Structure I Course Code: ARCH 208 Dr. Aeid A. Abdulrazeg

Beams Shear & Moment Diagrams

Contents Introduction Internal Forces in Members Various Types of Beam Loading and Support Shear and Bending Moment in a Beam Relations Among Load, Shear, and Bending Moment

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

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

Statically Indeterminate Beams Continuous Beam Propped Cantilever Beam

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

Beam Shear Magnitude (V) = sum of vertical forces on either side of the section can be determined at any section along the length of the beam Upward forces (reactions) = positive Downward forces (loads) = negative Vertical Shear = reactions – loads

Bending Moment Bending moment: tendency of a beam to bend due to forces acting on it Magnitude (M) = sum of moments of forces on either side of the section can be determined at any section along the length of the beam Bending Moment = moments of reactions – moments of loads

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!!! V M N Left Side of Cut Pb/L x

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!!! V M N Right Side of Cut 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.

Point 6 is just left of P and Point 7 is just right of P. 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.

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

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

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

Equation (1) states that the slope of the shear diagram at a point (dV/dx) is equal to the intensity of the distributed load w(x) at the point. Equation (2) states that the slope of the moment diagram (dM/dx) is equal to the intensity of the shear at the point. Equ. (1) Equ. (2)

By using integration Equation (3) states that the change in the shear between any two points on a beam equals the area under the distributed loading diagram between the points. Equation (4) states that the change in the moment between the two points equals the area under the shear diagram between the points Equ. (3) Equ. (4)

Common Relationships Load Constant Linear Shear Parabolic Moment Cubic

Common Relationships Load Constant Shear Linear Moment Parabolic M

Example Determine the shear and moment distributions produced in the simple beam by the 4-kN concentrated load.

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 

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

Example Draw the shear-force and bending-moment diagrams for the loaded beam and determine the maximum moment M and its location x from the left end. 2 kN/m 3 m 1.5 m 1.5 m

Example Draw the shear-force and bending-moment diagrams for the loaded beam and determine the maximum moment M and its location x from the left end.