Solving statically indeterminate structure slope deflection 10.01.03.019

Prestress Concrete Design Sessional (CE 416 ) Course Teacher: Galib Muktadir Sabrina Nasrin

Welcome to my Presentation

Name: Faris ImamId: 10.01.03.019Year: 4th

Semester: 2nd

Section: A

Topic

Solving Statically Indeterminate Structure: Slope deflection

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.

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.

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. 

Statically determinate Statically indeterminate

Equilibrium equations could be directly solved, and thus forces could be calculated

in an easy way

Equilibrium equations could be solved only when

coupled with physical law 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

CLASSIFICATION OF STRUCTURAL ANALYSIS PROBLEMS

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.

Slope Deflection EquationConsider a beam segment AB having end relations θA & θB

and relative displacement Δ as shown below-

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 = MF

AB + 2EI/L (2θA + θB)MBA = MF

BA + 2EI/L (2θB + θA)

Analysis Steps:1. Express the fixed end moment due to loads2. Express the end moment in terms of the end rotations

and relative displacement.3. Consider the condition of equilibrium of the joint

equations4. Solve for unknown rotations and displacements5. Find the end moments from slope deflection equation.

Formula

Calculation

Find the end moments of the continuous beam by slope deflection method.

MFAB = +.5* 102/ 12 = 4.17 k-ft

MFBA = -4.17 k-ft

MF BC = + (5*6*42)/ 102 = +4.80 k-ftMFCB = -(5*4* 62)/ 102 = -7.20 k-ftMF CD = + 1* 3 = 3 k-ft

Member EquationMAB = + 4.17 + 2EI/10 (2θA + θB) = .4θA + .2 θB + 4.17Similarly,MBA = .2 θA + .4θB -4.17MBC = .4 θB +.2θC + 4.80MCB = .2θB +.4θC -7.2

Joint EquationΣMA = 0 or, .4θA + .2θB + 4.17 = 0ΣMB = MBA + MBC = .2θA + .8θB + .2θC +.63 = 0ΣMC = 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

Now,Putting this value into member equation,MBA = .2 * (-9.88) + .4 *1.083 -4.17 = -6.57 k-ftMBC = .4 (-1.083) +.2 (11.04) + 4.80 =6.57 k-ftMCB = .2* (-1.083) +.4 * (11.04) -7.2 = -3 k-ft

THANK YOU