G.H.RAISONI COLLEGE OF ENGINEERING

CIVIL ENGINEERING DEPARTMENT

Presentation submitted by:Rahul Baraskar- 38Rajesh Sharma- 39Ritvik Chauhan- 40

Guided By: Pawade Sir

1

INTRODUCTION DETAILS OF SLOPE DEFLECTION METHOD PROCEDURE FOR SOLVING PROBLEMS SAMPLE PROBLEM

2

INTRODUCTIONINTRODUCTION

•In this method, unknown joint rotations and displacements are estimated or calculated.

•Final fixed end moments and final bending moment diagram can be estimated by calculating the unknown rotations and displacement.

3

Fig. 1 (a) Fig. 1 (b)

Fig. 1 (c)4

To clarify the concepts we will consider some examples:

• Beginning with the beam in Fig. 1(a). Here any load P applied to the beam will cause joint A only to rotate by amount A.

•While node B is completely restricted from moving as it if fixed. Hence the beam has only one unknown, A.

• The beam in Fig. 1(b) has joint at A, B, and C, and so has four unknowns, designed by the rotational displacements A, B, C, and the vertical displacement C.

5

Consider now the frame in Fig. 1(c). Again, an arbitrary loading P applied to the frame can cause nodes B and C to rotate nodes can be displaced horizontally by an equal amount. The frame therefore has three unknowns, A, B, C.

The unknowns here are also called as degree of freedom.

The degree of freedom is zero at fixed support.

The degree of freedom is present at simply support, roller support and hinged support.

6

For fig 1(a), the final moment at point A and B by slope deflection equation are:

MAB = MFAB +2EI/L [2A+ B ]

= MFAB +4EI A/L ( as B = 0)

MBA= MFBA+2EI/L [A+ 2B ]

= MFBA+2EI A/L ( as B = 0)

7

SLOPE AND DEFLECTION EQUATIONSSLOPE AND DEFLECTION EQUATIONS

For figure 1(b), the final moment at point B by slope deflection method is given as:

MBC = MFBC + 2EI/L [ 2B + C - 3 C/L]

(The term for displacement C is taken as negative if it is at the right side of support and sinks. And if it is to the left, it is taken as positive.)

8

STEP 1: Calculate fixed end moments for each joint considering each individual span as fixed from both ends.

STEP 2: Identify the unknowns or degree of freedom.

STEP 3: Construct the equation for condition of equilibrium about the point where degree of freedom is present.

9

STEP 4: Apply the slope and deflection equation equation at each point on the beam or structure.

STEP 5: Develop simultanious equations by using the equations for equilibrium.

STEP 6: Calculate the unknowns by solving the developed simultanious equations.

STEP 7: Substitute the values of unknowns in the original slope and deflection equations and calculate final fixed end moment.

10

PROBLEM: Find final fixed end moments for the given structure.

11

20 KN/m 80 KN

A B C4m 4m

(I) (I)

STEP 1:STEP 1: FIXED END MOMENTS-

1. MFAB = -WL2 /12 = -26.672. MFBA = WL2/ 12 = 26.673. MFBC = -WL/ 8 = -404. MFCB = WL/ 8 = 40

12

STEP 2:STEP 2: UNKNOWNS AND CONDITION OF EQUILIBRIUM-

A= C =0 (as A & C are fixed) B is unknown. (as B is simply supported)

CONDITION OF EQUILIBRIUM: ΣMB=0 MBA+MBC=0

13

STEP 3:STEP 3: SLOPE DEFLECTION EQUATION:

A. MAB= MFAB + 2EI/L [ 2A+ B+3 B/L] = -26.67 + 2EI/4 [B] (A = B = 0 ) = -26.67 + 0.5 B EISimilarly: B. MBA = 26.67 + B EI C. MBC = -40 + B EI D. MCB = 40 + 0.5B EI

14

STEP 4:STEP 4: CONSTRUCTION OF EQUATION OF EQUILIBRIUM AND ESTIMATION OF UNKNOWNS-

ΣMB=0 MBA+MBC=0 26.67 + B EI + -40 + B EI = 0 B=6.67/EI

15

STEP 5:STEP 5: SUBSTITUTING VALUES OF UNKNOWNS IN SLOPE DEFLECTION EQUATION:

MAB= -26.67 + .5 EI [ 6.67/EI] (as B = 6.67)

= -23.33 Similarly- MBA = 33.33 MBC= -33.33 MCB = 43.33

16

STEP 6: STEP 6: CALCULATION OF +VE BENDING MOMENT:

For span AB= WL2/8 = 40 KN-m For span BC = WL/4 = 80 KN-m

40 KN-m

80 KN-m

Hogging nature Hogging nature

23.33 KN-m 33.33 KN-m 43.33 KN-m

A B B C

17

Final Bending Moment Diagram:

-ve

+ve-ve -ve

+ve

-ve

23.33 KN-m

40 KN-m

80 KN-m

33.33 KN-m

43.33 KN-m

A B C

18

19