Slope Deflection Method

Slope eflection ethod Theory of Structures II Shahid ehmood epartment of ivil Engineering Swedish ollege of Engineering & Technology, Wah antt

Slope eflection ethod Previously we have discussed Force/Flexibility methods of analysis of statically indeterminate structures. In force method, the unknown redundant forces are determined first by solving the structure s compatibility equations; then other response characteristics ti of the structure are evaluated by equilibrium equations or superposition. n alternative approach can be used for analyzing is n alternative approach can be used for analyzing is termed the displacement or stiffness method.

Slope eflection ethod In displacement method, theunknown displacements are determined first by solving the structure s equilibrium equations; then the other response characteristics are evaluated through compatibility considerations and member force deformation relationships. The displacement methods includes Slope eflection ethod and oment istribution ethod. The slope deflection method was introduced by George The slope deflection method was introduced by George. aney in 1915. 3

Slope eflection ethod This method takes into account only the bending deformations. This method gives an understanding of the atrix Stiffness ethod, which forms the basis of most computer software currentlyused for structural t analysis. 4

Slope eflection Equations When a continuous beam or a frame is subjected to external loads, internal moments generally develop at the ends of its individual members. The slope deflection equations relate the moments at the ends of the member to the rotations and displacements of its end and the external loads applied to the member member. Let us consider an arbitrary member of the continuous beam. L 5

When the beam is subjected to external loads and support settlements, the member deforms as shown (exaggerated), and internal moments are induced at its ends. P w Un deformed position L eformed position (elastic curve) w P L EI constant 6

w L P ouble subscript notationisusedformemberendmoments, with the first subscript identifying the member end at which the moment acts and the second subscript indicating the other end of the member. denotes the moment at end of the member. denotes the moment at end of the member. 7

w P L Un deformed position Tangent at Elastic curve θ θ Tangent at θ & θ denote, respectively, the rotations of end and with respect to the un deformed (horizontal) position of the member. 8

w P L Un deformed position Tangent at Elastic curve θ θ Δ Tangent at Δ denotes the relative translation between the two ends of the Δ denotes the relative translation between the two ends of the member in the direction perpendicular to the un deformed axis of the member. 9

w P L Un deformed position Tangent at Elastic curve Ψ θ θ Δ Ψ ord Tangent at Ψ denotes the rotation of the member s chord (straight line Ψ denotes the rotation of the member s chord (straight line connecting the deformed positions of the member ends) due to the relative translation Δ. 10

w P L Un deformed position Tangent at Elastic curve Ψ θ θ Δ Ψ ord Tangent at Since the deformations are assumed to be small, the chord rotation can be expressed as ψ (1) 11 L

The sign convention used in thischapterh t isasfollows: The member end moments, end rotations, and chord rotation are positive when counterclockwise. Note that all the moments and rotations are shown in positive sense in figure on previous slide. The slope deflection equations can be derived by relating the member end moments to the end rotations and chord rotation by applying the second moment area theorem. 1

w P L Un deformed position Tangent at Δ θ Ψ ord Elastic curve Ψ θ Δ Δ Tangent at From figure we can see that θ + θ L + L () 13

y substituting Δ/LΨ into the preceding equation we have, θ ψ θ ψ L L () (3) Δ is tangential deviation of end from the tangent to the elastic curve at end and Δ is the tangential deviation of end from the tangent to the elastic curve at end. ccording to the second moment area theorem, the expressions for the tangential deviations Δ and Δ can be obtained by summing the moments about the ends and, respectively, of the area under /EI diagram between the two ends. 14

The bending moment diagrams for the member is constructed in parts by applying,, and the external loading separately on the member with simply supported ends. The three simple beam bending moment diagrams thus obtained are shown in Figure. L diagram (simple beam bending moment diagram due to external loads) 1 3 ending oment iagram 15

ssuming that the member is prismatic (EI is constant along the length of the member) we sum the moments of the area under the /EI diagram about the ends and, respectively, to determine the tangential deviations. 1 L L L L g EI 3 3 L L g (4a) 3EI 6EI EI 1 L L L L + + EI 3 3 L 6EI L 3EI + + g EI g (4b) 16

In which g and g are the moments about the ends and, respectively, of the area under the simple beam bending moment diagram due to external loading ( L diagram). The three terms in equations (4.a & 4.b) represent the tangential deviations due to,, and the external loading, acting separately on the member, with a negative term indicating that the corresponding tangential deviation i is in the directioni opposite to that shown on the elastic curve of the member. L 6EI L 3EI Tangential deviation due to 17

L 3EI L 6EI Tangential deviation due to w P g EI g EI Tangential deviation due to External Loading 18

y substituting the expressions for Δ and Δ into Eq. 3, we have L L g 3EI 6EI EIL L L g + + 6 EI 3 EI EIL θ ψ - - ( 5 a ) θ ψ (5b) To express the member end moments in terms of the end rotations, the chord rotation, and the external loading, we solve Eq. 5 simultaneously for and. Rewriting Eq. 5a as L L g 3EI 3EI EIL ( θ ψ ) 19

y substituting this equation into Eq. 5b and solving the resulting equation for,wehave EI θ + θ 3ψ + L L ( ) ( g g ) ( a) 6 and by substituting Eq. 6a into either Eq. 5a or 5b, we have EI θ + θ 3ψ + L L ( ) ( g g ) ( b) 6 It indicates that the moments develop at the ends of a member depend on the rotations and translations of member s ends as well as on the external loading applied between the ends. 0

Now, suppose that the member under consideration, instead being a part of a larger structure, was an isolated beam with both ends completely fixed against rotations and translations, as shown. w P FE θ θ Ψ 0 Fixed End oments FE The moments that would develop at the ends of such a fixed beam are referred to as fixed end end moments and their expression can be obtained by setting θ θ Ψ 0;thatis, FE FE ( g g ) ( a ) 7 L L ( g g ) ( 7b) 1

y comparing Eqs. 6 & 7, we find that the second terms on the right sides of Eqs. 6 are equal to the fixed end moments. EI θ + θ 3ψ L ( ) + FE ( a) 8 EI θ + θ 3ψ L ( ) + FE ( b) 8 Equations (8a &8b), which express the moments at the ends of a member in terms of its end rotations and translations for a specified external loading, are called slope deflections equations. These equations are valid for prismatic members, composed of linearly elastic material and subjected to small deformations. The deformations due to axial and shear forces are neglected.

The two slope deflection equations have the same form and either end of equations can be obtained from the other simply by switching the subscript and. EI nf n f ψ + L ( θ + θ 3ψ ) FE ( 9 ) nf in which the subscript n refers to the near end of the member where moment nf acts and the subscript f identifies the far (other) end of the member. 3

embers with One End Hinged The slope deflection equations derived previously are based on the condition that the member is rigidly connected to joints at both ends, so that the member end rotations θ and θ are equal to the rotations of the adjacent joints. When one of the member s ends is connected to the adjacent joint by a hinged connection, the moment at the hinged end must be zero. The slope deflections equations can be easily modified to reflect this condition. 4

With reference to the previous Figure of member, if the end of the member is hinged, then the moment at must be zero. y substituting 0into Equation (8), we write EI ψ L ( θ + θ 3 ) + FE ( a) 10 EI 0 ψ + L ( θ + θ 3ψ ) FE ( b ) 10 Solving Eq. (10) for θ, we obtain θ θ 3 L + ψ 4EI ( FE ) ( 11 ) 5

To determine θ from the slope deflection equations, we substitute btit t Eq. (11) into Eq. (10a), thus obtaining the modified slope deflection equations for member with a hinge at end. 3EI FE ψ L ( θ ) + FE ( a) 1 0 ( 1b) Similarly, it can be shown that for a member with a hinge at end, the rotation of the hinged end is given by θ θ 3 L + ψ 4EI ( FE ) ( 13 ) 6

ndthe modified d slope deflection equations can be expressed as 3EI ψ L FE ( θ ψ ) + FE ( a ) 14 0 ( 14b) ecause the modified slope deflection equations given by Eqs. (1) and(14)aresimilarinform,theycanbeconvenientlysummarized as 3 EI FE hr θr ψ + FE rh L ( ) ( a) rh 15 ( ) hr 0 15b 7

In which h the subscript r refers to the rigidly idl connected end of the member where the moment rh acts and the subscript h identifies the hinged end of the member. The rotation of the hinged end can now be written as θ h θr 3 + ψ L 4EI ( FE ) ( 16) hr 8

asic oncept of the Slope eflection ethod To illustrate the basic concept of the slope deflection method, consider the three span continuous beam shown in Figure below. 30 k 1.5 k/ft 0 ft 10 ft 10 ft 15 ft EI constant E 9,000 ksi I 500 in 4 lthough h the structure actually consists of a single continuous beam between the fixed supports and, for the purpose of analysis it is considered to be composed of three members,,, and, rigidly connected at joints,,, andlocated at the supports of the structure. 9

egrees of Freedom Identify the unknown independent d displacements (translations ti and rotations) of the joints of the structure. These unknown joint displacements are referred to as the degrees of freedom of the structure. From the qualitative deflected shape of the continuous beam shown in Figure below, we can see that none of its joints can translate. 1.5 k/ft θ 30 k θ θ 0 ft 10 ft 10 ft 15 ft θ The fixed joints and cannot rotate, t whereas joints and are free to rotate. 30

egrees of Freedom 1.5 k/ft 30 k θ θ θ 0 ft 10 ft 10 ft 15 ft θ This beam has two degrees of freedom, θ and θ, which represent the unknown rotations of joints and, respectively. The number of degrees of freedom is sometimes called the degree of kinematic indeterminacy of the structure. This beam is kinematically indeterminate to the second degree. structure without any degrees of freedom is termed kinematically determinate. 31

Equations of Equilibrium The unknown joint rotations ti are determined d by solving the equations of equilibrium of the joints that are free to rotate. The free body diagrams of the members and joints and of the continuous beam are shown. 0 ft 1.5 k/ft θ θ 30 k θ θ 10 ft 10 ft 15 ft 30 k 1.5 k/ft 3

Equations of Equilibrium Inaddition to the external loads, each member is subjected toan internal moment at each of its ends. The correct senses of the member end moments are not yet known, it is assumed that the moments at the ends of all the members are positive (counterclockwise). The free body diagrams of the joints show the member end moments acting in an opposite (clockwise) direction in accordance with Newton s law of action and reaction. 30 k 1.5 k/ft 33

Equations of Equilibrium ecause the entire structure t is in equilibrium, i each of its members and joints must also be in equilibrium. y applying the moment equilibrium equations 0and 0, respectively, to the free bodies of joints and, we obtain the equilibrium equations + + 0 0 (17a) (17b) 30 k 1.5 k/ft 34

Slope eflection Equations The equilibrium i equations Eqs. (17) can be expressed in terms of the unknown joint rotations, θ and θ, by using slope deflection equations that relate member end moments to the unknown joint rotations. efore we can write the slope deflection equations, we need to compute the fixed end d moments due to the external loads acting on the members of the continuous beam. To calculate the fixed endend moments, we apply imaginary clamps at joints and to prevent them from rotating. Or we generally provide fixed supports at the ends of each member to prevent the joint rotations as shown. 35

Slope eflection Equations 1.5 k/ft 30 k FE FE FE FE OR 1.5 k/ft 30 k FE FE FE FE FE FE The fixed end moments that develop at the ends of the members of this fully restrained or kinematically determinate structure can easily be evaluated by using the fixed end d moment expressions given inside the back cover of book. 36

Slope eflection Equations 1.5 k/ft 30 k 50 k ft 50 k ft FE FE OR 1.5 k/ft 30 k 50 k ft 50 k ft FE FE FE FE For member : wl 1.5( 0) FE FE 50k 1 1 ( 0) wl 1.5 50k 1 1 ft ft 37

Slope eflection Equations 1.5 k/ft 30 k 50 k ft 50 k ft 75 k ft 75 k ft OR 1.5 k/ft 30 k 50 k ft 50 k ft 75 k ft 75 k ft 0 0 For member : FE FE ( 0) PL 30 75k 8 8 PL 30( 0) 75k 8 8 ft ft 38

Slope eflection Equations The slope deflection equations for the three members of the continuous beam can now be written by using Eq. (9). Since none of the supports of the continuous beam translates, the chord rotations of the three members are zero (Ψ Ψ Ψ 0). lso, supports and are fixed, the rotations θ θ 0.y applying Eq. (9) for member, with as the near end and as the far end, we obtain the slope deflection equation EI θ 0 ( 0 + θ 0) + 50 0.1EI + 50 ( a) 18 Next, by considering as the near end and as the far end, we write EI θ + 0 0 50 0.EIθ 50 18b 39 0 ( ) ( )

Slope eflection Equations Similarly, il l by applying Eq. (9) for member,weobtain EI θ 18 0 ( θ + θ ) + 75 0. EI θ + 0.1 EI + 75 ( c ) EI θ 18 0 ( θ + θ ) 75 0. EI θ + 0.1 EI 75 ( d ) and for member, EI 18 15 ( θ ) 0.67 EI θ ( e ) EI 18 15 ( θ ) 0.133 EI θ ( f ) 40

Joint Rotations To determine the unknown joint rotations ti θ & θ,wesubstitute btit t the slope deflection equations Eqs. (18) into the joint equilibrium equations Eqs. (17) and solve the resulting systems of equations simultaneously for θ & θ. y substituting Eqs. (18b) and (18c) into Eq. (17a), we obtain ( 0. EIθ 50) + ( 0. EIθ + 0.1EIθ + 75) or 0. 4EIθ + 01. EIθ 5 0 ( 19a ) and by substituting Eqs. (18d) and (18e) into Eq. (17b), we get ( 0 EIθ + 01. EIθ 75). + 0.67EIθ 0 or 01. EIθ + 0. 467 EIθ 75 19 ( 19 b ) 41

Joint Rotations Solving Eqs. (19a) & (19b) simultaneously l for EIθ and EIθ,we obtain EIθ 108.46 k ft EIθ 183.8 k ft y substituting the numerical values of E 9,000 ksi 9,000(1) ksf and I 500 in. 4, we determine the rotations of joints and to be θ 0.011011rad or 0.011011rad θ 0.0018 rad 4

ember End oments The moments at the ends of the three members of the continuous beam can now be determined by substituting the numerical values of EIθ and EIθ into the slope deflection equations (Eqs. 18). ( 108. 46) + 50 39. ( 108. 46) 50 71.7 k-ft ( 108. 46) + 0.1( 183.8) + 75 71. 7 ( ) + 0.1( 108.46) 75 ( ) 49.1k-ft ( ) 4.4 0. 1 k-ft 0. 0. 0. 183.8 0. 67 183.8 0.133 183.8 k-ft or 717. k-ft k-ft 49.1k-ft or 49.1 k-ft 43

ember End oments To check thatt the solution of simultaneous equations (Eqs. 19) has been carried out correctly, the numerical values of member end moments should be substituted into the joint equilibrium equations (Eqs. 17). If the solution is correct, then the equilibrium equations should be satisfied. + 71.7 + 71.7 0 hecks + 49.1+ 49.1 0 hecks The member end moments just computed are shown on the free body diagrams of themembers and joints in Figure on next slide. 44

ember End oments 1.5 k/ft 71.7 30 k 49.1 49.1 4.4 39. 71.77 S 13.38 k S S S S S y ember End Shears The shear forces at the ends of members can now be determined by applying the equations of equilibrium to the free bodies of members. For member, y + 0 39. S ( 0) + 1.5( 0)( 10) 71.7 0 S 13.38 k 45

ember End Shears 1.5 k/ft 71.7 30 k 49.1 49.1 4.4 39. 71.77 S 16.6 k S S S S S 13.38 k For member, + F y 0 y 13.38 1.5 y ( 0) + S 0 S 16. 6 k 46

ember End Shears 1.5 k/ft 71.7 30 k 49.1 49.1 4.4 39. 71.77 S 16.6 k S 16.13 k S 13.87 k S S S 13.38 k For member, + 0 y 71.7 S ( 0) + 30( 10) y 49.1 0 S 16.13 13 k + Fy 0 16.13 30 + S 0 S 13.87 k 47

ember End Shears 1.5 k/ft 71.7 30 k 49.1 49.1 4.4 39. 71.77 S 16.6 k S 16.13 k S 13.87 k S 4.9 k S 4.9 k S 13.38 k For member, + 0 y 49.1 S ( 15) y + 4.4 0 S 4.9 k + Fy 0 4.9 + S 0 S 4.9 k 48

Support Reactions 1.5 k/ft 71.7 30 k 49.1 49.1 4.4 39. 16.66 k 16.13 k 71.77 S 16.6 k S 16.13 k S 13.87 k S 4.9 k S 4.9 k S 13.38 k y 3.75 k y From the free body diagram of joint, we can see that the vertical reaction at the roller support is equal to the sum of the shears at ends of member and ;thatis y S + S 16.6 + 16.13 3.75 k 49

Support Reactions 1.5 k/ft 71.7 30 k 49.1 49.1 4.4 39. 16.66 k 16.13 k 71.77 13.87 k 4.9 k S 16.6 k S 16.13 k S 13.87 k S 4.9 k S 4.9 k S 13.38 k y 3.75 k y 18.77 k The vertical reaction at the roller support equals the sum of shears at ends of members and. y S + S 13.87 + 4.9 18.77 k 50

Support Reactions 1.5 k/ft 30 k 3.75 k 18.77 k 51

Support Reactions 1.5 k/ft 30 k 39. k ft 13.38 k 3.75 k 18.77 k The reactions at the fixed support are equal to the shear and moment at the end of member. y S 13.38 k 39. k ft 5

Support Reactions 1.5 k/ft 30 k 4.4 k ft 39. k ft 13.38 k 3.75 k 18.77 k 4.9 k The reactions at the fixed support equal the shear and moment at end of the member. y S 4.9 k 4.44 k ft 53

Equilibrium heck 1.5 k/ft 30 k 4.4 k ft 39. k ft 13.38 k 3.75 k 18.77 k 4.9 k To check out computations of member end shears and support reactions, we apply the equations of equilibrium to the free body of fthe entire structure. t + Fy 0 ( 0 ) + 3.75 30 + 18.77 4.9 0 hecks 13.38 1.5 + 0 39. 13.38 18.77 15 ( 55 ) + 1.5 ( 0 )( 45 ) 3.75 ( 35 ) + 30 ( 5 ) ( ) + 4.4 0.1 0 hecks 54

Shear iagram 1.5 k/ft 30 k 4.4 k ft 39. k ft 13.38 k 3.75 k 18.77 k 4.9 k Using General sign conventions 13.38 16.13 4.9 F E 8.9 ft 16.6 13.87 55

oment iagram 1.5 k/ft 30 k 4.4 k ft 39. k ft 13.38 k 3.75 k 18.77 k 4.9 k Using General sign conventions 89.7 0.5 F 4.4 E 39. 71.7 49.1 56

1.5 k/ft 30 k 4.4 k ft 39. k ft 3.75 k 18.77 k 13.38 k 4.9 k 16.13 13.38 4.9 F E 89ft 8.9 16.6 89.7 13.87 0.5 F 4.4 E 39. 71.7 49.1 57

nalysis of ontinuous eam ased on the discussioni above, the procedure for the analysis of continuous beams can be summarized as follows: 1. Identify the degrees of freedom of structure.. ompute fixed end moments. 3. In case of support settlements, determine the chord rotations Ψ. 4. Write slope deflection equations. 5. Write equilibrium equations for each joint. 6. etermine the unknown joint rotations. 7. alculate member end moments by substituting the numerical values of joint rotations determined in step 6 into the slope deflection equations. 8. Satisfy the equilibrium equations for joints in step 5. 9. ompute member end shears. 10. etermine the support reactions by considering the equilibrium of joints. 11. Satisfy the equilibrium equations for end shears and support reactions. 1. raw shear and bending moment diagrams using the beam sign convention. 58

Structures with antilever Overhangs onsider a continuous beam with a cantilever overhang, as shown in the figure. w a ctual eam wa / w Statically eterminate antilever Portion S wa wa w wa / ctual eam 59

Example 1 etermine the reactions and draw the shear and bending moment diagrams for the two span continuous beam shown in Figure. 18 k k/ft 10 ft 15 ft 30 ft EI constant 60

Solution 1. egree of Freedom We can see that only joint of the beam is free to rotate. Thus, the structure has only one degree of freedom, which is the unknown joint rotation, θ. 18 k k/ft θ 10 ft 15 ft 30 ft EI constant 61

. Fixed End oments y using the fixed end moment expressions given inside the back cover of the book, we evaluate the fixed end moments due to the external loads for each member. ( )( 15 ) Pab 18 10 FE 64.8 k ft or + 64. 8 L 5 ( ) ( 15) Pa b 18 10 FE L 5 43. k ft or 64. 8 wl ( 30) FE 150 k ft 1 1 or + 150 k-ft FE 150 k ft or 150 k-ft kft k-ft k-ft kft ounterclockwise FE are positive, whereasclockwise FE are negative. 6

3. hord Rotations Since no support settlements occur, the chord rotations of both members are zero; that is, Ψ Ψ 0. ;, 4. Slope eflection Equations To relate the member end moments to the unknown joint rotation, θ,wewrite the slope deflection equation for the two members of the structure by applying Eq. (9). EI nf n f ψ + L ( θ + θ 3 ) FE ( 9) since the supports and are fixed, the rotations θ θ 0. nf 63

4. Slope eflection Equations Slope eflection Equation for ember EI ( θ ) + 64.8 0.08EIθ + 64.8 ( 1) 5 EI 5 ( θ ) 43. 0.16EIθ 43. ( ) Slope eflection Equation for ember EI 30 EI 30 ( θ ) + 150 0.133 EI θ + 150 ( 3 ) ( θ ) 150 0.06670667 EI θ 150 ( 4 ) 64

5. Equilibrium Equations The free body diagram of joint t isshown in Figure. 18 k k/ft 18 k k/ft ember end moments, which h are assumed to be in counterclockwise direction on the ends of members, must be applied in (opposite) clockwise direction on the free body of the joint in accordance with Newton s Third Law. 65

5. Equilibrium Equations The free body diagram of joint is shown in Figure. 18 k k/ft y applying the moment equilibrium equation 0to the free body of the joint, we obtain + + 0 (5) 66

6. Joint Rotations To determine the unknown joint rotations, θ, substitute the slope deflection equations (Eqs. & 3) into the equilibrium equation (Eq. 5). or from which ( 0.16 EI θ 43. ) + ( 0.133 EI θ + 150 ) 0 0.93EIθ 106.8 EIθ 364.5 k ft 67

7. ember End oments The member end moments can now be computed by substituting thenumericalvalueof EIθ back into the slope deflection equation (Eqs. 1 to 4). 0.08 0.16 0.133 0.0667 ( 364.5 ) + 64.8 35.6 k ft ( 364.5) 43. 101.5 k ft or 101. 5 ( 364.5) + 150 101.5 k ft ( 364.5) 150 174.3 k ft or 174.3 k-ft k-ft Positive answer for an end moment indicates that its sense is counterclockwise, whereasanegative answer implies a clockwise sense. s and are equal in magnitude but opposite in sense, the equilibrium equation + 0is satisfied. 68

18 k k/ft 35.6 101.5 101.5 101.5 101.5 174.3 8. ember End Shears The member end shears, obtained by considering the equilibrium of each member, are shown in figure below 18 k k/ft 101.5 35.6 101.5 101.5 101.5 174.3 8.16 9.84 7.57 3.43 69

9. Support Reactions The reactions at the fixed support and are equal to the forces and moments at the ends of the members connected to these joints. To determine the reaction at roller support, consider the equilibrium of the free body of joint in the vertical direction. y S + S 9.84 + 7.57 37.41k NS 18 k 9.84 7.57 k/ft 35.6 101.5 101.5 101.5 101.5 174.3 8.16 9.84 7.57 3.43 37.41 70

9. Support Reactions The support reactions are shown in figure below. 18 k k/ft 35.6 k ft 174.3 k ft 8.16 k 37.41 k 3.43 k 71

10. Equilibrium heck To check our calculations of member end shears and support reactions, we apply the equations of equilibrium to the free body of the entire structure. 18 k k/ft 35.6 k ft 174.3 k ft 37.41 k 8.16 k 3.43 k + Fy 0 8.16 18 + 37.41 ( 30 ) + 3.43 0 hecks 7

10. Equilibrium heck To check our calculations of member end shears and support reactions, we apply the equations of equilibrium to the free body of the entire structure. 18 k k/ft 35.6 k ft 174.3 k ft + 37.41 k 8.16 k 3.43 k 10 ft 15 ft 30 ft 0 ( 55 ) + 18 ( 45 ) 37.41 ( 30 ) + ( 30 )( 15 ) 174.3 0. 0 hecks 35.6 8.16 73

11. Shear Force iagram 18 k k/ft 35.6 k ft 174.3 k ft 8.16 k 37.41 k 3.43 k 7.57 8.16 16. ft E 9.84 3.43 74

11. ending oment iagram 18 k k/ft 35.6 k ft 174.3 k ft 8.16 k 37.41 k 3.43 k 88.7 46 E 35.6 101.5 174.3 75

Example etermine the reactions and draw the shear and bending moment diagrams for the continuous beam shown in Figure. 15 kn/m 60 kn 10 m 5m 5m I I I EI 00 GPa I 700 (10 6 )mm 4 76

Solution From figure we can see that all three joints of the beam are free to rotate. Thus the beam have 3 degrees of freedom, θ, θ, θ. The end supports and of the beam are simple supports at which no external moment is applied, the moments at the end of the member and at the end of the member must be zero. 15 kn/m 60 kn 0 0 77

Solution The ends and can be considered as hinged ends and the modified slope deflection equations can be used. 3EI FE rh r ψ rh L 15 hr 0 15b hr ( θ ) + FE ( a) ( ) The modified SE do not contain the rotations of the hinged ends, by using these equations the rotations θ,andθ of the simple supports can be eliminated, which will then involve only one unknown joint rotation, θ. 15 kn/m 60 kn 78

1. egree of Freedom. Fixed End oments θ ( ) 15 10 FE 15 kn m or 1 FE 15 kn m or 60( 10) 15( 10) FE + 00 kn m 8 1 + 15 kn-m 15 kn-m or + 00 FE 00 kn m or 00 kn-m kn-m 15 kn/m 60 kn 79

3. Slope eflection Equations Since both members of the beam have one end hinged, weuse Eqs. 15 to obtain the slope deflection equations for both members. 0 NS 3EI 10 3E 10 15 ( θ ) + 15 0.3 EI θ 187.5 ( 1 ) ( I ) 00 ( θ ) + 00 + 0.6 EI θ + 300 ( ) 0 NS 15 kn/m 60 kn 80

4. Equilibrium Equations y considering the moment equilibrium of the free body of joint, we obtain the equilibrium equation + 0 (3) 5. Joint Rotation To determine the unknown joint rotation θ we substitute the SE (Eqs. 1 &) into the equilibrium equations Eq.3to obtain 81

6. Joint Rotation or from which ( 0.3EIθ 187.5 ) + ( 0.6EIθ + 300 ) 0 0.9EIθ 11.5 EIθ 15 kn m 7. ember End oments The member end moments can now be computed by substituting the numerical value of EIθ into the slope deflection equations (Eqs. 1 & ). 8

8. ember End oments 0.3 0.6 ( 15 ) 187.5 5 kn m or 5 kn-m ( 15) + 300 5 kn m NS NS 9. ember End Shears and Support reactions 15 kn/m 97.5 17.5 60 kn 15 kn/m 5 5 5 5 5.5 97.5 17.5 8.5 y 5 15 kn/m 60 kn 5.5 kn 5 kn 8.5 kn 83

10. Equilibrium hecks 15 kn/m 60 kn 5.5 kn 5 kn 8.5 kn + Fy 0 ( 0 ) + 5 60 + 8.5 0 hecks 5.5 15 + 5.5 0 ( 0) + 15( 0)( 10) 5( 10) + 60( 5) 0 hecks 84

11. Shear Force & ending oment iagrams 60 kn 15 kn/m 5.5 kn 5 kn 8.5 kn 17.5 5.5 5.5 E 16. ft 3.5 m 7.5 97.5 8.5 85

11. Shear Force & ending oment iagrams 60 kn 15 kn/m 5.5 kn 5 kn 8.5 kn 5 91.88 E 5 86

Example 3 etermine the member end moments and reactions for the three span continuous beam shown, due to the uniformly distributed load and due to the support settlements of 5/8 in. at, and 1.5 in. at, and ¾ in. at. k/ft 0 ft 0 ft 0 ft EI 9,000 ksi I 7,800 in. 4 87

Solution 1. egree of Freedom Four joints of the beam are free to rotate, we will eliminate the rotations of simple supports at ends and and use the modified SE for member and respectively. The analysis will involve only two unknown joint rotations, θ and θ. k/ft 88

. Fixed End oments FE FE FE ( 0) 1 66.7 k ft or + 66.7 FE FE FE 66.7 k ft or 66. 7 k-ft kft k-ft 3. hord Rotations The specified support settlements are shown on a exaggerated scale. k/ft Ψ 5 in. 8 Ψ 1 1 in. Ψ 3 in. 4 89

3. hord Rotations k/ft 0 ft 0 ft 0 ft ψ 0.051051 0 0.006 0.079 ψ 0. 00365 0 ψ 1.5 0.75 ( 1) )0 0.00313 5 Ψ in. 8 1 1 i Ψ 1 in. Ψ 3 in. 4 90

4. Slope deflection Equations 0 3EI 10 EI 0 EI 0 NS ( θ + 0.006006 ) 100 0.15 EI θ + 0.0003900039 EI 100 ( 1 ) [ θ + θ 3 ( 0.00365 ) ] + 66. 7 0.EIθ + 0.1EIθ + 0. 0011EI [ θ + θ 3( 0.00365) ] 0.1 EI θ + 0. EI θ 3EI 0 0 + 0. 0011 EI + 66.7 66.7 66.7 ( ) ( 3 ) ( θ 0.00313) + 100 0.15EIθ 0.00047EI + 100 ( 4) NS 91

5. Equilibrium Equations + 0 (5) + 0 (6) () 6. Joint Rotations y substituting the slope deflection equations (Eqs. 1 4) intothe equilibrium equations (Eqs. 5 & 6), we obtain 0.35EIθ + 0.1EIθ 0.00149EI 01 0.1 EI θ + 0.35 EI θ 0.0006300063 EI + 33.3 33.33 substituting EI (9,000)(7,800)/(1) k ft into the right sides of the above equations yields 9

6. Joint Rotations 0.35EIθ 0.1EIθ + 0.1EIθ + 0.35EIθ,307.4 1,0.93 (7) (8) y solving Eqs. (7) and (8) simultaneously, we determine the values of EIθ and EIθ to be EIθ 6,68.81k ft EI θ 1131,131.57.81 k ft 7. ember End oments To compute the member end moments, substitute the numerical values of EIθ and EIθ back into the slope deflection equations (Eqs. 1 4) to obtain 93

7. ember End oments 47.7 k ft or 47 k-ft NS 47 k ft NS 808 k ft NS 808 k ft or 808 k-ft NS 8. ember End Shears and Support Reactions k/ft 41.38 81.79 k/ft 41.79 0.4 808 808 k/ft 47.7 47.7 1.38 41.38 81.79 41.79 y 13.17 17 y 6.19 y 0.4 60.4 94

8. ember End Shears and Support Reactions k/ft 41.38 81.79 k/ft 41.79 0.4 808 808 k/ft 47.7 47.7 1.38 41.38 81.79 41.79 y 13.17 17 y 6.19 y 0.4 60.4 k/ft 1.38 k 13.17 k 6.19 k 60.4 k 95

9. Shear and ending oment iagrams k/ft 1.38 k 13.17 k 6.19 k 60.4 k 81.79 41.79 1.38 0.4 41.38 60.4 96

9. Shear and ending oment iagrams k/ft 1.38 k 13.17 k 6.19 k 60.4 k 808. 47.6 97