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.

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