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# Shear Force And Bending Moment In Beams

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• Hi, Can anyone help me with writing 3 dimensional force balance of a curved beam in cylindrical coordinates? the forces are weight components, friction force which is a function of normal contact force, two tensile forces at both sides of beam. You can introduce reference also if you know any? thanks...

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### Shear Force And Bending Moment In Beams

1. 1. Mechanics of Materials Chaper4 Shear Force and Bending Moment in Beams HENAN UNIVERSITY OF SCIENCE & TECHNOLOGY Edited by YANG MIN-xian
2. 2. §4–1 Concepts of planar bending and calculation sketch of the beam §4–2 The shearing force and bending moment of the beam §4–3 The shearing-force and bending-moment equations · the shearing- force and bending-moment diagrams §4–4 Relations among the shearing force 、 the bending moment and the density of the distributed load and their applications §4–5 Plot the bending-moment diagram by the theorem of superpositiom §4–6 The internal-force diagrams of the planar rigid theorem frames CHAPTER 4 Shear Force and Bending Moment in Beams
3. 3. §4–1 CONCEPTS OF PLANAR BENDING AND CALCULATION SKETCH OF THE BEAM 1 、 CONCEPTS OF BENDING 1). BENDING : The action of the external force or external the couple vector perpendicular to the axis of the rod makes the axis of the rod change into curve from original straight lines, this deformation is called bending. 2).BEAM ： The member of which the deformation is mainly bending is generally called beam. Shear Force and Bending Moment in Beams
4. 4. 3). Practical examples in engineering about bending Shear Force and Bending Moment in Beams
5. 5. Shear Force and Bending Moment in Beams
6. 6. 4). Planar bending ： After deformation the curved axis of the beam is still in the same plane with the external forces. Symmetric bending （ as shown in the following figure ）— a special example of the planar bending. Shear Force and Bending Moment in Beams The plane of symmetry M P 1 P 2 q
7. 7. Unsymmetrical bending— if a beam does not possess any plane of symmetry, or the external forces do not act in a plane of symmetry of the beam with symmetric planes, this kind of bending is called unsymmetrical bending. In later chapters we will mainly discuss the bending stresses and deformations of the beam under symmetric bending. Shear Force and Bending Moment in Beams
8. 8. 2 、 Calculation sketch of the beam In general supports and external forces of the beam are very complex. We should do some necessary simplification for them for our convenient calculation and obtain the calculation sketch. 1). Simplification of the beams In general case we take the place of the beam by its axis. 2). Simplification of the loads The loads (including the reaction) acting on the beam may be reduced into three types ： concentrated force 、 concentrated force couple and distributed force. 3). Simplification of the supports Shear Force and Bending Moment in Beams
9. 9. ① Fixed hinged support 2 constraints ， 1 degree of freedom. Such as the fixed hinged support under bridges ， thrust ball bearing etc. ② Movable hinged support 1 constraint ， 2 degree of freedom. Such as the movable hinged support under the bridge ， ball bearing etc. Shear Force and Bending Moment in Beams
10. 10. ③ Rigidly fixed end 3 constraints ， 0 degree of freedom. Such as the support of diving board at the swimming pool ， support of the lower end of a wooden pole. 4) Three basis types of beams ① Simple beam(or simply supported beam) ② Cantilever beam Shear Force and Bending Moment in Beams A X A Y A M A M — Concentrated force couple q ( x ) — Distributed force
11. 11. ③ Overhanging beam 5). Statically determinate and statically indeterminate beams Statically determinate beams ： Reactions of the beam can be determined only by static equilibrium equations ， such as the above three kinds of basic beams. Statically indeterminate beams ： Reactions of the beam cannot be determined or only part of reactions can be determined by static equilibrium equations. Shear Force and Bending Moment in Beams — Concentrated force P q — Uniformly distributed force
12. 12. Example 1 A stock tank is shown in the figure. Its length is L =5m ， its inside diameter is D =1m ， thickness of its wall is t =10mm. Density of steel is 7.8g/cm³. Density of the liquid is 1g/cm³. Height of the liquid is 0.8m. Length of overhanging end is 1m. Try to determine the calculation sketch of the stock tank. Solution ： Shear Force and Bending Moment in Beams q — Uniformly Distributed force
13. 13. Shear Force and Bending Moment in Beams q — Uniformly Distributed force
14. 14. §4–2 THE SHEARING FORCE AND BENDING MOMENT OF THE BEAM 1 、 Internal force in bending ： Example Knowing conditions are P ， a ， l , as shown in the figure. Determine the internal forces on the section at the distance x to the end A . l A A B B Solution ：① Determine external forces Shear Force and Bending Moment in Beams P a P Y A X A R B
15. 15. ② Determine internal forces— method of section A Q M M Q Internal forces of the beam in bending 1). Bending moment ： M Moment of the internal force couple with the acting plane in the cross-section perpendicular to the section when the beam is bending. C C Shear Force and Bending Moment in Beams A B P Y A X A R B m m x Y A R B P Shearing force Bending moment
16. 16. 2). Shearing force ： Q Internal force which the acting line in the cross-section parallel to the section, when the beam is bending. 3) . Sign conventions for the internal forces: ① Shearing force Q : It is positive when it results in a clockwise rotation with respect to the object under consideration, otherwise it is negative. ② Bending moment M ： It is positive when it tends to bend the portion concave upwards, otherwise it is negative. Q (+) Q (–) Q (–) Q (+) M ( + ) M ( + ) M ( – ) M ( – ) Shear Force and Bending Moment in Beams
17. 17. Example 2 ： Determine the internal forces acting on sections 1—1 and 2—2 section as shown in fig.(a). Solution ： Determine internal forces by the method of section . Free body diagram of the left portion of section 1—1 is shown in fig. （ b ） . Fig. （ a ） 2 、 Examples Q 1 A M 1 Fig. （ b ） Shear Force and Bending Moment in Beams q qL a b 1 1 2 2 qL x 1
18. 18. Free body diagram of the left portion of section 2—2 is shown in fig. （ b ） . 图（ a ） q Q 2 B M 2 图（ c ） Shear Force and Bending Moment in Beams x y qL a b 1 1 2 2 qL x 2
19. 19. 1. Internal-force equations : Expressions that show the internal forces as functions of the position x of the section. . 2. The shearing-force and bending-moment diagrams: Shearing-force diagram sketch of the shearing-force equation Bending Moment diagram sketch of the bending-moment equation §4–3 THE SHEARING-FORCE AND BENDING-MOMENT EQUATIONS THE SHEARING-FORCE AND BENDING-MOMENT DIAGRAMS Shear Force and Bending Moment in Beams ) ( x Q Q  Shearing force equation ) ( x M M  Bending moment equation ) ( x Q Q  ) ( x M M 
20. 20. Example 3 Determine the internal-force equations and plot the diagrams of the beam shown in the following figure. Solution ：① Determine the reactions of the supports ② Write out the internal- force equations P ③ Plot the internal- force diagrams Q ( x ) M ( x ) x x P – PL Shear Force and Bending Moment in Beams Y O L M ( x ) x Q ( x ) M O ⊕ ○
21. 21. Solution ：① Write out the internal-force equations ② Plot the internal- force diagram L q Q ( x ) x – qL Shear Force and Bending Moment in Beams M ( x ) x Q ( x ) M ( x ) x ⊕ ○
22. 22. Solution ：① Determine the reactions of the supports ② Write out the internal- force equations q 0 R A ③ Plot the internal- force diagrams R B x Shear Force and Bending Moment in Beams ⊕ ⊕ L Q ( x ) x M ( x ) ○
23. 23. 1 、 Relations among the shearing force 、 the bending moment and the the distributed load By analysis of the equilibrium of the infinitesimal length d x ， we can get §4–4 RELATIONS AMANG THE SHEARING FORCE, THE BENDING MOMENT AND THE INDENSITY OF THE DISTRIBUTED LOAD AND THEIR APPLICATIONS q ( x ) q ( x ) M ( x )+d M ( x ) Q ( x )+d Q ( x ) Q ( x ) M ( x ) d x A y Shear Force and Bending Moment in Beams Slope of the tangential line at a point in the shearing-force diagram is equal to the intensity of the distributed load at the same point. d x x
24. 24. q ( x ) M ( x )+d M ( x ) Q ( x )+d Q ( x ) Q ( x ) M ( x ) d x A y Slope of the tangential line at a point in the bending-moment diagram is equal to the magnitude of the shearing force at the same point. Relation between the bending moment and the indensity of the distributed load ： Shear Force and Bending Moment in Beams
25. 25. 2 、 Relations between the shearing force 、 the bending moment and the external load External force No external-force segment Uniform-load segment Concentrated force Concentrated couple Characteristics of Q-diagram Characteristics of M-diagram Horizontal straight line Inclined straight line Sudden change from the left to right No change Inclined straight line curves Flex from the left to the right Sudden change from the left to the right Opposite tom Shear Force and Bending Moment in Beams q= 0 q> 0 q< 0 C P C m x Q Q> 0 Q Q< 0 x Increasing function x Q x Q Decreasing function x Q C Q 1 Q 2 Q 1 – Q 2 = P x Q C x M Increasing function x M Decreasing function x M Tomb-like x M Basin-like x M Flex opposite to P M x M 1 M 2
26. 26. Simple method to plot the diagram : The method to plot the diagrams by using the relation between the internal forces and the external forces and values of the internal forces at some special points. Example 4 Plot the internal force diagrams of the beams shown in the following figures by the simple method to plot the diagram. Solution : Special points : Plot the diagram by using the relation between the internal forces and the external forces and the internal force values at some special points of the beam . End point 、 partition point （ the point at which external forces changed ） and stationary point etc. Shear Force and Bending Moment in Beams a a qa q A
27. 27. Left end ： Shape of the curve is determined according to And the law of the point acted by concentrated force. Partition point A ： Stationary point of M ： Right end ： Q x x M Shear Force and Bending Moment in Beams a a qa q A ； ； qa 2 – qa –
28. 28. Example 5 Plot the internal-force diagrams of the beams shown in the following figures by the simple method to plot the diagram. Solution ： Determine reactions Left end A ： Right of point B ： Left of point C ： Stationary point of M ： Right of point C ： Right end D ： q qa 2 qa R A R D Q x qa/ 2 qa/ 2 qa/ 2 A B C D qa 2 /2 x M qa 2 /2 qa 2 /2 3 qa 2 /8 Left of point B ： Shear Force and Bending Moment in Beams – – + – +
29. 29. §4–5 PLOT THE DIAGRAM OF BENDING MOMENT BY THE THEOREM OF SUPERPOSITIOM 1 、 Theorem of superposition ： Internal forces in the structure due to simultaneous action of many forces are equal to algebraic sum of the internal forces due to separate action of each force. Applying condition ： Relation between the parameters (internal forces 、 stresses 、 displacements ） and the external forces must be linear, that is they satisfy Hooke’s law. Shear Force and Bending Moment in Beams
30. 30. 2 、 Structural members in mechanics of material is of small deformation and linear elasticity, and must obey this principle —— method of superposition Steps ： ① Plot respectively the diagram of the bending moment of the beam under the separate action of each external load ； ② Sum up the corresponding longitudinal coordinates (Attention: do not simply piece together figures. ） Shear Force and Bending Moment in Beams
31. 31. Example 6 Plot the diagram of bending moment by the principle of superposition. ( AB =2 a ， force P is acting at the middle point of the beam AB. ） P q P = + A A A B B B = + Shear Force and Bending Moment in Beams q x M 2 x M 1 x M + + +
32. 32. 3 、 Applications of symmetry and antisymmetry ： For the symmetric structure under the action of symmetric loads the diagram of its shearing stress Q is antisymmetric and the diagram of the bending moment M is symmetric. For the symmetric structure under the action of antisymmetric loads the diagram of its shearing stress Q is symmetric and the diagram of the bending moment M is antisymmetric. Shear Force and Bending Moment in Beams
33. 33. Example 7 Plot internal-force diagrams of the beams shown in the following figure. P PL PL 0.5 P 0.5 P 0.5 P 0.5 P P 0 0.5 P 0.5 P 0.5 P P Shear Force and Bending Moment in Beams P L L L L L L Q x Q 1 x Q 2 x – – + –
34. 34. P PL PL 0.5 P 0.5 P 0.5 P 0.5 P P 0 M x M 1 x M 2 x 0.5 PL PL 0.5 PL 0.5 PL Shear Force and Bending Moment in Beams P L L L L L L – + + +
35. 35. Example 8 Correct the mistakes in the following internal-force diagrams. a 2 a a q qa 2 A B Q x x M qa/ 4 qa/ 4 3 qa/ 4 7 qa/ 4 qa 2 /4 49 qa 2 /32 3 qa 2 /2 5 qa 2 /4 Shear Force and Bending Moment in Beams R A R B – – + +
36. 36. Example 9 Knowing Q-diagram, determine external loads and M-diagram (Therefore no concentrated force couples acted on the beam). M (kN·m) Shear Force and Bending Moment in Beams Q (kN) x 1m 1m 2m 2 3 1 5kN 1kN q =2kN/m + – + x + 1 1 1.25 –
37. 37. §4–6 THE INTERNAL-FORCE DIAGRAMS OF THE PLANAR RIGID FRAMES 1 、 Planar rigid frame 1). Planar rigid frame ： Structure made from rods of different direction that are mutually connected in rigidity at their ends in the same plane. Characteristics ： There are internal forces Q, M and N in each rod . 2). Conventions to plot diagram of internal forces ： Bending-moment diagram ： Plot it at the side where fibers are elongated and not mark the sign of positive or negative. Shearing-force and axial-force diagrams ： May be plotted at any side of the frame （ In common the diagram with positive value is plotted outside the frame ）， but must mark the signs of positive and negative. Shear Force and Bending Moment in Beams
38. 38. Example 10 Try to plot the internal-force diagrams of the rigid frame shown in the figure. P 1 P 2 a l A B C N- diagram Q - diagram P 1 M - diagram Shear Force and Bending Moment in Beams – P 2 + P 1 + P 1 a P 1 a P 1 a+ P 2 l
39. 39. 1 、 Method to determine directly the internal forces ： When we determine the internal forces in an arbitrary section A , we can take the left part of section A as our study object and use the following formulas to calculate internal forces. where P i and P j are respectively upward and downward external forces acted on the left part. DIAGRAMS OF SHEARING STRESSES AND BENDING MOMENTS EXERCISE LESSONS ABOUT INTERNAL FORCES OF BENDING Shear Force and Bending Moment in Beams
40. 40. Relations among the shearing force 、 the bending moment and the external load ： 2 、 Simple method to plot the diagram : The method to plot the diagrams by using the relation between the internal forces and the external forces and using values of the internal forces at some special points. Shear Force and Bending Moment in Beams q ( x )
41. 41. 3 、 Principle of superposition ： Internal forces in the structure due to simultaneous action of many forces are equal to the algebra sum of the internal forces due to separate action of each force. 4 、 Applications of symmetry and antisymmetry ： For the symmetric structure under the action of symmetric loads the diagram of its shearing stress is antisymmetric and the diagram of bending moment is symmetric. For the symmetric structure under the action of antisymmetric loads the diagram of its shearing stress is symmetric and the diagram of bending moment is antisymmetric Shear Force and Bending Moment in Beams
42. 42. Shear Force and Bending Moment in Beams 5 、 Relations between the shearing force, the bending moment and the external load External force No external-force segment Uniform-load segment Concentrated force Concentrated couple Characteristics of Q-diagram Horizontal straight line Inclined straight line Sudden change from the left to right No change Inclined straight line curves Flex from the left to the right Sudden change from the left to the right Opposite tom q= 0 q> 0 q< 0 C P C m x Q Q> 0 Q Q< 0 x Increasing function x Q x Q Decreasing function x Q C Q 1 Q 2 Q 1 – Q 2 = P x Q C x M Increasing function x M Decreasing function x M Tomb-like x M Basin-like x M Flex opposite to P M x M 1 M 2
43. 43. Example 1 Plot the bending-moment diagrams of the beam shown in the following figure . (1) Shear Force and Bending Moment in Beams 2 P a a P = 2 P P + x M x M 1 x M 2 = + – + + 2 Pa 2 Pa Pa
44. 44. (2) q q q q = + = + 3 qa 2 /2 qa 2 /2 qa 2 Shear Force and Bending Moment in Beams a a x M 1 x M – + – x M 2
45. 45. (3) PL /2 = + = + PL /2 PL /4 PL /2 PL /2 Shear Force and Bending Moment in Beams P L/2 L/2 P x M 2 x M x M 1 – + –
46. 46. (4) 50kN 20kNm = + = + 20kNm 50kNm 20kNm 20kNm 20kNm 20kNm 30kNm 20kNm Shear Force and Bending Moment in Beams 2m 2m x M 2 x M x M 1 50kN + + –
47. 47. y z h b Solution : （ 1 ） Shearing stress on the cross section is Example 2 The structure is shown in the figure. Try to prove ： ( 1 ） resultant of the shearing stresses in an arbitrary cross section is equal to the shearing force in the same section ； （ 2 ） Resultant moment of the normal stresses in an arbitrary cross section is equal to the bending moment in the same section ； （ 3 ） which force can balance the resultant of the shearing stress in the longitudinal section at middle height balanced ?. q Shear Force and Bending Moment in Beams Normal stress on the cross section is
48. 48. (2) Resultant shearing force in the cross section is ： (3 ) Resultant force couple Shear Force and Bending Moment in Beams
49. 49. (4) Shearing stress in the middle longitudinal section is ： Resultant of the shearing stress in the longitudinal section is balanced by resultant of the normal stress in the right-side section. (5) Resultant of the shearing stress in the longitudinal section is ： Shear Force and Bending Moment in Beams x L  max  
50. 50. Thanks a lot Shear Force and Bending Moment in Beams