Solving statically indeterminate structure slope deflection 10.01.03.019
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This document summarizes the slope deflection method for analyzing statically indeterminate structures. It defines statically indeterminate structures as those where the equilibrium equations are insufficient to determine internal forces. The slope deflection method relates member end moments to end rotations and displacements using slope deflection equations. The analysis process involves expressing fixed end moments, developing member equations relating end moments to rotations/displacements, establishing joint equilibrium equations, solving the equations to find unknown rotations and displacements, and determining end moments. An example problem applies these steps to solve for end moments in a continuous beam using the slope deflection method.
Solving statically indeterminate structure slope deflection 10.01.03.019
- 1. Prestress Concrete Design Sessional (CE 416 ) Course Teacher: Galib Muktadir Sabrina Nasrin
- 2. Welcome to my Presentation Name: Faris Imam Id: 10.01.03.019 Year: 4th Semester: 2nd Section: A
- 3. Topic Solving Statically Indeterminate Structure: Slope deflection
- 4. What is Statically Indeterminate Structure: In statics, a structure is statically indeterminate when the static equilibrium equations are insufficient for determining the internal forces and reactions on that structure. Based on Newton's laws of motion, the equilibrium equations available for a two-dimensional body are : the vectorial sum of the forces acting on the body equals zero. This translates to Σ H = 0: the sum of the horizontal components of the forces equals zero; Σ V = 0: the sum of the vertical components of forces equals zero; : the sum of the moments (about an arbitrary point) of all forces equals zero.
- 5. In the beam, the four unknown reactions are VA, VB, VC and HA. The equilibrium equations are: Σ V = 0: VA − Fv + VB + VC = 0 Σ H = 0: HA − Fh = 0 Σ MA = 0: Fv · a − VB · (a + b) - VC · (a + b + c) = 0.
- 6. Since there are four unknown forces (or variable) (VA, VB, VC and HA) but only three equilibrium equations, this system of simultaneous equations does not have a unique solution. The structure is therefore classified as statically indeterminate.
- 7. CLASSIFICATION OF STRUCTURAL ANALYSIS PROBLEMS Statically determinate Statically indeterminate Equilibrium equations could Equilibrium equations could be directly solved, and thus be solved only when forces could be calculated coupled with physical law in an easy way and compatibility equations Not survivable, moderately used in modern aviation (due to damage tolerance requirement) Survivable, widely used in modern aviation (due to damage tolerance property) Easy to manufacture Hard to manufacture
- 8. What is Slope Deflection Method? In the slope-deflection method, the relationship is established between moments at the ends of the members and the corresponding rotations and displacements. This method was developed by Axel Bendexon in Germany in 1934. This method is applicable for the analysis of statically indeterminate beams or rigid frames.
- 9. Slope Deflection Equation Consider a beam segment AB having end relations θA & θB and relative displacement Δ as shown below-
- 10. Slope deflection equation relates the moment acting on the ends of a member with the end rotations and relative displacement . MAB = MFAB + 2EI/L (2θA + θB + 3Δ/L) MBA = MFBA + 2EI/L (2θB + θA + 3Δ/L) If relative displacement Δ is zero, then – MAB = MFAB + 2EI/L (2θA + θB) MBA = MFBA + 2EI/L (2θB + θA)
- 11. Analysis Steps: 1. Express the fixed end moment due to loads 2. Express the end moment in terms of the end rotations and relative displacement. 3. Consider the condition of equilibrium of the joint equations 4. Solve for unknown rotations and displacements 5. Find the end moments from slope deflection equation.
- 12. Formula
- 13. Calculation Find the end moments of the continuous beam by slope deflection method.
- 14. MFAB = +.5* 102/ 12 = 4.17 k-ft MFBA = -4.17 k-ft MF BC = + (5*6*42)/ 102 = +4.80 k-ft MFCB = -(5*4* 62)/ 102 = -7.20 k-ft MF CD = + 1* 3 = 3 k-ft
- 15. Member Equation MAB = + 4.17 + 2EI/10 (2θA + θB) = .4θA + .2 θB + 4.17 Similarly, MBA = .2 θA + .4θB -4.17 MBC = .4 θB +.2θC + 4.80 MCB = .2θB +.4θC -7.2
- 16. Joint Equation ΣMA = 0 or, .4θA + .2θB + 4.17 = 0 ΣMB = MBA + MBC = .2θA + .8θB + .2θC +.63 = 0 ΣMC = MCB +MCD = (.2θB +.4 θC -7.2) +3 = 0 = .2 θB + .4 θC -4.2 = 0 Solving these equation by calculator, θA = -9.88 ; θB = 1.083 ; θC = 11.04
- 17. Now, Putting this value into member equation, MBA = .2 * (-9.88) + .4 *1.083 -4.17 = -6.57 k-ft MBC = .4 (-1.083) +.2 (11.04) + 4.80 =6.57 k-ft MCB = .2* (-1.083) +.4 * (11.04) -7.2 = -3 k-ft
- 18. THANK YOU