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M2l11

1. 1. Module 2 Analysis of StaticallyIndeterminate Structures by the Matrix Force Method Version 2 CE IIT, Kharagpur
2. 2. Lesson 11The Force Method of Analysis: Frames Version 2 CE IIT, Kharagpur
3. 3. Instructional ObjectivesAfter reading this chapter the student will be able to1. Analyse the statically indeterminate plane frame by force method.2. Analyse the statically indeterminate plane frames undergoing support settlements.3. Calculate the static deflections of a primary structure (released frame) underexternal loads.4. Write compatibility equations of displacements for the plane deformations.5. Compute reaction components of the indeterminate frame.6. Draw shear force and bending moment diagrams for the frame.7. Draw qualitative elastic curve of the frame.11.1 IntroductionThe force method of analysis can readily be employed to analyze the indeterminateframes. The basic steps in the analysis of indeterminate frame by force method arethe same as that discussed in the analysis of indeterminate beams in the previouslessons. Under the action of external loads, the frames undergo axial and bendingdeformations. Since the axial rigidity of the members is much higher than thebending rigidity, the axial deformations are much smaller than the bendingdeformations and are normally not considered in the analysis. The compatibilityequations for the frame are written with respect to bending deformations only. Thefollowing examples illustrate the force method of analysis as applied to indeterminateframes.Example 11.1Analyse the rigid frame shown in Fig.11.1a and draw the bending moment diagram.Young’s modulus E and moment of inertia I are constant for the plane frame.Neglect axial deformations. Version 2 CE IIT, Kharagpur
4. 4. The given one- storey frame is statically indeterminate to degree one. In the presentcase, the primary structure is one that is hinged at A and roller supported at D .Treat horizontal reaction at D , R Dx as the redundant. The primary structure (which isstable and determinate) is shown in Fig.11.1.b.The compatibility condition of theproblem is that the horizontal deformation of the primary structure (Fig.11.1.b) due toexternal loads plus the horizontal deformation of the support D , due to redundant Version 2 CE IIT, Kharagpur
5. 5. RDx (vide Fig.11.1.b) must vanish. Calculate deformation a11 due to unit load at Din the direction of R Dx . Multiplying this deformation a11 with R Dx , the deformation dueto redundant reaction can be obtained. Δ = a11 RDx (1)Now compute the horizontal deflection Δ L at D due to externally applied load. Thiscan be readily determined by unit load method. Apply a unit load along R Dx asshown in Fig.10.1d.The horizontal deflection Δ L at D in the primary structure due to external loading isgiven by D δM v M ΔL = ∫ dx (2) A EI Version 2 CE IIT, Kharagpur
6. 6. where δM v is the internal virtual moment resultant in the frame due to virtual loadapplied at D along the resultant R Dx and M is the internal bending moment of theframe due to external loading (for details refer to Module 1,Lesson 5).Thus, ΔL = ∫ 6 (12 x − x )x dx 2 +∫ (36 − 9 x)6 6 6 0( x ) 0 EI 0 EI dx + ∫ 0 EI dx (span AB, origin at A) (span BC, origin at B) (span DC, origin at D ) 864 ΔL = (3) EIIn the next step, calculate the displacement a 1 1 at D when a real unit load isapplied at D in the direction of RDx (refer to Fig.11.1 d). Please note that the sameFig. 11.1d is used to represent unit virtual load applied at D and real unit loadapplied at D . Thus, D δm v m a11 = ∫ dx A EI 6 6 6 x 2 dx 36dx x2 =∫ EI ∫ EI + + ∫ dx 0 0 0 EI 360 = (4) EINow, the compatibility condition of the problem may be written as Δ L + a11 R Dx = 0 (5)Solving, RDx = −2.40 kN (6)The minus sign indicates that the redundant reaction R Dx acts towards left.Remaining reactions are calculated from equations of static equilibrium. ∑F x = 0 ⇒RAx = −12 + 2.40 = −9.60 kN (towards left) ∑M D = 0 ⇒ RAy = −9 kN (dowwards) RDy = +9 kN (upwards) Version 2 CE IIT, Kharagpur
7. 7. The bending moment diagram for the frame is shown in Fig. 11.1e Version 2 CE IIT, Kharagpur
8. 8. Example 11.2Analyze the rigid frame shown in Fig.11.2a and draw the bending moment and shearforce diagram. The flexural rigidity for all members is the same. Neglect axialdeformations.Five reactions components need to be evaluated in this rigid frame; hence it isindeterminate to second degree. Select Rcx (= R1 ) and Rcy ( = R2 ) as the redundantreactions. Hence, the primary structure is one in which support A is fixed and thesupport C is free as shown in Fig.11.2b. Also, equations for moments in variousspans of the frame are also given in the figure. Version 2 CE IIT, Kharagpur
9. 9. Calculate horizontal (Δ L )1 and vertical (Δ L ) 2 deflections at C in the primary structuredue to external loading. This can be done by unit load method. Thus, Version 2 CE IIT, Kharagpur
10. 10. Version 2 CE IIT, Kharagpur
11. 11. (96 + 24 x)(3 + x) 3 3 96 x ( Δ L )1 = ∫ dx +∫ dx +0 0 EI 0 EI (Span DA, origin at D) (Span BD, origin at B) (span BC, Origin B) 2268 = (1) EI (96 + 24 x)(−4)dx 96(−4)dx 48 x(−2 − x)dx 3 3 2 (Δ L ) 2 = ∫ +∫ +∫ +0 0 EI 0 EI 0 EI (Span DA, origin at D) (Span BD, origin at B) (Span BE, origin at E)(Span EC, Origin C) 3056 ( Δ L )2 = − (2) EI Version 2 CE IIT, Kharagpur
12. 12. In the next step, evaluate flexibility coefficients, this is done by applying a unit loadalong, R1 and determining deflections a11 and a21 corresponding to R1 and R2respectively (vide, Fig .11.2 c). Again apply unit load along R2 and evaluatedeflections a 22 and a12 corresponding to R2 and R1 and respectively (ref.Fig.11.2d). 6 x2 72 a11 = ∫ dx = (3) 0 EI EI x(−4) 6 a12 = a 21 = ∫ dx + 0 0 EI 72 = (4) EIand 6 4 16 ( x) 2 a 22 = ∫ dx + ∫ dx 0 EI 0 EI 117.33 = (5) EIIn the actual structure at C, the horizontal and vertical displacements are zero.Hence, the compatibility condition may be written as, (Δ L )1 = a11 R1 + a12 R2 = 0 (Δ L ) 2 = a12 R1 + a22 R2 = 0 (6)Substituting the values of (Δ L )1 , ( Δ l ) 2 , a11 , a12 and a 22 in the above equations andsolving for and R1 , R2 we get R1 = −1.056 kN (towards left) R2 = 27.44 kN (upwards)The remaining reactions are calculated from equations of statics and they are shownin Fig 11.2e. The bending moment and shear force diagrams are shown in Fig. 11.2f.. Version 2 CE IIT, Kharagpur
13. 13. Version 2 CE IIT, Kharagpur
14. 14. 11.2 Support settlementsAs discussed in the case of statically indeterminate beams, the reactions areinduced in the case of indeterminate frame due to yielding of supports even whenthere are no external loads acting on it. The yielding of supports may be either lineardisplacements or rotations of supports (only in the case of fixed supports) .Thecompatibility condition is that the total displacement of the determinate frame(primary structure) due to external loading and that due to redundant reaction at agiven support must be equal to the predicted amount of yielding at that support. If thesupport is unyielding then it must be equal to zero.Example 11.3A rigid frame ABC is loaded as shown in the Fig 11.3a, Compute the reactions if thesupport D settles by 10 mm. vertically downwards. Assume EI to be constant for allmembers. Assume E = 200 GPa and I = 10−4 m 2 . Version 2 CE IIT, Kharagpur
15. 15. This problem is similar to the previous example except for the support settlement.Hence only change will be in the compatibility equations. The released structure isas shown in Fig.11.3b .The deflections (Δ L )1 and (Δ L ) 2 at C in the primary structuredue to external loading has already been computed in the previous example. Hence, 2052 ( Δ L )1 = (1) EI Version 2 CE IIT, Kharagpur
16. 16. − 3296 (Δ L ) 2 = (2) EITherefore, (Δ L )1 = 0.1026 m (Δ L ) 2 = −0.1635 mThe flexibility coefficients are, 72 a11 = (3) EI − 72 a12 = a 21 = (4) EI 117.33 a 22 = (5) EINow, the compatibility equations may be written as, (Δ L )1 + a11 R1 + a12 R2 = 0 (Δ L ) 2 + a 21 R1 + a 22 R2 = −10 × 10 −3 (6)Solving which, R1 = −2.072 kN (towards left) R2 = +26.4 kN (upwards) (7)The reactions are shown in Fig.11.3c. Version 2 CE IIT, Kharagpur
17. 17. Example 11.4Compute the reactions of the rigid frame shown in Fig.11.4a and draw bendingmoment diagram .Also sketch the deformed shape of the frame. Assume EI to beconstant for all members.Select vertical reaction at C, R1 as the redundant .Releasing constraint againstredundant, the primary structure is obtained. It is shown in Fig.11.4b. Version 2 CE IIT, Kharagpur
18. 18. The deflection (Δ L )1 in the primary structure due to external loading can becalculated from unit load method. (12 x)(−4)dx 3 ( Δ L )1 = ∫ 0 EI (span DA, origin at D) − 216 = (Downwards) (1) EINow, compute the flexibility coefficient, 4 6 x2 16 a11 = ∫ dx + ∫ dx 0 EI 0 EI 117.33 = (2) EIThe compatibility condition at support C is that the displacement at C in the primarystructure due to external loading plus the displacement at C due to redundant mustvanish. Thus, − 216 117.33 + R1 = 0 (3) EI EISolving, R1 = 1.84 kN (4)The remaining reactions are calculated from static equilibrium equations. They areshown in Fig.11.4d along with the bending moment diagram. Version 2 CE IIT, Kharagpur
19. 19. Version 2 CE IIT, Kharagpur
20. 20. To sketch the deformed shape/elastic curve of the frame, it is required to computerotations of joints B and C and horizontal displacement of C . These joint rotationsand displacements can also be calculated from the principle of superposition .Thejoint rotations are taken to be positive when clockwise. Towards this end firstcalculate joint rotations at B (θ BL ) and C (θ CL ) and horizontal displacement at C in thereleased structure (refer to Fig.11.4b).This can be evaluated by unit load method. 12( x)(−1) − 54 3 θ BL = ∫ dx = (5) 0 EI EI (12 x)(−1) − 54 3 θ CL = ∫ dx = (6) 0 EI EI 12 x(3 + x) 3 270 Δ CL = ∫ dx = (7) 0 EI EINext, calculate the joint rotations and displacements when unit value of redundant isapplied (Fig.11.4c). Let the joint rotations and displacements be θ BR ,θ CR and Δ CR . 6 4dx 24 θ BR = ∫ = (8) 0 EI EI (− x)(−1) (−4)(−1) 4 6 32 θ CR = ∫ dx + ∫ dx = (9) 0 EI 0 EI EI (−4) x − 72 6 Δ CR = ∫ dx = (10) 0 EI EINow using the principle of superposition, the actual rotations and displacements atthe joints may be obtained. θ B = θ BL + θ BR R1 (11) −54 24 × 1.84 9.84 = + = − EI EI EI(Clockwise rotation) θ C = θ CL + θ CR × R1 (12) − 54 32 × 1.84 4.88 = + = EI EI EI(Counterclockwise rotation) Δ C = Δ CL + Δ CR R1 (13) Version 2 CE IIT, Kharagpur
21. 21. 270 72 × 1.84 137.52 = − = (towards right) EI EI EIThe qualitative elastic curve is shown in Fig. 11.4h. Version 2 CE IIT, Kharagpur
22. 22. SummaryIn this lesson, the statically indeterminate plane frames are analysed by forcemethod. For the purpose of illustrations only bending deformations of the frame areconsidered as the axial deformations are very small. The problem of yielding ofsupports in the case of plane frames is also discussed. The procedure to drawqualitative elastic curve of the frame is illustrated with the help of typical example.The bending moment and shear force diagrams are also drawn for the case of planeframe. Version 2 CE IIT, Kharagpur