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- 1. CE 255 Theory of Structures Dr.-Ing. J. W. Ampofo
- 2. CE 255 β THEORY OF STRUCTURES Chapter 1- The Conjugate Beam Method Proposed by Professor Mueller-Breslau in 1865, the Conjugate Beam Method is a slightly modified form of the Moment-Area Method (Chapter 2). Conjugate Beam is defined as the imaginary beam with the same dimensions as that of the original beam but load at any point on the conjugate beam is equal to the bending moment at that point of the original beam divided by EI. The Conjugate Beam Method is an engineering method used to find the slope and deflection of beams.
- 3. There are two theorems related to the conjugate beam Theorem 1 The slope at a point in the real beam is equal to the shear at the corresponding point in the conjugate beam. Theorem 2 The displacement of a point in the real beam is equal to the moment at the corresponding point in the conjugate beam. The two theorems follow from the equations of internal loading and beam theory. If the intensity of loading on a beam is w, V the shear force and M the bending moment then ππ ππ₯ = π€ πππ π = β« π€ ππ₯ (1.1) But π = ππ ππ π2 π ππ₯2 = π€ πππ π = β« (β« π€ππ₯) ππ₯ (1.2) If π πππ π¦ are slope and deflection at a point on the beam, then from the beam theory ππ ππ₯ = π πΈπΌ πππ π = β« π πΈπΌ ππ₯ (1.3) But π = ππ¦ ππ₯ π2 π¦ ππ₯2 = π πΈπΌ πππ π¦ = β« [β« π πΈπΌ ππ₯] ππ₯ (1.4)
- 4. Comparing equations (1.1) and (1.3) π = β« π€ππ₯ π = β« π πΈπΌ ππ₯ We see that if the load w is replaced by π πΈπΌ , then the shear V corresponds to the slope π. Similarly, comparing eqns. (1.2) and (1.4) π = β« (β« π€ππ₯) ππ₯ π¦ = β« [β« π πΈπΌ ππ₯] ππ₯ We see that if the w is replaced with π πΈπΌ then the bending moment corresponds to the deflection
- 5. Conjugate Beam Supports When the conjugate beam is drawn, it is important that the shear and moment developed in the conjugate beam correspond to the slope and displacement conditions in the real beam.
- 6. PROCEDURE FOR ANALYSIS 1. Draw the Bending Moment Diagram for the given beam. 2. Construct the conjugate beam with the π πΈπΌ loading, Remember that when the π πΈπΌ diagram is positive the loading is upwards and when the π πΈπΌ diagram is negative, the loading is downwards. 3. Use the equations of equilibrium to solve for the reactions of the conjugate beam. 4. Solve for the shear and moment at the points or points where the slope and displacement are desired. If the values are positive, the slope is counter clockwise and the displacement is upward. Example of how a real beam is transformed into a conjugate beam
- 7. Area Properties The area properties are well known for triangular and rectangular areas. For parabolic curves, we have
- 8. Example 1.1 Determine the slope and deflection at B of the cantilever below EI=constant Solution First draw the bending moment diagram. We note that for this case. The diagram can be drawn without first calculating the reactions Next, draw the conjugate beam and load it with the π πΈπΌ diagram, remembering that loading is downward because the π πΈπΌ diagram is negative.
- 9. Cut a section through the beam near B and consider the free body diagram to the left of the section. For the vertical equilibrium of forces, we have π + 1 2 β ππΏ πΈπΌ β πΏ = 0 π = β ππΏ2 2πΈπΌ The rotation ππ΅ of the real beam ππ΅ = β ππΏ2 2πΈπΌ The minus sign means that it is a clockwise rotation. Taking moments about a point at the section
- 10. π + 1 2 β ππΏ πΈπΌ β πΏ β 2 3 πΏ = 0 π = β ππΏ3 3πΈπΌ The deflection π¦π΅ = β ππΏ3 3πΈπΌ The negative sign indicates that deflection is downward. Example 1.2 Calculate the maximum deflection of the simply supported beam loaded with udl of intensity w. Solution The determination of reactions and drawing the bending moment diagram is trivial and left as an exercise for the student
- 11. ο· Construct and load the conjugate beam with the π πΈπΌ diagram ο· Calculation of Reactions We note that the loading is symmetrical and therefore each support will take half of the total load. The total load is equal to the area under the parabola. π π΄ = π π΅ = 1 2 β 2 3 β πΏ β π€πΏ2 8πΈπΌ = π€πΏ3 24πΈπΌ For the given system, we know that the maximum deflection occurs at mid-span. We therefore make a section at the mid-span of the beam and compute the initial moment.
- 12. π₯Μ = 3 8 β πΏ 2 = 3πΏ 16 Taking moments at a point on the section, we have π + ππΏ3 24πΈπΌ β πΏ 2 β ππΏ3 24πΈπΌ β 3πΏ 16 = 0 π = ππΏ3 24πΈπΌ ( 3πΏ 16 β πΏ 2 ) = β 5ππΏ4 384πΈπΌ Max deflection π¦πππ₯ = 5ππΏ4 384πΈπΌ (πππ€ππ€πππ)
- 13. Example 1.3 For the simply supported beam below, determine the deflection at C and the maximum deflection. Solution The maximum moment occurs under the load, with a value π = 2 Γ 6 Γ 3 6+3 = 4πππ (Student should confirm this by calculating reactions) Conjugate Beam and Loading Reactions
- 14. β ππ΄ = 0 9π π΅ = 1 πΈπΌ ( 1 2 β 4 β 6 β 4) + 1 πΈπΌ ( 1 2 β 4 β 3 β 7) = 90 πΈπΌ π π΅ = 10 πΈπΌ π π΄ = 1 πΈπΌ ( 1 2 β 4 β 9) β 10 πΈπΌ = 8 πΈπΌ Bending moment at C β ππΆ = 0 ππΆ + 8 πΈπΌ β 6 β 1 πΈπΌ ( 1 2 β 4 β 6) β 2 = 0 ππΆ = β 48 πΈπΌ + 24 πΈπΌ = β 24 πΈπΌ π·ππππππ‘πππ ππ‘ πΆ = 24 πΈπΌ (downward)
- 15. Maximum Deflection We note that the maximum deflection ο· Occurs near mid-span ο· Occurs where the slope of the elastic curve or shear force is zero The maximum deflection occurs between A and C, therefore we calculate the shear force at a section x from support A. At a section x metres from A, The intensity of load y can be calculates using similar triangles π¦ π₯ = 4 6 β π¦ = 2 3 π₯ Summing vertical forces, π(π₯) + 8 πΈπΌ β 1 πΈπΌ β 1 2 π₯ β 2 3 π₯ = 0 π(π₯) + 8 πΈπΌ β π₯2 3πΈπΌ = 0 π(π₯) = β 8 πΈπΌ + π₯2 3πΈπΌ
- 16. At maximum deflection, π(π₯) = 0 π₯2 3πΈπΌ = 8 πΈπΌ π₯2 = 24 β π₯ = 4.899π ππππ₯ = 8 πΈπΌ β 4.899 + 1 πΈπΌ β 1 2 β 4.899 6 β 4 β 4.899 β 4.899 3 = β26.13 πΈπΌ πππ₯πππ’π πππππππ‘πππ ππ 26.13 πΈπΌ πππ ππππ’ππ 4.899π ππππ π΄ BEAMS OF VARIABLE RIGIDITY The conjugate beam can be also be used to calculate the slope and deflection of beam of variable rigidity. This is illustrated in the example below. Example 1.4 Calculate the slope and deflection at the free end of the cantilever beam of Example 1.1 if the moment of inertia (2nd moment of area) varies as shown below.
- 17. Solution The fact that the beam has variable rigidity has no effect on the bending moment diagram. We note that, using similar triangles, the moment at C is -PL/2 When loading the conjugate beam, the moments between A and C are divided by 1 2πΈπΌ and those between C and B are divided by 1 πΈπΌ . This gives the following loaded conjugate beam. The slope at B is the shear force at B, the total load on the beam. ππ΅ = (π΄πππ ππ π‘πππππ§ππ’π πππ‘π€πππ π΄ πππ πΆ + π΄πππ ππ π‘πππππππ πππ‘π€πππ πΆ πππ π΅)
- 18. = β [ 1 2 ( ππΏ 2πΈπΌ + ππΏ 4πΈπΌ ) β πΏ 2 + 1 2 β ππΏ 2πΈπΌ β πΏ 2 ] = β 3ππΏ3 16πΈπΌ β ππΏ3 8πΈπΌ = β 5ππΏ3 16πΈπΌ π¦π΅ = β(ππππππ‘ ππ π‘πππππ§πππππ ππππ + ππππππ‘ ππ π‘ππππππ’πππ ππππ) ππππ’π‘ π πππππ‘ ππ π π πππ‘πππ π‘βπππ’πβ π΅ Since the centroid of the trapezium is not known the trapezium can be divided into a. 2 triangles b. A triangle and a rectangle And their moments combined β the principle of superposition Moment of trapezoidal load using 2 traingles Trapezium is divided into an upper triangle of area 1 2 β ππΏ 2πΈπΌ β πΏ 2 and the distance of its centroid from B is 2 3 β πΏ 2 + πΏ 2
- 19. Theref6re πππ’ = β 1 2 β ππΏ 2πΈπΌ β πΏ 2 ( 2 3 β πΏ 2 + πΏ 2 ) = β5ππΏ3 48πΈπΌ And a lower triangle of area 1 2 β πΏ 2 β ππΏ 4πΈπΌ with distance of centroid from B 1 3 β πΏ 2 + πΏ 2 Therefore πππΏ = β 1 2 β ππΏ 4πΈπΌ β πΏ 2 ( 1 3 β πΏ 2 + πΏ 2 ) = βππΏ3 24πΈπΌ Total moment ππ = πππ’ + πππΏ = β5ππΏ3 48πΈπΌ β ππΏ3 24πΈπΌ = β7ππΏ3 48πΈπΌ Alternatively we may calculate the moment of the trapezoidal load using a rectangle and a triangle Following the same procedure as before πππ’ = β 1 2 β ππΏ 4πΈπΌ β πΏ 2 ( 2 3 β πΏ 2 + πΏ 2 ) = β5ππΏ3 96πΈπΌ
- 20. πππΏ = β ππΏ 4πΈπΌ β πΏ 2 ( 1 2 β πΏ 2 + πΏ 2 ) = β3ππΏ3 32πΈπΌ Total Moment ππ = πππ’ + πππΏ = β5ππΏ3 96πΈπΌ β 3ππΏ3 32πΈπΌ = β14ππΏ3 96πΈπΌ β7ππΏ3 48πΈπΌ As we see, two methods yield the same results. Finally we calculate the moment of the triangular load between C and B about B. π2 = β 1 2 β ππΏ 2πΈπΌ β πΏ 2 β 2 3 β πΏ 2 = β ππΏ3 24πΈπΌ
- 21. Deflection at B π¦π΅ = π1 + π2 = β7ππΏ3 48πΈπΌ β ππΏ3 24πΈπΌ = β9ππΏ3 48πΈπΌ
- 22. EXERCISES P1.1 Determine the slope and deflection at the free end of the cantilever beam below P1.2 For the beam below, calculate the maximum deflection within the span A-B P1.3 Calculate the rotation just to the left and just to the right of B. Compute the deflection at D
- 23. CHAPTER 2 THE MOMENT-AREA METHOD The Moment-Area Method was developed by Otto Mohr in 1868 and presents powerful tool for finding the deflection of structures subjected primarily to bending. Its ease of finding deflections of statically determinate structures makes it ideal for solving statically indeterminate structures, using compatibility of displacements. THEORETICAL BASIS We consider the length AB of abeam in its undeformed state (page 22). From the diagram, we note. 1. AB is the original unloaded length of the beam and AβBβ is deflected position of AB when loaded. 2. The angle subtended at the centre of arc AβOBβ is Σ¨ and is the change in slope from Aβ to Bβ 3. PQ is a very short length of the beam, measured as dS along the curve and dx along the x-axis 4. dΣ¨ is the change in slope from P to Q and is also the angle subtended at the centre of the arc dS. 5. M is the average bending moment over the portion dx between P and Q. 6. The distance β is known as the vertical intercept and is the distance from Bβ to the produced tangent to the curve at Aβ which crosses under Bβ at C. It is measured perpendicular to the undeformed neutral axis.
- 25. MOHRβS FIRST THEOREM (Mohr 1) Noting that the angles are small and measured in radians, we have ππ = π β ππ π = ππ ππ From the Euler-Bernoulli Theory of Bending, we know 1 π = π πΈπΌ Hence ππ = π πΈπΌ ππ But for small deflections, the chord and arc length are similar, i.e. ππ β ππ₯, giving ππ = π πΈπΌ ππ₯ The total change in slope between A and B is thus β« ππ π΅ π΄ = β« π πΈπΌ ππ₯ π΅ π΄ πππ΅π΄ = ππ΅ β ππ΄ = β« π πΈπΌ π΅ π΄ ππ₯ [πΆβππππ ππ π ππππ]π΄π΅ = [π΄πππ ππ π πΈπΌ ππ’πππππ] π΄π΅ Usually the beam is prismatic and so E and I do not change over the length AB, whereas the bending moment M will change thus ππ΄π΅ = 1 πΈπΌ β« π ππ₯ π΅ π΄
- 26. [πΆβππππ ππ π ππππ]π΄π΅ = [π΄πππ ππ π πΈπΌ ππ’πππππ] π΄π΅ πΈπΌ This is Mohrβs First theorem (Mohr 1) ο· The change in slope between any two points on an elastic curve is equal to the area of the bending moment diagram between the points divided by EI. MOHRβS SECOND THEOREM (Mohr II) From the diagram we can see that πβ= π₯ β ππ But ππ = π πΈπΌ ππ₯ Therefore πβ= π πΈπΌ β π₯ β ππ₯ And for the portion AB, we have β« πβ π΅ π΄ = β« π πΈπΌ β π₯ β ππ₯ π΅ π΄ βπ΅π΄= [β« π πΈπΌ β ππ₯ π΅ π΄ ] π₯Μ = πΉπππ π‘ ππππππ‘ ππ π πΈπΌ πππππππ ππππ’π‘ π΅ This can be interpreted as [ ππππ‘ππππ πΌππ‘ππππππ‘ ] π΅π΄ = [ π΄πππ ππ π πΈπΌ ππ’πππππ ] π΅π΄ β [ π·ππ π‘ππππ ππππ π΅ π‘π πππ‘ππππ ππ ( π πΈπΌ ) π΅π΄ πππππππ ]
- 27. This is Mohrβs second theorem and may be stated as: ο· For an originally straight beam subjected to bending moment, the vertical intercept between one terminal and the tangent to the curve of another terminal is the first moment of the π πΈπΌ diagram about the terminal where the intercept is measured. Two crucial things must be noted from this definition; 1. Vertical intercept is not deflection. It is the distance from the deformed position of the beam to the tangent of the deformed shape of the beam at another location Thus ββ πΏ 2. The moment of the curvature diagram must e taken about the point where the vertical intercept is required. That is βπ΅π΄β βπ΄π΅
- 28. Example 2.1 For the cantilever beam shown, find the slope and deflection at B Solution
- 29. For Mohr 1, [πΆβππππ ππ π ππππ]π΄π΅ = [π΄πππ ππ π πΈπΌ ππ’πππππ] π΄π΅ ππ΅ β ππ΄ = 1 2 β πΏ β ππΏ πΈπΌ But ππ΄ = 0 Therefore ππ΅ = ππΏ2 2πΈπΌ In this case, because the tangent from A is horizontal, the vertical intercept βπ΅π΄ is equal to the deflection S at B Since the intercept is measured at B, the first moment of the π πΈπΌ diagram must be taken from this point
- 30. Thus from Mohr II, we have βπ΅π΄= ( 1 2 β πΏ β ππΏ πΈπΌ ) β 2πΏ 3 And so the deflection at B is πΏπ΅ = ππΏ3 3πΈπΌ Example 2.2 For the following simply supported beam, calculate the slope at A using Mohrβs theorems Solution In the Moment-Area Method, the deflected shape diagram is used to identify relationships between vertical intercepts and slopes
- 31. The key to the solution here is that we calculate βπ΅π΄ using Mohr II but from the diagram we see that we can use the formula π = π π for small angles βπ΅π΄= πΏππ΄ Once we know βπ΅π΄ from Mohr II, we can find ππ΄ = βπ΅π΄ πΏ β To calculate βπ΅π΄ using Mohr II, we need the π πΈπΌ diagram
- 32. π = 2 3 β πΏ 3 = 2πΏ 9 π = πΏ 3 + 1 2 ( 2πΏ 3 ) = 5πΏ 9 Thus from Mohr II βπ΅π΄= 1 2 β 2πΏ 3 β 2ππΏ 9πΈπΌ β 5πΏ 9 + 1 2 β πΏ 3 β 2ππΏ 9πΈπΌ β 2πΏ 9 = 10ππΏ3 243πΈπΌ + 2ππΏ3 243πΈπΌ = 12ππΏ3 243πΈπΌ = 4ππΏ3 81πΈπΌ But βπ΅π΄= πΏππ΄ ππ΄ = βπ΅π΄ πΏ β = 4ππΏ3 81πΈπΌ
- 33. GENERAL PROCEDURE FOR FINDING DEFLECTIONS To find deflection at any location x from a support, use the following relationships between slopes and vertical intercepts. 1. Use Mohr II to find the slope at the support 2. For the location x and from the diagram we have πΏπ₯ = π₯ β ππ΅ β βππ΅ MAXIMUM DEFLECTION We know that the maximum deflection occurs at a slope π = 0
- 34. To find where the slope is zero 1. Calculate slope at some point, say support A, using Mohr II 2. Using Mohr I determine at what distance from the point of known slope (ππ΄) the change in slope (Mohr I), πππ΄ equals the known slope, (ππ΄) 3. This is the point of maximum deflection since ππ΄ β πππ΄ = ππ΄ β ππ΄ = 0 Example 2.3 For the following beam of constant EI a. Determine ππ΄, ππ΅ and πΏπΆ b. What is the maximum deflection and where is it located?
- 35. The first step is to draw the bending moment diagram and draw the deflected shape diagram with slopes and tangents as indicated. Slopes at A and B πΈπΌβπ΄π΅= ( 2 3 β 2) ( 1 2 β 2 β 53.4) + (2 + 4 3 ) ( 1 2 β 4 β 53.4) = 53.4 ( 4 3 + 20 3 ) = 427.2 Therefore βπ΄π΅= 427.2 πΈπΌ β But βπ΄π΅= 6ππ΅ 6ππ΅ = 427.2 πΈπΌ Therefore ππ΅ = 71.2 πΈπΌ Similarly for the slope at A πΈπΌβπ΅π΄= ( 2 3 β 4) ( 1 2 β 4 β 53.4) + (4 + 1 3 β 2) ( 1 2 β 2 β 53.4) = 53.4 ( 16 3 + 14 3 ) = 534 βπ΅π΄= 534 πΈπΌ β
- 36. But βπ΅π΄= 6ππ΄ Therefore ππ΄ = 89.2 πΈπΌ Deflection at C To find the deflection at C, we use the vertical intercept βπΆπ΅ and ππ΅ From the diagram above πΏπΆ = 4ππ΅ β βπΆπ΅ From the bending moment diagram and slope at B πΈπΌπΏπΆ = 4 β 71.2 β ( 1 2 β 4 β 53.4) ( 4 3 ) = 142.4 Therefore πΏπΆ = 142.4 πΈπΌ
- 37. MAXIMUM DEFLECTION The first step in finding the maximum deflection is to locate its position. We know two things; 1. Maximum deflection occurs where there is zero slope 2. Maximum deflection is always close to the centre of span Based on these facts, we work with Mohr I to find the point of zero slope which will be located between B and C, as follows. πΆβπππππ πππ‘ππ‘πππ = ππ΅ β 0 = ππ΅ Since we know that the change in slope is also the area of the π πΈπΌ diagram we need to find the point x where the area of the π πΈπΌ diagram is equal to ππ΅ πβπ’π πΈπΌ(ππ΅ β 0) = (53.4 β π₯ 4 ) β 1 2 β π₯ πβπ’π πΈπΌ(ππ΅) = (53.4 β π₯2 8 ) π΅π’π‘ ππ΅ = 71.2 πΈπΌ , βππππ
- 38. πΈπΌ ( 71.2 πΈπΌ ) = 53.4 β π₯2 8 π₯2 = 10.66 π₯ = 3.265π ππππ π΅ πΏπππ₯ = π₯ππ΅ β βππ΅ πΈπΌπΏπππ₯ = 3.265 β 71.2 β (53.4 β 3.2652 8 ) β 3.265 8 = 155 πΏπππ₯ = 155 πΈπΌ
- 39. APPLICATION TO STATICALLY INDETERMINATE STRUCTURES ο· Moment-Area Method for Built-in-Beams A beam is said to be built in or encastre when both ends are rigidly fixed so that the slope at the ends remain horizontal It follows that, if EI is constant ο· Change in slope from end to end is zero and Mohr I gives β π΄π = 0 ο· The vertical intercept are zero at both ends and Mohr II gives β π΄π π₯Μ π = 0 Where π΄π are the areas under the bending moment diagram and π₯Μ π the distances of the centroids to the end where the intercepts are being measured. It is convenient to show the bending moment diagram to any load such as below as the algebraic sum of 2 parts β one due to loads, treating the beam as simply supported and the other due to end moments, introduced to bring the slope back to zero.
- 40. Mohr I states that the sum of the areas, positive and negative is zero π΄1 β π΄2 = 0 β π΄1 = π΄2 And Mohr II gives π΄1π₯Μ 1 = π΄2π₯Μ 2 These equations can be used to solve for the unknown fixed-end moments ππ΄ and ππ΅. We illustrate this by the following examples.
- 41. Example 2.4 For the built-in beam shown, determine the end moments and the deflection under the load (maximum deflection) Solution The bending moment diagram is of the form Due to symmetry, ππ΄ = ππ΅ = π Draw the free moment and fixed moment diagrams
- 42. From Mohr I, the area of the free moment diagram is equal to that of the fixed moment diagram πβπ’π πΏ 2 β ππΏ 4 = π β πΏ π‘βπππππππ π = ππΏ 8 Since ππ΄ = ππ΅ = π , application of Mohr II s redundant, However for unsymmetrical loading, ππ΄ β ππ΅ and a second equation is obtained from Mohr II to solve for the two unknowns ππ΄, ππ΅
- 43. Deflection Taking moments at C where the intercepts is measured πΈπΌβπΆπ΄= ( 1 2 β πΏ 2 β ππΏ 4 ) β 1 3 β πΏ 2 β ( πΏ 2 β ππΏ 8 ) β 1 2 β πΏ 2 = ππΏ3 96 β ππΏ3 64 = β ππΏ3 192 πΏπππ₯ = βπΆπ΄= β ππΏ3 192πΈπΌ Note that the same results are obtained measuring the intercept βπ΄πΆ
- 44. APPLICATION TO INDETERMINATE STRUCTURES β USE OF PRINCIPLE OF SUPERPOSITION AND DISPLACEMENT COMPATIBILITY Generally, we can use the principle of superposition to separate indeterminate structures into primary and reactant structures. For the built-in beams of the previous section, the primary structures is the statically determinate simply-supported beam with the given load (Free moment) and the reactant structure is the simply-supported beam loaded with the moment reactions (Fixed Moments). ο· For these structures, we can calculate the deflections at a point for which the deflection is known for the original structure. ο· We then use compatibility of displacement to equate the two calculated deflections to the known deflection of the original structure. ο· Doing so will yield the redundant reaction chosen for the reactant structure. Once this is known, all other load effects (bending moment, shear force, deflections and rotations) can be calculated. We illustrate this in the following example: Example 2.5 For the propped cantilever shown, determine the maximum deflection.
- 45. Solution The system is statically indeterminate to the first degree so one of the reactions must ne release to obtain a statically- determinate primary structure. There are 2 possibilities β either the moment at A or the reaction at B. Letβs settle for the latter. πΉππππ = πππππππ¦ + π ππππ‘πππ‘ π π΅ ππ π‘βπ π£πππ’π ππ π‘βπ πβππ ππ ππππ’πππππ‘ In the final structure we know that the deflection at B,πΏπ΅, must be zero as it is a roller support. Therefore from the bending moment diagram that results from the superposition of the primary and reactant structures, we calculate πΏπ΅ in terms of RB and solve π πππΏπ΅ = 0Ξ΄ From the deflected shape and bending moment diagram of the primary structure. βπ΅π΄= πΏπ΅ π = 1 πΈπΌ β 1 3 (β256) β 8 β ( 3 4 β 8) = β 4096 πΈπΌ and from those of the reactant structure βπ΅π΄= πΏπ΅ π = 1 πΈπΌ β 1 2 β 8π π΅ β (β256) β 8 β ( 2 3 β 8) = 512 3πΈπΌ π π΅
- 46. The displacement compatibility at B gives πΏπ΅ = πΏπ΅ π + πΏπ΅ π = β 4096 πΈπΌ + 512 3πΈπΌ π π΅ = 0 512 3πΈπΌ π π΅ = 4096 π π΅ = 24 ππ With π π΅ known, we can draw the final bending moment diagram by adding the BMDs of the primary and reactant structures LOCATION OF THE MAXIMUM DEFLECTION Assume the maximum deflection is at a location x from support B. The slope at the maximum deflection is zero. This means that between B and x, the slope changes by
- 47. ππ΅ and therefore by Mohr I, the area of the π πΈπ diagram between πΏπππ₯ and B is equal to ππ΅. To calculate the required area of the π πΈπ diagram, it is convenient to use the primary and reactant diagram rather than the final diagram. πΈπΌππ΅π = β 1 3 β 4π₯2 β π₯ + 1 2 β 24π₯ β π₯ = β 4 3 π₯3 + 12π₯2 πΈπΌππ΅ = β 1 3 β 256 β 8 + 1 2 β 8 β 24 β 8 = 256 3 πΈπΌππ΅π = πΈπΌππ΅ β 4 3 π₯3 + 12π₯2 = 256 3 4π₯3 β 36π₯2 + 256 = 0 π₯ = 3.37 πΈπΌπΏπππ₯ = β 1 3 β 4 β 3.372 β 3.37 β 3 4 β 3.37 + 1 2 β 24 β 3.372 β 2 3 β 3.37 = 177.2
- 48. πΏπππ₯ = 177.2 πΈπΌ EXERCISE P2.1 For a simply-supported beam of span L with a concentrated Load P at mid-span, shows that the deflection under the load is πΏ = ππΏ3 48πΈπΌ P2.2 For a simply-supported beam of span L loaded with UDL of intensity w, show that the mid-span deflection is πΏ = 5ππΏ4 384πΈπΌ P2.3 Apply Mohrβs Theorem to calculate the end moments of the built-in beams below.
- 49. P2.4 Assume a simply-supported beam as a primary structure for the propped cantilever loaded with UDL as shown below. Use the appropriate displacement compatibility to calculate the moment at A. P2.5 Find the maximum deflection in span AB and the deflection at C in terms of EI
- 50. CHAPTER 3 CONTINUOUS BEAMS AND THE THREE MOMENT EQUATION A beam is used to be continuous when supported on more than 2 supports. Continuous beams are statically indeterminate. The intermediate supports of a continuous beam are always subjected to some bending moment. The analysis of a continuous beam is similar to that of fixed end beam. The bending moment diagram of a continuous beam, under any set of loading, may be drawn in two stages: 1.) By considering the beam as a series of discontinuous beams (with simple supports) and drawing the bending moment diagram for the given loads, (Free Moment Diagram). 2.) By superimposing the Free Moment diagram to the end moments (Fixed Moment Diagram)
- 51. In Fig 3.1, ABC is a continuous beam in which πΏ1 = π πππ ππ π΄π΅ πΏ2 = π πππ ππ π΅πΆ πΌ1 = ππππππ‘ ππ πΌππππ‘ππ ππ ππππ ππ ππππ π΄π΅ πΌ2 = ππππππ‘ ππ πΌππππ‘ππ ππ ππππ ππ ππππ π΅πΆ ππ΄ = ππ’πππππ‘ ππππππ‘ ππ‘ π΄ ππ΅ = ππ’πππππ‘ ππππππ‘ ππ‘ π΅ ππΆ = ππ’πππππ‘ ππππππ‘ ππ‘ πΆ π΄1 = π΄πππ ππ π‘βπ ππππ ππππππ‘ πππππππ ππ π πππ π΄π΅ π΄2 = π΄πππ ππ π‘βπ ππππ ππππππ‘ πππππππ ππ π πππ π΅πΆ Fig 3.2 shows the deflected shape. Draw a tangent to the deflected shape (Elastic curve) at B meeting the vertical through A at D and the vertical through C at E. We note that
- 52. tan π = βπ΄ πΏ1 = β βπΆ πΏ2 β¦ β¦ β¦ β¦ β¦ ππ. 3.1 From Mohr II βπ΄ = βπ΄π΅= 1 πΈπΌπ [(π΄1π₯Μ 1 + ππ΄ β πΏ1 2 β πΏ1 3 ) + (ππ΅ β πΏ1 2 β 2πΏ1 3 )] = 1 6πΈ ( 6π΄1π₯Μ 1 πΌ1 + ππ΄πΏ1 2 πΌ1 + 2ππ΅πΏ1 2 πΌ1 ) βπ΄ πΏ1 = 1 6πΈ ( 6π΄1π₯Μ 1 πΌ1πΏ1 + ππ΄πΏ1 πΌ1 + 2ππ΅πΏ1 πΌ1 ) ππ 3.2 βπΆ πΏ2 = 1 6πΈ ( 6π΄2π₯Μ 2 πΌ2πΏ2 + ππΆπΏ2 πΌ2 + 2ππ΅πΏ2 πΌ2 ) ππ 3.3 Substituting eq. 3.2 and eq 3.3 in eq 3.1 we obtain ππ΄ πΏ1 πΌ1 + 2ππ΅ ( πΏ1 πΌ1 + πΏ2 πΌ2 ) + ππΆ πΏ2 πΌ2 = β6 ( π΄1π₯Μ 1 πΌ1πΏ1 + π΄2π₯Μ 2 πΌ2πΏ2 ) This is known as the three Moment Equation If the moment of inertia of the beam is constant, the equation reduces to ππ΄πΏ1 + 2ππ΅(πΏ1 + πΏ2) + ππΆπΏ2 = β6 ( π΄1π₯Μ 1 πΏ1 + π΄2π₯Μ 2 πΏ2 ) The Three Moment Equation was derived by CLAPEYRON in 1857 and relates the unknown reactant moments to the free bending moment diagram for each two
- 53. consecutive spans of a continuous beam. By writing this equation for each adjacent pair of spans, a sufficient number of equations are obtained pair of spans, a sufficient number of equations are obtained to solve for the unknown moments. Example 3.1 A continuous beam ABCD, simply supported at A, B, C and D is loaded as shown below. Find the support moments and draw the bending moment and shear force diagram. Solution Draw the continuous beam as a series of simply-supported beams and draw the free moment diagram EI is constant.
- 54. The degree of redundancy is 2 - ππ΅ πππ ππΆ We consider the first 2 spans, AB and BC And calculate the terms π΄1π₯Μ 1 πππ π΄2π₯Μ 2 in the three moment equation. The moment π΄1π₯Μ 1 is taken at A and π΄2π₯Μ 2 is taken at C. The moments are taken at the exterior supports and NEVER at the middle supports. The areas π΄1 πππ π΄2 are each divided into two as indicated by the dotted lines so that the centroids of the resulting triangles can be easily located. π΄1π₯Μ 1 = 1 2 β 12 β 2 β 2 3 β 2 + 1 2 β 12 β 4 (2 + 1 3 β 4) = 16 + 80 = 96 π΄2π₯Μ 2 = 1 2 β 9.6 β 3 β 2 3 β 3 + 1 2 β 9.6 β 2 (3 + 1 3 β 2) = 28.8 + 35.2 = 64 We can now write the three moment equation for span AB ππ΄πΏ1 + 2ππ΅(πΏ1 + πΏ2) + ππΆπΏ2 = β6 ( π΄1π₯Μ 1 πΏ1 + π΄2π₯Μ 2 πΏ2 ) We note that the continuous beam has a simple support at A, i.e. ππ΄ = 0 0 + 2ππ΅(6 + 5) + ππΆ β 5 = β6 ( 96 6 + 64 5 )
- 55. 22 ππ΅ + 5ππΆ = β172.8 Next, we move to spans BC and CD In calculating π΄1π₯Μ 1, we note that while π΄1 is the same area π΄2 shown in the spans AB and BC, the moment is taken at the same point. For spans AB, BC the moment was taken at C while for the present spans, it is taken at B. π΄1π₯Μ 1 = 1 2 β 9.6 β 2 β 2 3 β 2 + 1 2 β 9.6 β 3 (2 + 1 3 β 3) 12.8 + 43.2 = 56 π΄2π₯Μ 2 = 2 3 β 4 β 6 β 4 2 = 32 We now write the 3 Moment Equations, noting that the unknown moments are now ππ΅, ππΆ and ππ·. However, ππ· is zero because D is an exterior simple supports ππ΅ β 5 + 2ππΆ(5 + 4) = β6 ( 56 5 + 32 4 ) 5ππ΅ + 18ππΆ = β115.2 The Equations (a) and (b) are solved simultaneously to obtain ππ΅ = β6.84πππ
- 56. ππΆ = β4.48 πππ REACTIONS With the support moments, calculated, the reactions can be calculated from the simply-supported beams as follows: Considering span AB β ππ΅ = 0 6π π΄ β 9 β 4 + 6.84 = 0 π π΄ = 36 β 6.84 6 = 4.86 ππ π π΅ πΏ = 9 β 4.86 = 4.14 ππ
- 57. Span BC β ππΆ = 0 5π π΅ π β 6.84 β 8 β 3 + 4.48 = 0 π π΅ π = 5.27 ππ π πΆ πΏ = 8 β 5.27 = 2.73 ππ Span CD β ππΆ = 0 6π π· β 3 β 4 β 2 + 4.48 = 0 π π· = 4.88 ππ
- 58. π πΆ π = 3 β 4 β 4.88 = 7.12 ππ π π΅ = π π΅ πΏ + π π΅ π = 4.14 + 5.27 = 9.41 ππ π π = π πΆ πΏ + π πΆ π = 2.73 + 7.12 = 9.85 ππ The bending moment diagram can be drawn by superimposing the free moment diagram on the fixed moment diagram. The shear force diagram can be constructed from the reactions of the simply- supported beams.
- 59. CONTINUOUS BEAM WITH OVERHANG If a continuous beam has an overhang that end remains statically determine since the support moment is known. No additional equation needs to be formulated due to the overhang. For the above beams, we have only one redundant moment, ππ΅ , A is an end simple support, ππ΄ = 0 , and at C we observe that ππΆ = βππΏ3, the negative sign indicating a hogging moment. For such a structure, we first formulate the three moment equation involving ππ΄, ππ΅ and ππΆ and substitute the known values of ππ΄ and ππΆ. CONTINUOUS BEAM WITH FIXED END SUPPORT For a continuous beam with fixed end supports, end moments are developed at the supports. The rotation at the fixed end does not take place and the support moment is required to be calculated.
- 60. In order to analyse such as beam by the Theorem of Three Moment, an additional equation is required for each fixed end. In such a case, whenever an exterior support is fixed, an additional imaginary span of length zero with no loads is appended to the end. In formulating equation 2 in the above sketch, all terms involving the appended zero length span are zero. Thus for span BC and CCβ Three Moment Equation ππ΅πΏ1 + 2ππΆ(πΏ1 + 0) = β6 ( π΄1π₯Μ 1 πΏ1 + 0) ππ΅πΏ1 + 2ππΆπΏ1 = β6 π΄1π₯Μ 1 πΏ1
- 61. Example 3.2 For the continuous beam with constant EI, determine the supports moments and draw the bending moment and shear force diagrams. π΄πΌ π΄ ππ π‘βπ πππππππππ¦ π§πππ π πππ ππππ π΄πΌ π΄ πππ π΄π΅ π΄1π₯Μ 1 = 0 π΄2π₯Μ 2 = 1 2 β 24 β 3 (2 + 1 3 β 3) + 1 2 β 24 β 2 β 2 3 β 2 = 108 + 32 = 140
- 62. 3 Moment Equation: 2ππ΄(5) + ππ΅(5) = β6 ( 140 5 ) 10ππ΄ + 5ππ΅ = β168 Spans AB and BC π΄1π₯Μ 1 = 36 β 2 3 β 3 + 24 (3 + 1 3 β 2) = 160 π΄2π₯Μ 2 = 2 3 β 5 β 31.25 β 5 2 = 260.42 3 Moment Equation ππ΄(5) + 2ππ΅(5 + 5) + ππΆ(5) = β6 ( 160 5 + 260.42 5 ) 5ππ΄ + 20ππ΅ + 5ππΆ = β504.5 π΅π’π‘ ππΆ = β10 β΄ 5ππ΄ + 20ππ΅ = β454.5 ππππ£πππ ππππ (1)πππ (2) π πππ’ππ‘πππππ’π ππ¦ ππ΄ = β6.21 πππ ππ΅ = β21.17 πππ
- 63. Reactions Left Free Body Diagram β ππ΅ = 0 5π π΄ + 21.17 β 6.21 β 20 β 2 = 0 5π π΄ = 40 β 21.17 + 6.21 π π΄ = 5 ππ π π΅ πΏ = 20 β 5 = 15 ππ Right Free Body Diagram β ππ΅ = 0 5π πΆ + 21.17 β 10 β 50 β 5 2 = 0 5π πΆ = 125 + 10 β 21.17
- 64. π πΆ = 22.77 ππ π π΅ π = 50 β 22.77 = 27.23 ππ π π΅ = π π΅ πΏ + π π΅ π = 15 + 27.23 = 42.23 ππ
- 65. EXERCISES P3.1 Analyse the propped cantilever beam below and draw the bending moment and shear force diagrams. EI is constant. P3.2 Analyse the fixed beam with internal hinge and draw the bending moment diagram EI=constant.
- 66. P3.3 For the continuous beam below, use the three moment theorem to calculate the unknown moments and draw the bending moment and shear force diagrams.
- 67. CHAPTER 4 DEFLECTION BY ENERGY METHODS Energy Methods belong to a class of procedures employing the Principle of Conservation of Energy to determine the deflections of structures. To this class belong, among others, ο· The STRAIN ENERGY METHODS ο· The PRINCIPLE OF VIRTUAL WORK ο· CASTIGLIANOβS METHOD In this course, we shall limit ourselves to the principle of virtual work. 4.1 PRINCIPLE OF VIRTUAL WORK ο· A virtual displacement is an imaginary displacement imposed on a structure, compatible with the boundary conditions. ο· Virtual Force is a set of imaginary forces in equilibrium The work done by the real force during virtual displacement or the work done by the virtual force during real displacement is called virtual work. ο· For virtual work to be done, either the displacement or force must be virtual. Both cannot be real nor can both be virtual.
- 68. The Principle of virtual work states that: ο· A body is in equilibrium if, and only if the virtual work of all forces acting on the body is zero INTERNAL AND EXTERNAL VIRTUAL WORK When a structure deforms, work is done both by the applied loads moving through a displacement, as well as by the internal stresses moving through corresponding strains (increase in strain energy) in the structure. Thus when virtual displacements or forces are causing virtual work, we have: πΏπ = 0 πΏππΌ β πΏππΈ = 0 πΏππΌ = πΏππΈ Where; ο· Virtual work is denoted πΏπ and is zero for a body in equilibrium ο· External Work done is πΏππΈ , and ο· Internal Work is πΏππΌ The Material can be linear or non-linear The external virtual work is πΏππΈ = β(πΏπππ β ππππππ πππππππ π£πππ‘π’ππ πππ πππππππππ‘π ) The internal virtual work is the work done by the internal forces on their corresponding virtual displacements.
- 69. We shall consider only the axial forces N and moment M among the internal forces. The effects of shear and torsion are insignificant. For a length ππ₯ of a member under constant axial load N and constant moment M, with extension ππΏ and rotation ππ, the internal virtual work. πΏππΌ = πππΏ + πππ ππΌ = β« π ππΏ + β« π ππ
- 70. THE VIRTUAL WORK METHOD In the Virtual Work Method, also called the Unit Load or Dummy Load Method, two sets of loading are considered. SET 1: Structure subjected to actual external loads or temperature changes SET 2: Unit Load acting on , structure. This is a fictitious or dummy load introduced for the purpose of calculating the displacements in the structure. The unit load is placed at the point of the structure and in the direction where displacement is required. Impose the displacements due to the first set (real loadings) as virtual displacement for the unit load. The virtual work done by the unit load. ππΈ = 1 β β Where β is the real displacement being sought 1 β β= β« π1 ππΏ + β« π1 ππ Where β , ππΏ πππ ππ πππ π‘βπ π πππ πππ πππππππππ‘π And 1, π1 πππ π1 πππ π‘βπ ππππ‘π’ππ πΏπππππππ Using Hookeβs Law (Refer to Fig 4.1) ππΏ = ππ πΈπ΄ ππ₯, πππ ππ = ππ πΈπΌ ππ₯
- 71. Where: ππ β ππ₯πππ πππππ ππ ππππππ ππππ π‘βπ ππππ π‘ π ππ‘ ππ πππππππ(ππ₯π‘πππππ πππππ ) ππ β πππππππ ππππππ‘ ππ ππππππ ππππ π‘βπ ππ₯π‘πππππ πππππ ππΌ β ππ₯πππ πππππ ππ ππππππ ππππ π‘βπ π’πππ‘ ππππ ππΌ β πππππππ ππππππ‘ ππ ππππππ ππππ π‘βπ π’πππ‘ ππππ β= β« ππΌππ πΈπ΄ ππ₯ + β« ππΌππ πΈπΌ ππ₯ From trusses, the bending effects are negligible and β= β« ππΌππ πΈπ΄ ππ₯ Since the axial forces in truss members are constant the above simplifies to β= β ππΌπ πππ πΈπ΄π πΏπ π π=1 Where the summation goes over the number of members For the beam and for many frames, the axial effect on displacements is very small and can be omitted; Thus β= β« ππΌ ππ πΈπΌ πΏ 0 ππ₯ In most cases, for statically determinate structures, the moment diagrams ππΌ and ππ can be drawn without writing the equations. The Integral β« ππΌ ππ ππ₯ can be computed graphically.
- 72. β« ππΌ ππ ππ₯ = β« ππ ππΌ ππ₯ = π΄ β π πΏ 0 πΏ 0 β΄ β= π΄ β π πΈπΌ
- 73. Example 4.1 Use the unit load method to calculate the mid-span deflection and the slope at the supports of a simply-supported beam loaded with UDL of intensity w. Solution Method 1: Integration Set1 Loading ππ = π€πΏ 2 π₯ β π€π₯2 2 Set 2 loading ππ = 1 2 π₯ ; (0 β€ π₯ β€ πΏ 2 )
- 74. Due to symmetry, the integration can be performed over half the span and the results multiplied by 2. β= 1 πΈπΌ β« ππΌ ππ ππ₯ πΏ 0 = 2 πΈπΌ β« 1 2 π₯ ( π€πΏπ₯ 2 β π€π₯2 2 ) πΏ 2 0 ππ₯ β= 2 πΈπΌ β« ( π€πΏπ₯2 4 β π€π₯3 4 ) ππ₯ πΏ 2 0 = π πΈπΌ [( π€πΏπ₯3 12 β π€π₯4 16 )] π π³ π = 2 πΈπΌ [ π€πΏ4 96 β π€πΏ4 256 ] = π€πΏ4 πΈπΌ ( 1 48 β 1 128 ) = 5π€πΏ4 384πΈπΌ Method 2 β Graphical (Preferred) Set 1 Loading
- 75. Set 2 loading Due to symmetry we can deal with the diagrams up to mid-span and double the results β΄ β = 2 β π΄ β π πΈπΌ = 2 β 2 3 β πΏ 2 β π€πΏ2 8 β 5 8 β πΏ 4 β 1 πΈπΌ
- 76. = 5π€πΏ4 384πΈπΌ To calculate the rotation or slope at the support, we apply a unit moment to support A and draw the ππΌ diagram ππ΄ = 1 πΈπΌ β 2 3 β πΏ β π€πΏ2 8 β 1 2 = π€πΏ3 24πΈπΌ
- 77. Example 4.2 For the frame below, determine the vertical deflection at node D. EI is constant for all members Solution For the set 1 loading, letβs use the principle of superposition letting the UDL and concentrated load act independently. For the systems on the right, the bending moment diagrams can easily be drawn.
- 78. πΈπΌ βπ· π» = 80 + 10 + 20 = 110 βπ· π» = 110 πΈπΌ
- 79. Example 4.3 For the truss above, determine the following: a. The vertical deflection of node 5 b. The rotation of member 1-5 c. The change in distance between nodes 2 and 3 Solution Sub questions (a), (b) and (c) all make use of the same set 1 Loading. The axial forces in the members must be determined for the given external loading.
- 80. a. To calculate the vertical deflection of node 5, put a vertical unit load at node 5 and calculate the axial forces in all members. Now make and compute the following table Member Length(L) Area (A) ππΌ ππ ππΌπππΏ/π΄ 1-3 πΏ 2 β A 0 P 0 2-5 πΏ 2 β A -1 -2P PL/A 4-5 πΏ 2 β A 0 -P 0 1-2 πΏ 2 β A -1 -P PL/2A 1-5 πΏβ2 2 β π΄β2/2 β2 β2P 2PL/A 3-5 πΏ 2 β A 0 -P 0 3-4 πΏβ2 2 β π΄β2/2 0 β2P 0 β ππΌπππΏ π΄ = 7ππΏ 2π΄ Therefore Vertical deflection of node 5 β5 π£ = 7ππΏ 2πΈπ΄
- 81. (b) We need to subject member 1-5 to a unit moment. Since trusses do not take moments at the joints, we use a couple (pair of equal but opposite forces) at the ends of member 1-5 to produce a unit moment. The force at each end, F, is calculated as πΉ β πΏβ2 2 = 1 πΉ = 2 β2πΏ = β2 πΏ To calculate the rotation of member 1-5, replace the column containing ππΌ in (a) with those above and compute the last column in the table as follows:
- 82. Member Length(L) Area (A) ππΌ ππ ππΌπππΏ/π΄ 1-3 πΏ 2 β A 0 P 0 2-5 πΏ 2 β A -2/L -2P 2P/A 4-5 πΏ 2 β A 0 -P 0 1-2 πΏ 2 β A 0 -P 0 1-5 πΏβ2 2 β π΄β2/2 β2/L β2P 2P/A 3-5 πΏ 2 β A 0 -P 0 3-4 πΏβ2 2 β π΄β2/2 0 β2P 0 β ππΌπππΏ π΄ = 4π π΄ β1β5= 4π πΈπ΄ (c) The nodes 2 and 3 either approach each other or move away from each other. The unit loads (P=1) act at nodes 2 and 3 along the line joining them but directed opposite each other. The forces are self-equilibrating and produce no reactions.
- 83. Member Length(L) Area (A) ππΌ ππ ππΌπππΏ/π΄ 1-3 πΏ 2 β A ββ2/2 P ββ2PL/4A 2-5 πΏ 2 β A ββ2/2 -2P β2PL/2A 4-5 πΏ 2 β A 0 -P 0 1-2 πΏ 2 β A ββ2/2 -P β2PL/4A 1-5 πΏβ2 2 β π΄β2/2 1 β2P β2PL/A 3-5 πΏ 2 β A ββ2/2 -P β2PL/4A 3-4 πΏβ2 2 β π΄β2/2 0 β2P 0 β ππΌπππΏ π΄ = 7β2 ππΏ 4π΄ β2β3= 7β2 ππΏ 4π΄ Since the value is positive, the nodes approach each other as indicated.
- 84. EXERCISES P4.1 Use the unit load method to calculate the vertical deflection at D P4.2 Calculate the slope at A and the deflection at C P4.3 For the cantilever truss, calculate the vertical deflection at joints 5 and 3. EA=80,000 kN for all members.