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# 10.01.03.029

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### 10.01.03.029

1. 1. Powerpoint Templates Page 2
2. 2. BENDING MOMENT DIAGRAM BMD show how the applied loads to a beam create a moment variation along the length of the beam.
3. 3. Mt Mt Bending Moment Bending Moment = moments of reactions – moments of loads
4. 4. Distributed load acts downward on beam.
5. 5. Internal shear force causes a clockwise rotation of the beam segment; and the internal moment causes compression in the top fibers of the segment.
6. 6. SIGN CONVENTIONS • A force that tends to bend the beam downward is said to produce a positive bending moment. A force that tends to shear the left portion of Positive Bending Negative Bending the beam upward with respect to the right portion is said to produce a positive shearing force. Positive Shear Negative Shear
7. 7. PROCEDURE 1. Draw the free-body-diagram of the beam with sufficient room under it for the shear and moment diagrams (if needed, solve for support reactions first). 1. Draw the shear diagram under the free-body-diagram. • The change in shear ∆V equals the negative area under the distributed loading. • Label all the loads on the shear diagram ∆V = − ∫ w( x )dx
8. 8. PROCEDURE 3. Draw the moment diagram below the shear diagram. • The shear load is the slope of the moment and point moments result in jumps in the moment diagram. • The area under the shear diagram equals the change in moment over the segment considered (up to any jumps due to point moments). • Label the value of the moment at all important points on the moment diagram. ∆M = ∫ V ( x )dx
9. 9. Relations between Distributed Load, Shear and Moment Distributed Load Slope of the shear diagram Slope of the shear diagram dV = − w(x) dx dM =V dx Negative of distributed load intensity Shear moment diagram Change in shear ∆M BC = ∫ Vdx Area under shear diagram Change in moment ∆VBC = − ∫ w( x)dx Area under shear diagram
10. 10. Procedure for Analysis Support Reactions • • Find all reactive forces and couple moments acting on the beam Resolve them into components Shear and Moment Reactions • • • • Specify separate coordinates x Section the beam perpendicular to its axis V obtained by summing the forces perpendicular to the beam M obtained by summing moments about the sectioned end Shear and Moment Reactions • • Plot (V versus x) and (M versus x) Convenient to plot the shear and the bending moment diagrams below the FBD of the beam
11. 11. Shear & Moment Diagram
12. 12. Common Relationships 0 Constant Linear Constant Linear Parabolic Linear Parabolic Cubic Load Shear Moment
13. 13. Common Relationships 0 Constant Constant Constant Linear Linear Load 0 Linear Parabolic M Shear Moment
14. 14. Rules of Thumb/Review • Moment is dependent upon the shear diagram  the area under the shear diagram = change in the moment (i.e. A shear diagram = ΔM) • Straight lines on shear diagrams create sloping lines on moment diagrams • Sloping lines on shear diagrams create curves on moment diagrams • Positive shear = increasing slope • Negative shear = decreasing slope
15. 15. BEAMS IN BENDING P1 w P2 Ra P3 Rb Point of zero SFD Point of maximum BMD Point of contra-flexure Point of maximum
16. 16. Concentrated Load • Find reactions • Cut through beam to the left of the load P (a distance x from the left end), FBD – Equilibrium yields V and M for the left side of the beam • Cut through the beam to the right of P, FBD – Equilibrium yields V and M for the right side of the beam.
17. 17. Concentrated Load Moment Diagram Bending Moment Diagram • The bending moment in the left side increases linearly from zero at the support to P(ab/L) at the concentrated load x=a • In the right side, the bending moment is again a linear function of x, varying from P(ab/L) at x=a to zero at the support x=L. • The maximum bending moment is therefore P(ab/L), which occurs at the concentrated load. In this example a=b=L/2
18. 18. a b A R Ay = y P Example 1 : C P ⋅b ( a + b) R By = x P a ∑M z Mxz A P ⋅b ( a + b) Qxy =0 ; + Mxz + P ( x − a ) ∴ Mxz Where Pb ( x) = 0 − ( a + b) Pb ( x ) − P ( x − a) = ( a + b) ( x − a) x B can only be +VE or ZERO. P⋅a ( a + b)
19. 19. y P a b A Pb ( a + b) x B Pa ( a + b) x (i) When x ≤ a : (ii) When x > a : BMD: C Mxz = Mxz Mxz Pab ( a + b) Pb ( x ) − P ( x − a) ( a + b) 0 Pb ( x ) − P ( x − a) = ( a + b) Eq. 1 2 Eq. 2 +ve 0 A C 1 B
20. 20. Example 2: P y Mxz=P.L A x B L RAy=P P.L Mxz P Mxz Qxy Mxz Qxy Q & M are POSITIVE Qxy x P Mxz Qxy ∑F = 0 ; ∑M = 0 ; y z ∴ Q xy = P ∴ Mxz = −P( L − x )
21. 21. Q xy = P; Mxz = −P( L − x ) P L y P.L Mxz A B x Qxy x P P Qxy Mxz +ve Shear Force Diagram (SFD) 0 0 -P.L -ve Bending Moment Diagram (BMD)
22. 22. V & M Diagrams 8 kips 12 ft P = 20 kips 12 kips 20 ft 8 kips V x (kips) What is the area of the blue rectangle? 96 ft-kips M (ft-kips) -12 kips 96 ft-kips b What is the area of the green rectangle? -96 ft-kips a c x
23. 23. BMD for simply supported beam with UDL: Parabolic, max moment at mid span of value WL2/8 , where w is the distributed load and L the length of the beam.
24. 24. UNIFORM LOAD • The beam and its loading is symmetric, the reactions are equal to wL/2 • The slope of the shear diagram at each point equals the negative distributed load intensity at each point dV = −w( x ) dx
25. 25. UNIFORM LOAD • Therefore, the shear force and bending moment at a distance x from the left end are: dV dM =− ( x ), w =V dx dx • These equations are valid through the length of the beam and can be plotted as shear and bending moment diagrams.  The maximum value at the midpoint where dM = V = 0. dx  Mmax= wL2/8
26. 26. Distributed Load w per unit length Example : Distributed Load y x Mxz= wL2 A 2 B L x RAy=wL wL2 Mxz wx 2 Qxy wL Mxz Qxy ∑F = 0 ; ∑M = 0 ; y z wL − wx − Q xy = 0 ⇒ Q xy = w ( L − x ) wL2 x − wL ( x ) + wx   = 0 Mxz + 2 2
27. 27. ⇒ Q xy = w ( L − x ) ⇒ Mxz wx 2 wL2 = wLx − − 2 2 Mxz BMD: L 0 -ve x -wL2 2 wL2 =− 2 @ x = 0; Mxz @ x = L; Mxz = 0 L @x = ; 2 Mxz wL2 =− 8
28. 28. 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.