Lec7 MECH ENG STRucture
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1. The document discusses linear elastic springs and examples of springs connected in series and in parallel. It introduces concepts of force-deformation relationships, compatibility, and using these together with equilibrium equations to solve statically indeterminate problems. 2. For springs in series, the total displacement is the sum of the individual spring displacements. The effective spring constant is calculated from the individual spring constants. 3. Solving examples involves setting up free body diagrams, writing equilibrium equations, using force-deformation relationships, and adding compatibility equations to solve for unknown forces and displacements.
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Lec7 MECH ENG STRucture
- 1. 2.001 - MECHANICS AND MATERIALS I Lecture #7 10/2/2006 Prof. Carol Livermore Recall: 3 Basic Ingredients 1. Forces, Moments, and Equilibrium 2. Displacements, Deformations, and Compatibility 3. Forces-Deformation Relationships Linear Elastic Springs Linear: k is a constant not a function of P or 隆. If it were non-linear: Elastic: Loading and unloading are along the same curve. 1
- 2. EXAMPLE: Springs in series Q: What are the reactions at the supports? Q: How are P and 隆 related? FBD Fx = 0 RAx + P = 0 RAx = P Fy = 0 Ry = 0 2
- 3. FBD of Spring 1 Fx = 0 F1 = P FBD of Spring 2 Fx = 0 F2 = P Look at compatibility: Undeformed 3
- 4. Deformed Compatibility: 隆1 + 隆2 = 隆 De鍖ne: 隆 - How much things stretch u - How much things move So: ux = 隆 Force-Deformation: F1 = k1 隆1 F2 = k2 隆2 Put it all together: 隆1 = F1 /k1 隆2 = F2 /k2 F1 F2 1 1 隆 = 隆1 + 隆 2 = + =P + k1 k2 k1 k2 So: k2 + k1 隆=P k2 k1 4
- 5. k1 k2 P = 隆 k1 + k2 k1 k2 kef f = k1 + k2 Plot Sanity Check 1. k1 = k2 = k k2 kef f = 2k = k , P 2 隆 2 2. k1 >> k2 kef f = k2 2 k2 All 鍖exibility is due to weaker spring. k 1+ k 1 EXAMPLE Q: How far does C displace? Q: What are the forces in the spring? 1. FBD 5
- 6. Note: No forces in y. Fx = 0 P + R Ax + R Bx = 0 FBD: Spring 1 RAx + F1c = 0 F1c = RAx FBD: Pin P F1c + F2c = 0 FBD: Spring 2 6
- 7. RBx F2c = 0 F2c = RBx So: P + R Ax + R Bx = 0 Note: We already have this equation. We need more than just equilibrium! We cannot 鍖nd the reactions at the supports using equilibrium alone. This is statically indeterminate. Test for Static Indeterminancy # Unknowns # Equations of Equilibrium If #Unknowns > #Equations Static Indeterminancy. 2. Lets try Force Deformation relationships. F1 = k1 隆1 F2 = k2 隆2 3. Now add compatibility. 7
- 8. 隆1 + 隆2 = 0 隆1 = uc x 隆2 = uc x 4. Solve equations. F1 = k1 隆1 = k1 uc x F2 = k2 隆2 = k2 uc x P k2 uc k1 uc = 0 x x P = (k1 + k2 )uc x P uc = x k1 + k2 What is the loadsharing? k1 P F1 = k1 + k2 k2 P F2 = k1 + k2 Check EXAMPLE 8
- 9. Q: Forces in springs A, B, C? Q: Find y(x). Assumes small deformations. Is this statically indeterminate? What are the unconstrained degrees of freedom (D.O.F)? 1. Vertical displacements Fy = 0 2. Rotation about z Mz = 0 What are the unknowns? FA ,FB ,FC 3 Unknowns. #Unknowns > #Equations This is statically indeterminate. 1. Equations of Equilibrium Fy = 0 P FA FB FC = 0 MA = 0 L P a FC FB L = 0 2 9
- 10. 2. Force Deformation Relationship FA = k隆A FB = k隆B FC = k隆C 3. Compatibility 隆A = uA Y 隆B 隆A tan = L 隆C 隆A tan = L/2 10