Beams Shear force & bending moment A beam is defined as a structural member designed primarily to support forces acting perpendicular to the axis of the member. The principal difference between beams and the axially loaded bars and torsionally loaded shafts is in the direction of the applied load . Definition of beam :-

Beams:

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

Types of Beams:

Types of Beams Depends on the support configuration M F v F H Fixed F V F V F H Pin Roller Pin Roller F V F V F H

Statically Indeterminate Beams:

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

Internal Reactions in Beams:

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 a b

Internal Reactions in Beams:

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!!!

Internal Reactions in Beams:

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!!!

Finding Internal Reactions:

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.:

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 2 3 9 4 5 8 Point 6 is just left of P and Point 7 is just right of P.

PowerPoint Presentation:

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

PowerPoint Presentation:

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

PowerPoint Presentation:

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

Draw Some Conclusions:

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.

PowerPoint Presentation:

The Relationship Between Load, Shear and Bending Moment

PowerPoint Presentation:

Load 0 Constant Linear Shear Constant Linear Parabolic Moment Linear Parabolic Cubic Common Relationships

PowerPoint Presentation:

Load 0 0 Constant Shear Constant Constant Linear Moment Linear Linear Parabolic Common Relationships M

Example: Draw Shear & Moment diagrams for the following beam:

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

PowerPoint Presentation:

3 m 1 m 1 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

PowerPoint Presentation:

Atrium Structure Excavator Car Chassis Yacht Examples of Devices under Bending Loading

PowerPoint Presentation:

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