P4 Stress and Strain Dr. A.B. Zavatsky MT07 Lecture 3 Statically Indeterminate Structures

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1 4 Stress and Strain Dr... Zavatsky MT07 ecture 3 Statically Indeterminate Structures Statically determinate structures. Statically indeterminate structures (equations of equilibrium, compatibility, and force-displacement; use of displacement diagrams) olts and turnbuckles. Temperature effects. Misfits and pre-strains. 1

2 Statically Determinate Structures Reactions and internal forces can be determined solely from free-body diagrams and equations of equilibrium. Results are independent of the material from which the structure has been made. 5 kn Unknowns reaction forces + bar forces ( + 1) Independent equations [equilibrium in x & y directions at each joint] (number of joints) (8) kn Double-check structure for internal mechanisms, etc.

3 Statically Indeterminate Structures Reactions and internal forces cannot be found by statics alone (more unknown forces than independent equations of equilibrium). Results are dependent on the material from which the structure has been made. R C a unknown forces Only 1 useful equation of equilibrium b R + R 0 R Need to find another equation 3

4 R Flexibility or Force Method 1 a R static redundant C + b R 1 released structure Equation of compatibility expresses the fact that the change in length of the bar must be compatible with the conditions at the supports

5 Write the force-displacement relations. These take the mechanical properties of the material into account. b and E 1 R E Substituting into the equation of compatibility gives: R b R + 0 E E b *Note that flexibilities (b/e) and (/E) appear in this equation. Hence, this approach is called the flexibility method. Substituting into the equilibrium equation gives: R R a *Note that we have solved for forces. Hence, this approach is also called the force method. 5

6 Stiffness or Displacement Method R R C a C + R b R C1 C R Equation of equilibrium (forces at C must balance) R + R Equation of compatibility (at point C) C C1 C R 6

7 Write the force-displacement relations and solve for the forces. R C1 Ra E C1E a and and C R Rb E CE b Substituting into the equilibrium equation gives: E a E b C 1 C + *Note that stiffnesses (E/a) and (E/b) appear in this equation. Hence, this approach is called the stiffness method. Using the compatibility condition (displacements equal) gives: ab E a C ab ( + b) E *Note that we have solved for displacement. Hence, this approach is also called the displacement method. 7

8 Finally, substituting into the expressions for forces gives: R R ab E ab E E a E b b a So, both the flexibility method and the stiffness method give the same result. The choice of approach will depend on the problem being solved. 8

9 Use of Displacement Diagrams in Statically Indeterminate roblems (based on Example 3, page 70, Gere & Timoshenko) C CD C 5 sinα sinα α 1 D α ar D is supported by two wires, CD and C. load is applied at. The wires have axial rigidity E. Disregarding the weight of the bar, find the forces in the wires. 9

10 R y T CD T C α 1 α R x Equilibrium Horizontal Direction F x Rx TCD cosα1 TC cosα 0 Vertical Direction F y Ry + TCD sinα1 + TC sin α 0 Moments about (ccw +) ( T α ) + ( T sinα ) ( ) M CD sin 1 C 0 4 unknown forces, only 3 equations 10

11 Compatibility We can relate the tensions in the two wires by considering the extensions of the wires. D D C D CD Displacement diagram CD C D sinα sinα 1 D sinα 11

12 Force-displacement relations TCDCD and E CD C TC E C Using the equilibrium moment equation, the compatibility equation, and the force-displacement relations, it is possible to solve for the forces in the wires. We find that T C 1.15 and T CD

13 Elasto-plastic nalysis of a Statically Indeterminate Structure (based on Example -19 from Gere) ar 1 ar (a) (b) b b b Horizontal beam is rigid. Supporting bars 1 and are made of an elastic perfectly plastic material with yield stress σ Y, yield strain ε Y, and Young s modulus E σ Y / ε Y. Each bar has cross-sectional area. (c) Find the yield load Y and the corresponding yield displacement Δ Y at point. Find the plastic load and the corresponding plastic displacement Δ at point. Draw a load-displacement diagram relating the load to the displacement Δ of point. 13

14 R y F 1 F R x b b b Moment Equilibrium (ccw +) M 1 F + F 3 1 Compatibility 1 ( F ) b + ( F ) b ( 3b) Force-displacement 1 1 and E F F E 0 F F ar will yield first, since F > F 1. 14

15 F Y 6Y σ Y 5 5σ Y (a) 6 The corresponding elongation of bar is: F F σ Y E E E The downward displacement of the bar at point is: Δ Y 3 3σ Y E When the plastic load is reached, both bars will be stretched to the yield stress, and F 1 F σ Y. From equilibrium, F 1 + F 3 (b) σ Y (a) 15

16 The corresponding elongation of bar 1 (which has just reached yield) is: F F σ Y E E E The downward displacement of the bar at point is: ar yields Δ 3 1 ar 1 yields 3σ E Y (b) Note that / Y 6/5 and Δ /Δ Y Y oth bars plastic ar plastic, bar 1 elastic Δ Y Δ (c) Δ 16

17 olts and Turnbuckles The simplest way to produce a change in length. Nut and olt Distance travelled by the nut n p n number of turns (not necessarily an integer) p pitch of the screw (units mm / turn) Turnbuckle Right-hand screw eft-hand screw Distance travelled n p Often used to tension cables 17

18 ased on Gere, Example -9 Steel cable Turnbuckle Rigid plate Copper tube The slack is removed from the cables by rotating the turnbuckles until the assembly is snug but with no initial stresses (do not want to stretch the cables and compress the tube). Find the forces in the tube and cables when the turnbuckles are tightened by n turns, and determine the shortening of the tube. 18

19 If the turnbuckles are rotated through n turns, the cables will shorten by a distance 1 n p s c s The tensile forces in the cables s and the compressive force in the tube c must be such that the final lengths of the cables and tube is the same. 19

20 Equilibrium (forces must balance) s 0 c Compatibility (shortening of tube must equal shortening of cable) 3 1 Force-displacement 1 np s E 3 s c s c E c With these equations, we can solve for the forces in the tube and cables and for the shortening of the tube. 0

21 Temperature Effects Changes in temperature produce expansion or contraction of structural materials. When heated, the block expands in all three directions: x, y, z. For most structural materials, ε T α (ΔT) ε T thermal strain α coefficient of thermal expansion (HT, units 1/K or 1/ C) ΔT change in temperature The change in length of the block in NY direction can be found using T ε T α (ΔT), where is one of the block s dimensions. 1

22 Expansion is positive; contraction is negative. relatively modest change in temperature produces about the same magnitude of strain as that caused by ordinary working stresses. This shows that temperature effects can be important in engineering design. For mild steel, α 11 x 10-6 K -1 (HT, page 39) Thermal strain: ε T α (ΔT) 11 x 10-6 K -1 (40 K) με Yield strain: ε Y σ Y / E (40 x 10 6 a)/(10 x 10 9 a) με Ordinary working stress might be 50% of the yield stress, giving 570 με Thermal strains are USUY reversible.

23 Free expansion or contraction occurs when an object rests on a frictionless surface or hangs in open space. Then no stresses are produced by a uniform temperature change, but there are strains. In statically determinate structures, uniform temperature changes in the members produce thermal strains (and corresponding changes in length) without producing any corresponding stresses. statically indeterminate structure may or may not develop thermal stresses, depending on the character of the structure and the nature of the temperature changes. C D Since D can move horizontally, no stresses are developed when the entire structure is heated uniformly. If only some bars are heated, thermal stresses will develop. 3

24 Example: The temperature of the bar is raised by ΔT. Find the reactions at the supports. R R R T ΔT + ΔT R Equilibrium: R + R 0 Compatibility: T + R 0 or T - R 4

25 Using temperature-displacement and force-displacement relations: T R α ( ΔT ) R E Substituting in to the compatibility condition gives: α R ( ΔT ) R E E α ( ΔT ) (compression) Substituting into the equilibrium equation gives: R R E α ( ΔT ) 5

26 Misfits and re-strains When a member is manufactured with a length slightly different from its prescribed length, the member will not fit into the structure as planned and the geometry of the structure will be different from what was planned. Such situations are misfits. Some misfits are created intentionally to introduce strains into the structure at the time it is built. ecause these strains exist before any loads are applied, they are called pre-strains. long with the pre-strains are usually pre-stresses. Examples: spokes in bicycle wheels, pre-tensioned faces of tennis racquets, shrink-fitted machine parts, pre-stressed concrete beams 6

27 If a structure is statically determinate, small misfits in one or more members will not produce strains or stresses, although the initial configuration will depart from the theoretical. Here, having CD longer than expected will not induce pre-strains or pre-stresses. can rotate to accommodate the length change. C D In a statically indeterminate structure, misfits cannot be accommodated without pre-stresses. 7

28 C E D F ssume that CD is slightly longer than prescribed. Then to assemble the structure, CD must be compressed by external forces (or EF must be stretched by external forces). The bars can then be fitted into place and the external loads released. s a result, the beam will deform and rotate. If CD is put into compression, EF will be in tension. So, pre-strains will exist and the structure will be pre-stressed, although no external loads are acting. When a load is added, additional stresses and strains will result. 8