STATICALLY INDETERMINATE STRUCTURES

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1 STATICALLY INDETERMINATE STRUCTURES

2 INTRODUCTION Generally the trusses are supported on (i) a hinged support and (ii) a roller support. The reaction components of a hinged support are two (in horizontal and vertical directions), and for a roller support reaction component is only one (in vertical direction). For any truss, if the number of members forming the truss is m and the number of joints is j, then for the frame to be rigid m = 2 j 3. The equations of static equilibrium are sufficient to determine the internal forces in members of a rigid truss. If members are less than 2j 3, the truss becomes a collapsible truss and if m > (2j 3), the truss becomes a redundant truss.

3 Equations from statics, F x 0, F 0, M 0 y O i.e. summation of forces in x- and y-directions, respectively, is zero and summation of moments of forces about any point is zero, are not sufficient to determine the internal forces in members of a redundant truss. A unit load method is used to analyze forces in members of a redundant truss. Structures used in bridges are generally redundant with multiple degree of redundancy. The study of such structures can be done by numerical methods; we will analyze the trusses with single degree of redundancy. There are many redundant structures in which strain compatibility condition can be used for analyzing the forces in members of trusses or frames.

4 ANALYSIS OF REDUNDANT FRAMES WITH STRAIN COMPATIBILITY CONDITION Let us consider a system of three wires of steel, brass, and aluminium supporting a rigid bar OABCD, which is hinged at end O. A load P is applied at end D of the bar as shown in Figure 1. Let s say the internal forces in bars of steel, brass, and aluminium are F A, F B, and F C, respectively. Then from the conditions of static equilibrium: P = F A + F B + F C - R O Moreover, taking moments about O, where R O is reaction at hinged end O, 4aP = a F A + 2a F B + 3a F C or 4P = F A + 2F B + 3F C Figure 1 (a) A rigid bar supported by three wires (b) Deformations in wires

5 There are four unknowns, i.e. R O, F A, F B, and F C, but only two equations, so conditions of static equilibrium cannot analyze the forces in this indeterminate structure. We have to consider the deformation produced in each bar after the application of load, or in other words strain compatibility condition has to be used. The rigid bar is going to take the new position OA B C D. From this diagram, AA = δ A deformation in bar A BB = δ B deformation in bar B CC = δ C deformation in bar C but from the strain compatibility, as is obvious from Figure 1 (b). δ B = 2 δ A δ C = 3 δ A

6 where From strain compatibility condition Let us take E steel = 2E brass = 3E al So, F B = F A and F C = F A

7 F B = F A and F C = F A Putting these values in 4P = F A + 2F B + 3F C From P = F A + F B + F C - R O, reaction R O 3F A P 2 3 P P 3 P R 0 reaction is downward.

8 Example 7.1 A frame ABCD is shown in the figure. There are three bars AD, BD, and CD hinged at D as shown. Bars AD and CD are of copper while bar BD is of steel. Length of middle bar BD is L. Area of cross-section of each bar is the same. A load P is applied at end D. Determine the forces in the three bars. Given E steel = 2E copper. From the condition of static equilibrium: P = F BD + F AD cos45 + F CD cos45 (i) Due to symmetry F AD = F CD, because both bars are identical and made of copper. So, P = F BD + 2F AD cos45 F BD = F BD F AD (ii) F AD F CD

9 We have to use conditions of strain compatibility. After the application of load P, point D moves down to D and all the bars are extended. Extension in AD is D D and extension in BD is DD. But DD sin45 = D D (iii) and where A is the area of the cross-section. Using Eq. (iii),

10 From Eqs. (iv) and (v), Force F BD F AD F CD Now from Eq. (ii)

11 Let us verify the equilibrium of forces at joint D P = F BD F AD = ( ) P = ( ) P = P F BD F AD F CD

12 Exercise 7.1 A rigid bar is suspended by three bars of steel and aluminium as shown in the figure. The bars are of equal length and equal area of cross-section. A load W is suspended on rigid bar so that the rigid bar remains horizontal after the application of load. Determine forces in aluminium and steel bars. E steel = 3E aluminium

13 DEGREE OF REDUNDANCY The total degree of redundancy or degree of indeterminacy of a frame is equal to the number by which the unknown reaction components exceed the condition of equation of equilibrium. The excess members are called redundants. Total degree of redundancy, T T = m (2 j R) where m = j = R = total number of members in a frame, total number of joints in frame, and total number of reaction components. Following criteria are applied for reaction components at supports of the frame: 1. For a roller support reaction component is For a hinged support reaction components are For a fixed support as in cantilever reaction components are 3.

14 Let us consider the following example: A frame ABCDEF shown in Figure 2 is hinged at end A and roller supported at end E. There are 11 members in frame. Number of members, m = 11 Number of joints, j = 6 Figure 2 Redundant frame Reaction components, R = 2 (for hinged support at A) + 1 (for roller support at E) = 3 Degree of redundancy, T = m (2j R) = 11 (2 6 3) = 11 9 = 2

15 Similarly, let us consider another frame ABCD, as shown in Figure 3, having six members and four joints. Frame is roller supported at A and hinged at end C. Let us determine its degree of redundancy. m = 6, number of members j = 4, number of joints Figure 3 Triangular frame R = number of reaction components = 2 (for hinged end C) + 1 (for roller support A) = 3 Degree of redundancy T = m (2j R) = 6 (2 4 3) = 6 5 = 1

16 Exercise 7.2 Determine the degree of redundancy for the frames shown in the figure. T = m (2j R) [For all the cases, degree of redundancy is 1].

17 ANALYSIS OF STATICALLY INDETERMINATE TRUSSES Method of virtual work or unit load method is one of the several methods available for solution of forces in members of an indeterminate truss. The following relationship is used to calculate the displacement in a member subjected to tensile or compressive load. Displacement SUL AE where S = internal force in truss members due to applied external loads (external loads on truss plus a unit load) U = axial load in truss members due to unit load applied externally at a point, E = Young s modulus of elasticity of member, L = length of member, and A = area of cross-section of member.

18 We will consider trusses with only single degree of redundancy. The following steps are taken in order to determine the forces in members of an indeterminate truss. From the indeterminate truss, remove the redundant member so as to obtain a statically determinate structure as shown in Figures 4 (a) and (b). Obtain the forces in members of the statically determinate truss as shown in Figure 4(b) by equilibrium equations of statics. Say these forces are S 0 in each member. Figure 4 (a) Statically indeterminate (b) Member BD removed statically determinate

19 Consider the same truss with one redundant member BD cut. Let the force X a = 1 is applied on this member (from both ends, no support). Find the forces in members without the external force P (Figure 5). Net force in each member S = S 0 + X a u a Now let X a be different from unity. Figure 5 Unit load along BD Let the relative displacement in redundant member be δ a. Say the area of cross-section of each member is A and Young s modulus is E.

20 or Put (rigid structure) and get the value of X a, or or

21 Forces in members of truss (putting the value of Xa) Member S Magnitude of S AB 0.707X a + 0.5P BC P 0.707X a 0.5P CD P 0.707X a 0.5P DA 0.707X a + 0.5P BD X a 0.707P AC P Figure 6 shows the forces in the members of the truss. Figure 6 Forces in members of truss Equilibrium of Truss at joints: Let us check the equilibrium of forces at joints of the truss. For that, we have to first calculate the reactions. Taking moments about A, P a = R D a Reaction, R D = P Reaction at A, R AH = P (horizontal component) R AV = P (vertical component)

22 Exercise 7.2 A redundant frame ABCD is shown in the figure. Material of all the bars is the same and the area of cross-section is also the same. Determine forces in all the members of the frame.

23 Example 7.3 For the truss shown in the figure, determine the support reactions and forces in all members. Area of cross-section of members AB, BC, and BD is 25 cm 2 and for members AD and DC it is 22.5 cm 2. Material is the same for all the members. Number of members, m = 5 Number of joints, j = 4 Both the supports are hinged. So, the number of reaction components, R = 2 2 = 4 Degree of redundancy T = m (2j R) = 5 (2 4 4) = 1

24 Vertical reactions at A and C R AV = R CV = 100 kn (due to symmetry). Let us say horizontal reaction at A and C is X a. Remove X a from both ends and calculate forces in members. Due to symmetry, we will solve for half the truss. Angles, s = 45 b = tan = sin s = coss = sin b = cos b = Forces R CV F AB cos45 = F AD cos26.56 (x direction) F AB = F AD F AB = F AD

25 Now F AD sin = F AB sin 45 F AD = F AB putting the value of F AB = (1.265 F AD ) = F AD Force, (To have equilibrium of forces at joint A, F AD is tensile and F AB will be compressive as shown). F AB = F BC = 283 kn Joint B F AB sin 45 = F BC sin 45

26 But = F BD Force, = F BD F BD = +200 kn (tensile) A Horizontal reactions Now applying the unit load X a = 1 at ends A and C as shown in the figure. F AD sin = F AB sin 45 F AD = F AB F AD = F AB (y direction) F AB = F AD cos = 0.894F AD (x direction) = F AB (putting the value of F AD ) = 1.414F AD F AB = (tension) F AD = = (comp.)

27 Joint B F BD = 2 F AB = = 2 (compressive) Total force in a member, S = S 0 + X a u a = Force in a member due to external load after removing redundant reaction + force due to load X a Also,

28 If we tabulate the values of S, u a, L, A, etc., E is the same for all members:

29 Now Reactions and forces in members or The forces are balanced at all the joints. All members are in compression.

30 Example 7.4 A triangular truss ABCD is shown in the figure. It is subjected to a horizontal load P at joint B. Determine the support reactions and forces in the members of truss. The cross-sectional area of outer members AB, BC, and CA is twice the cross-sectional area of inner members AD, BD, and CD. The material of all members is the same. Taking moments of forces about point A P 0.866L = R CV L R CV, reaction = 0.866P R AV = 0.866P R AH = P as shown.

31 For degree of redundancy: Number of members, m = 6 Number of joints, j = 4 Reactions components, R = 2 (for hinged end) + 1 (roller supported end) = 3 Degree of redundancy, T = m (2j R) = 6 (2 4 3) = 1

32 Let us remove one member AC from the frame, and find forces S 0 in members. Member AC is removed (figure). Joint C F BC cos 60 = F DC cos 30 F BC 0.5 = 0.866F DC F BC = F DC Moreover, F DC sin P = F BC sin F DC P = F BC 120 Putting the value of F BC 0.5 F DC P = F DC = 1.5 F DC Force, F DC = 0.866P F BC = = 1.5P

33 Joint D ADC = CDB = BDA = 120 so F DC = F DB = F DA = 0.866P (tensile) Joint A F AB cos60 + F AD cos30 = P 0.5F AB P = P 0.5F AB = (P 0.75P) F AB = 0.5P 120 Checking for vertical loads. F AB sin60 + F AD sin30 = P P = 0.866P (verified)

34 If we tabulate these values: Now consider only X a = 1 along member AC as shown in the figure. F AB sin60 = F AD sin30 F AB = F AD 0.5 F AD = 1.732F AB F AD cos30 = F AB cos F AD = 0.5F AB + 1 Putting the value of F AD F AB = 0.5F AB + 1 F AB = 1 = F BC F AD = (comp) = F DC = F DB

35 If we tabulate these u a values also Length AB = BC = CA = L, area of cross-section 2a

36 Material of all members is the same. Let us tabulate the different parameters.

37 Now or Resultant Forces in members

38 Example 7.5 Find the bending moment at any point of a semi-circular arch shown in the figure. Both the supports are hinged. Flexural rigidity EI is constant throughout. For the arch, number of members, m = 1 number of joints, j = 2 number of reaction components, R = 2 2= 4 (both ends are hinged). Degree of redundancy, T = m (2j R) = 1 (2 2 4) = 1

39 There will be reactions X and at each end as shown. At a particular section a-a, bending moment where k = R sin θ. Applying the principle of virtual work or unit load method, X M O M ds EI X 2 M i ds EI Taking the advantage of symmetrical loading,

40 or

41 Therefore, Bending moment at any section

42 Example 7.6 A rigid bar EF of negligible weight is suspended by four wires A, B, C, and D of the same length, same area of cross-section and same material. A load P is suspended at C as shown in the figure. Determine the forces in the four wires. From the conditions of static equilibrium, P = P A + P B + P C + P D (i) Moments about point E P A 0 + P B a + P C 2a + P D 3A = 2aP or 2P = P B + 2P C + 3P D (ii)

43 From the two equations, the values of four forces P A, P B, P C, and P D cannot be determined. We have to take the help of a compatibility condition, i.e. considering the deformation of the bar as shown in the figure. Deformation in bar A =δ A Deformation in bar B = δ A + δ Deformation in bar C = δ A + 2δ Deformation in bar D = δ A + 3δ Rigid bar EF will take the position A B C D after the application of load. Using Hooke s law, in the elastic region, deformation is directly proportional to load applied. δa δ A A Say δ A P A (load on wire A) δ P (load due to additional deformation δ)

44 So, P B = P A + P P C = P A + 2P P D = P A + 3P Total reactions = P A + P B + P C + P D = P A + P A + P + P A + 2P + P A + 3P = 4P A + 6P = P (iii) δ A Putting the values in (ii) 2P = P A + P + 2 (P A + 2P ) + 3 (P A + 3P ) 2P = P A + 2P A + 3P A + P + 4P + 9P 2P = 6P A + 14P or P = 3P A + 7P (iv)

45 From Eqs. (iii) and (iv), 4P A + 6P = 3P A + 7P P A = P (v) So, P B = 2P P C = 3P P D = 4P Total reaction P + 2P + 3P + 4P = P P = 0.10P (vi) δ A or forces in wire A, P A = P = 0.1P in wire B, P B = 0.2P in wire C, P C = 0.3P in wire D, P D = 0.4P

46 Exercise 7.3 A rigid bar EF of negligible weight is suspended by four wires A, B, C, and D of the same material, same length and same area of crosssection. A load P is suspended as shown in the figure. Determine the forces in wires A, B, C, and D.

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