips 12 kips 12 ft 20 ft Point 6 is just left of P and Point 7 is just right of P. P = 20 kips 1 2 3 4 5 8 9 10 6 7 8 kips 12 kips 12 ft 20 ft 8 kips V x (kips) -12 kips M (ft-kips) 16 32 48 64 80 96 72 48 24 x V & M Diagrams 8 kips P = 20 kips 12 kips 12 ft 20 ft 8 kips V x (kips) What is the slope of this line? 96 ft-kips/12’ = 8 kips M (ft-kips) -12 kips 96 ft-kips b What is the slope of this line? -12 kips a c x V & M Diagrams 8 kips P = 20 kips 12 kips 12 ft 20 ft 8 kips V x (kips) What is the area of the blue rectangle? 96 ft-kips -12 kips 96 ft-kips b What is the area of the green rectangle? -96 ft-kips M (ft-kips) a c x 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. The Relationship Between Load, Shear and Bending Moment w( x ) the load function M( x ) V( x )dx V( x ) w( x )dx Common Relationships 0 Constant Linear Constant Linear Parabolic Linear Parabolic Cubic Load Shear Moment Common Relationships Load 0 0 Constant Constant Constant Linear Linear Linear Parabolic M Shear Moment Example: Draw Shear & Moment diagrams for the following beam 12 kN 8 kN A C D B 1m RA = 7 kN 3m 1m RC = 13 kN 12 kN 8 kN A C D B 1m 3m 8 7 V 1m 8 7 -15 (kN) -5 7 M (kN-m) 2.4 m -8