686 686
CHAPT HAPTER ER 16
Slop Slopee-De Defl flec ecti tion on Meth Method od
FIG. 16.16 (contd.)
frame to an arbitrary horizontal displacement D and draw a qualitative deflected shape of the frame, which is consistent with its support conditions as well as with our assumption that the members of the frame are inextensible. To draw the deflected shape, which is shown in Fig. 16.1 16.16(b 6(b), ), we first first imagi imagine ne that that the the me memb mber erss BD and CD are disdisconnected at joint D. Since member AC is assumed to be inextensible, joint C can move only in an arc about point A. Furthermo Furthermore, re, since the translation of joint C is assumed to be small, we can consider the arc to be a straight line perpendicular to member AC . Thus, in order to move joint C horizontally by a distance D, we must displace it in a direction perpendicular to member AC by a distance CC 0 (Fig. (Fig. 16.16(b 16.16(b)), )), so that the horiz horizont ontal al compon componen entt of CC 0 equals D. Note that although joint C is free to rotate, its rotation is ignored at this stage of the analysis, and the elastic curve AC 0 of member is drawn with the tangent at C 0 parallel to the undeformed direction AC is of the member. The member CD remains horizontal and translates as a rigid body into the position C 0 D1 with the displacement DD1 equal to CC 0 , as shown in the figure. Since the horizontal member CD is assumed to be inextensible and the translation of joint D is assumed to be small, the end D of this member can be moved from its deformed position D 1 only in the vertical direction. Similarly, since member BD is also
SECT SECTIO ION N 16.5 16.5
Anal Analys ysis is of Fram Frames es with with Side Sidesw sway ay
687 687
assumed to be inextensible, its end D can be moved only in the direction perpendicular to the member. Therefore, to obtain the deformed position of joint D, we move the end D of member CD from its deformed position D 1 in the vertical direction and the end D of member BD in the direction direction perpendi perpendicular cular to BD, until the two ends meet at point D 0 , where where they they are reconn reconnect ected ed to obtain obtain the displa displaced ced positio position n D 0 of joint D. By assuming that joint D does not rotate, we draw the elastic curves C 0 D 0 and BD 0 , respectively, of members CD and BD, to complete the deflected shape of the entire frame. The chord rotation of a member can be obtained by dividing the relative displacement between the two ends of the member in the direction perpendicular to the member, by the member’s length. Thus we can see from Fig. 16.16(b) that the chord rotations of the three members of the frame are given by CC 0 cAC ¼ L1
DD 0 cBD ¼ L2
D 1 D 0 cCD ¼ L
(16.26)
in which the chord rotations of members AC and BD B D are considered to be negative because they are clockwise (Fig. 16.16(c)). The three chord rotations can be expressed in terms of the joint displacement D by considering the displacement diagrams of joints C and D, shown in Fig. 16.16(b). Since CC 0 is perpendicular to AC , which is inclined at an angle b 1 with the vertical, CC 0 must make the same angle b 1 with the horizontal. Thus, from the displacement diagram of joint C (triangle (triangle CC 0 C 2 ), we can see that CC 0 ¼
D
(16.27)
cos b 1
Next, Next, let us consid consider er the displa displacem cement ent diagra diagram m of joint joint D (triangle 0 DD1 D ). It has been shown previously that DD1 is equal in magnitude and parallel to CC 0 . Therefore, DD2 ¼ DD 1 cos b 1 ¼ D
Since DD 0 is perpendicular to member BD, it makes an angle b 2 with the horizontal. Thus, from the displacement diagram of joint D , DD 0 ¼
DD2 D ¼ cos b 2 cos b 2
(16.28)
and D1 D 0 ¼ DD 1 sin b 1 þ DD 0 sin b 2 ¼
D
cos b 1
sin b 1 þ
D
cos b 2
sin b 2
or D1 D 0 ¼ D ðtan b 1 þ tan b 2 Þ
(16.29)
688
CHAPTER 16
Slope-Deflection Method
By substituting Eqs. (16.27) through (16.29) into Eq. (16.26), we obtain the chord rotations of the three members in terms of D: cAC ¼ cBD ¼ cCD ¼
D
L
D
(16.30a)
L1 cos b 1 D
(16.30b)
L2 cos b 2
ðtan b 1 þ tan b 2 Þ
(16.30c)
The foregoing expressions of chord rotations can be used to write the slope-deflection equations, thereby relating member end moments to the three unknown joint displacements, y C yD , and D. As in the case of the rectangular frames considered previously, the three equilibrium equations necessary for the solution of the unknown joint displacements can be established by summing the moments acting on joints C and D and by summing the horizontal forces acting on the entire frame. However, for frames with inclined legs, it is usually more convenient to establish the third equilibrium equation by summing the moments of all the forces and couples acting on the entire frame about a moment center O , which is located at the intersection of the longitudinal axes of the two inclined members, as shown in Fig. 16.16(d). The location of the moment center O can be determined by using the conditions (see Fig. 16.16(d)) ;
a1 cos b 1 ¼ a 2 cos b 2 a1 sin b 1 þ a 2 sin b 2 ¼ L
(16.31a) (16.31b)
By solving Eqs. (16.31a) and (16.31b) simultaneously for a1 and a2 , we obtain a1 ¼
L cos b 1 ðtan b 1 þ tan b 2 Þ
(16.32a)
a2 ¼
L cos b 2 ðtan b 1 þ tan b 2 Þ
(16.32b)
Once the equilibrium equations have been established, the analysis can be completed in the usual manner, as discussed previously.
Multistory Frames The foregoing method can be extended to the analysis of multistory frames subjected to sidesway, as illustrated by Example 16.12. However, because of the considerable amount of computational e¤ort involved, the analysis of such structures today is performed on computers using the matrix formulation of the displacement method presented in Chapter 18.
SECTION 16.5
Analysis of Frames with Sidesway
689
Example 16.10 Determine the member end moments and reactions for the frame shown in Fig. 16.17(a) by the slope-deflection method.
Solution Degrees of Freedom The degrees of freedom are y C yD , and D (see Fig. 16.17(b)). ;
FIG. 16.17 continued
690
CHAPTER 16
Slope-Deflection Method
FIG. 16.17 (contd.) continued
SECTION 16.5
Analysis of Frames with Sidesway
691
Fixed-End Moments By using the fixed-end moment expressions given inside the back cover of the book, we obtain
FEMCD ¼ FEMDC ¼
40 ð3Þð4Þ 2
ð7Þ
:
40 ð3Þ 2 ð4Þ
ð7Þ 2
’
¼ 39 2 kN m
2
¼ 29 4 kN m @ :
or
þ39 2 kN m
or
29 4 kN m
:
:
FEMAC ¼ FEM CA ¼ FEM BD ¼ FEM DB ¼ 0 Chord Rotations From Fig. 16.17(b), we can see that cAC ¼
D
cBD ¼
7
D
cCD ¼ 0
5
Slope-Deflection Equations
¼ 0 286 2 2 3 ¼ ¼ 0 571 7 7 2 ¼ 3 ¼ 0 4 5 5 2
2 EI D M AC ¼ yC 3 7 7 M CA
M BD
EI
EI
M DB ¼
EI
5
EI yC þ 0 122EI D
:
D
yC
EI yC þ 0 122EI D
:
D
yD
:
2yD 3
D
(1)
:
(2)
:
EI yD þ 0 24EI D
(3)
:
¼ 0 8EI yD þ 0 24EI D :
5
(4)
:
M CD ¼
2 EI ð2yC þ yD Þ þ 39 2 ¼ 0 571EI yC þ 0 286EI yD þ 39 2 7
(5)
M DC ¼
2 EI ðyC þ 2yD Þ 29 4 ¼ 0 286EI yC þ 0 571EI yD 29 4 7
(6)
:
:
:
:
:
:
:
:
Equilibrium Equations By considering the moment equilibrium of joints C and D, we obtain the equilibrium
equations M CA þ M CD ¼ 0
(7)
M DB þ M DC ¼ 0
(8)
P To establish the third equilibrium equation, we apply the force equilibrium equation
F X ¼ 0 to the free body of the
entire frame (Fig. 16.17(c)), to obtain
S AC þ S BD ¼ 0
in which S AC and S BD represent the shears at the lower ends of columns AC and BD, respectively, as shown in Fig. 16.17(c). To express the column end shears in terms of column end moments, we draw the free-body diagrams of the two columns (Fig. 16.17(d)) and sum the moments about the top of each column: S AC ¼
M AC þ M CA
7
and
S BD ¼
M BD þ M DB
5 continued
692
CHAPTER 16
Slope-Deflection Method
By substituting these equations into the third equilibrium equation, we obtain M AC þ M CA
7
þ
M BD þ M DB
5
¼ 0
which can be rewritten as 5ðM AC þ M CA Þ þ 7ðM BD þ M DB Þ ¼ 0
(9)
Joint Displacements To determine the unknown joint displacements yC yD , and D, we substitute the slopedeflection equations (Eqs. (1) through (6)) into the equilibrium equations (Eqs. (7) through (9)) to obtain ;
1 142EI yC þ 0 286EI yD þ 0 122EI D ¼ 39 2
(10)
0 286EI yC þ 1 371EI yD þ 0 24EI D ¼ 29 4
(11)
4 285EI yC þ 8 4EI yD þ 4 58EI D ¼ 0
(12)
:
:
:
:
:
:
:
:
:
:
:
Solving Eqs. (10) through (12) simultaneously yields EI yC ¼ 40 211 kN m 2 :
EI yD ¼ 34 24 kN m 2 :
EI D ¼ 25 177 kN m 3 :
Member End Moments By substituting the numerical values of EI yC EI yD , and EI D into the slope-deflection ;
equations (Eqs. (1) through (6)), we obtain M AC ¼ 14 6 kN m M CA ¼ 26 kN m M BD ¼ 7 7 kN m :
Ans.
:
26 kN m @
or
Ans.
’
Ans.
’
M DB ¼ 21 3 kN m :
M CD ¼ 26 kN m
14 6 kN m @
or
:
Ans.
’
M DC ¼ 21 3 kN m :
Ans.
21 3 kN m @
or
Ans.
:
To check that the solution of the simultaneous equations (Eqs. (10) through (12)) has been carried out correctly, we substitute the numerical values of member end moments back into the equilibrium equations (Eqs. (7) through (9)): M CA þ M CD ¼ 26 þ 26 ¼ 0
Checks
M DB þ M DC ¼ 21 3 21 3 ¼ 0
Checks
:
:
5ðM AC þ M CA Þ þ 7ðM BD þ M DB Þ ¼ 5 ð14 6 26Þ þ 7ð7 7 þ 21 3Þ ¼ 0 :
:
:
Checks
Member End Shears The member end shears, obtained by considering the equilibrium of each member, are shown
in Fig. 16.17(e). Member Axial Forces With end shears known, member axial forces can now be evaluated by considering the equilibrium of joints C and D . The axial forces thus obtained are shown in Fig. 16.17(e). Support Reactions See Fig. 16.17(f). Equilibrium Check The equilibrium equations check.
Ans.
SECTION 16.5
Analysis of Frames with Sidesway
693
Example 16.11 Determine the member end moments and reactions for the frame shown in Fig. 16.18(a) by the slope-deflection method.
Solution Degrees of Freedom Degrees of freedom are y C yD , and ;
D.
Fixed-End Moments Since no external loads are applied to the members, the fixed-end moments are zero.
FIG. 16.18
continued
694
CHAPTER 16
Slope-Deflection Method
FIG. 16.18 (contd.) continued
SECTION 16.5
Analysis of Frames with Sidesway
695
FIG. 16.18 (contd.)
Chord Rotations From Fig. 16.18(b), we can see that
5
CC 0
4 ¼ cAC ¼ 20 20 cBD ¼
cCD ¼
DD 0
16
C 0 C 1
20
¼
¼
D
16
3 4 20
D
¼ 0 0625D :
¼ 0 0625D :
D
¼ 0 0375D :
continued
696
CHAPTER 16
Slope-Deflection Method
Slope-Deflection Equations M AC ¼
2 EI ½yC 3ð0 0625DÞ ¼ 0 1EI yC þ 0 0188EI D 20
(1)
M CA ¼
2 EI ½2yC 3ð0 0625DÞ ¼ 0 2EI yC þ 0 0188EI D 20
(2)
M BD ¼
2 EI ½yD 3ð0 0625DÞ ¼ 0 125EI yD þ 0 0234EI D 16
(3)
M DB ¼
2 EI ½2yD 3ð0 0625DÞ ¼ 0 25EI yD þ 0 0234EI D 16
(4)
M CD ¼
2 EI ½2yC þ yD 3ð0 0375DÞ ¼ 0 2EI yC þ 0 1EI yD 0 0113EI D 20
(5)
M DC ¼
2 EI ½2yD þ yC 3ð0 0375DÞ ¼ 0 2EI yD þ 0 1EI yC 0 0113EI D 20
(6)
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
Equilibrium Equations By considering the moment equilibrium of joints C and D, we obtain the equilibrium equations M CA þ M CD ¼ 0
(7)
M DB þ M DC ¼ 0
(8)
The third equilibrium equation is established by summing the moments of all the forces and couples acting on the free body of the entire frame about point O , which is located at the intersection of the longitudinal axes of the two columns, as shown in Fig. 16.18(c). Thus
þ ’
P
M O ¼ 0
M AC S AC ð53 33Þ þ M BD S BD ð42 67Þ þ 30ð26 67Þ ¼ 0 :
:
:
in which the shears at the lower ends of the columns can be expressed in terms of column end moments as (see Fig. 16.18(d)) S AC ¼
M AC þ M CA
S BD ¼
and
20
M BD þ M DB
16
By substituting these expressions into the third equilibrium equation, we obtain 1 67M AC þ 2 67M CA þ 1 67M BD þ 2 67M DB ¼ 800 :
:
:
:
(9)
Joint Displacements Substitution of the slope-deflection equations (Eqs. (1) through (6)) into the equilibrium
equations (Eqs. (7) through (9)) yields 0 4EI yC þ 0 1EI yD þ 0 0075EI D ¼ 0
(10)
0 1EI yC þ 0 45EI yD þ 0 0121EI D ¼ 0
(11)
0 71EI yC þ 0 877EI yD þ 0 183EI D ¼ 800
(12)
:
:
:
:
:
:
:
:
:
By solving Eqs. (10) through (12) simultaneously, we determine EI yC ¼ 66 648 k-ft 2 :
EI yD ¼ 125 912 k-ft 2 :
EI D ¼ 5 233 6 k-ft 3 ;
:
continued
SECTION 16.5
Analysis of Frames with Sidesway
697
Member End Moments By substituting the numerical values of EI yC EI yD , and EI D into the slope-deflection ;
equations (Eqs. (1) through (6)), we obtain M AC ¼ 91 7 k-ft
’
M CA ¼ 85 1 k-ft
’
:
:
M BD ¼ 106 7 k-ft ’ M DB ¼ 91 k-ft :
Ans. Ans.
’
Ans. Ans.
M CD ¼ 85 1 k-ft
or
:
M DC ¼ 91 k-ft
or
85 1 k-ft @ :
Ans.
91 k-ft @
Ans.
Back substitution of the numerical values of member end moments into the equilibrium equations yields M CA þ M CD ¼ 85 1 85 1 ¼ 0
Checks
M DB þ M DC ¼ 91 91 ¼ 0
Checks
:
:
1 67M AC þ 2 67M CA þ 1 67M BD þ 2 67M DB ¼ 1 67ð91 7Þ þ 2 67ð85 1Þ :
:
:
:
:
:
:
:
þ 1 67ð106 7Þ þ 2 67ð91Þ :
:
¼ 801 5 & 800 :
:
Checks
Member End Shears and Axial Forces See Fig. 16.18(e). Support Reactions See Fig. 16.18(f).
Ans.
Equilibrium Check The equilibrium equations check.
Example 16.12 Determine the member end moments, the support reactions, and the horizontal deflection of joint F of the two-story frame shown in Fig. 16.19(a) by the slope-deflection method.
Solution Degrees of Freedom From Fig. 16.19(a), we can see that the joints C D E , and F of the frame are free to rotate, and translate in the horizontal direction. As shown in Fig. 16.19(b), the horizontal displacement of the first-story joints C and D is designated as D1 , whereas the horizontal displacement of the second-story joints E and F is expressed as D1 þ D2 , with D2 representing the displacement of the second-story joints relative to the first-story joints. Thus, the frame has six degrees of freedom—that is, y C yD yE yF D1 , and D2 . ;
;
;
;
;
;
Fixed-End Moments The nonzero fixed-end moments are
FEMCD ¼ FEM EF ¼ 200 k-ft FEMDC ¼ FEM FE ¼ 200 k-ft
continued
698
CHAPTER 16
Slope-Deflection Method
Chord Rotations See Fig. 16.19(b). cAC ¼ cBD ¼ cCE ¼ c DF ¼
D1
20 D2
20
cCD ¼ c EF ¼ 0 Slope-Deflection Equations Using I column ¼ I and I girder ¼ 2 I , we write M AC ¼ 0 1EI yC þ 0 015EI D1
(1)
M CA ¼ 0 2EI yC þ 0 015EI D1
(2)
M BD ¼ 0 1EI yD þ 0 015EI D1
(3)
M DB ¼ 0 2EI yD þ 0 015EI D1
(4)
:
:
:
:
:
:
:
:
M CE ¼ 0 2EI yC þ 0 1EI yE þ 0 015EI D2
(5)
M EC ¼ 0 2EI yE þ 0 1EI yC þ 0 015EI D2
(6)
:
:
:
:
:
:
FIG. 16.19 continued
SECTION 16.5
Analysis of Frames with Sidesway
699
FIG. 16.19 (contd.) continued
700
CHAPTER 16
Slope-Deflection Method
M DF ¼ 0 2EI yD þ 0 1EI yF þ 0 015EI D2
(7)
M FD ¼ 0 2EI yF þ 0 1EI yD þ 0 015EI D2
(8)
M CD ¼ 0 2EI yC þ 0 1EI yD þ 200
(9)
M DC ¼ 0 2EI yD þ 0 1EI yC 200
(10)
M EF ¼ 0 2EI yE þ 0 1EI yF þ 200
(11)
M FE ¼ 0 2EI yF þ 0 1EI yE 200
(12)
:
:
:
:
:
:
:
:
:
:
:
:
:
:
Equilibrium Equations By considering the moment equilibrium of joints C , D , E , and F , we obtain M CA þ M CD þ M CE ¼ 0
(13)
M DB þ M DC þ M DF ¼ 0
(14)
M EC þ M EF ¼ 0
(15)
M FD þ M FE ¼ 0
(16)
To establish the remaining two equilibrium equations, we successively pass a horizontal section just above the lower ends of the columns of each story of the frame and apply the equation of horizontal equilibrium ð F X ¼ 0 Þ to the free body of the portion of the frame above the section. The free-body diagrams thus obtained are shown in Fig. 16.19(c) and (d). By applying the equilibrium equation F X ¼ 0 to the top story of the frame (Fig. 16.19(c)), we obtain
P
P
Similarly, by applying
P
S CE þ S DF ¼ 10
F X ¼ 0 to the entire frame (Fig. 16.19(d)), we write S AC þ S BD ¼ 30
By expressing column end shears in terms of column end moments as S AC ¼ S CE ¼
M AC þ M CA
S BD ¼
20 M CE þ M EC
S DF ¼
20
M BD þ M DB
20 M DF þ M FD
20
and by substituting these expressions into the force equilibrium equations, we obtain M CE þ M EC þ M DF þ M FD ¼ 200
(17)
M AC þ M CA þ M BD þ M DB ¼ 600
(18)
Joint Displacements Substitution of the slope-deflection equations (Eqs. (1) through (12)) into the equilibrium equations (Eqs. (13) through (18)) yields
0 6EI yC þ 0 1EI yD þ 0 1EI yE þ 0 015EI D1 þ 0 015EI D2 ¼ 200
(19)
0 1EI yC þ 0 6EI yD þ 0 1EI yF þ 0 015EI D1 þ 0 015EI D2 ¼ 200
(20)
:
:
:
:
:
:
:
:
:
:
0 1EI yC þ 0 4EI yE þ 0 1EI yF þ 0 015EI D2 ¼ 200
(21)
0 1EI yD þ 0 1EI yE þ 0 4EI yF þ 0 015EI D2 ¼ 200
(22)
0 3EI yC þ 0 3EI yD þ 0 3EI yE þ 0 3EI yF þ 0 06EI D2 ¼ 200
(23)
0 1EI yC þ 0 1EI yD þ 0 02EI D1 ¼ 200
(24)
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
continued
SECTION 16.5
Analysis of Frames with Sidesway
701
By solving Eqs. (19) through (24) by the Gauss-Jordan elimination method (Appendix B), we determine EI yC ¼ 812 988 k-ft 2 :
EI yD ¼ 241 556 k-ft 2 :
EI yE ¼ 789 612 k-ft 2 :
EI yF ¼ 353 248 k-ft 2 :
EI D1 ¼ 15 272 728 k-ft 3
or
D1 ¼ 0 :0758
ft ¼ 0 91 in !
EI D2 ¼ 10 787 878 k-ft 3
or
D2 ¼ 0 :0536
ft ¼ 0 643 in !
;
:
;
:
:
:
:
:
Thus, the horizontal deflection of joint F of the frame is as follows: DF
¼ D 1 þ D2 ¼ 0 91 þ 0 643 ¼ 1 553 in ! :
:
:
:
Ans.
Member End Moments By substituting the numerical values of the joint displacements into the slope-deflection equations (Eqs. (1) through (12)), we obtain ’ M AC ¼ 147 8 k-ft Ans. :
’
M CA ¼ 66 5 k-ft :
Ans.
M BD ¼ 204 9 k-ft
’
Ans.
M DB ¼ 180 8 k-ft
’
Ans.
:
:
M CE ¼ 79 7 k-ft :
M EC ¼ 77 4 k-ft :
or
79 7 k-ft @
Ans.
or
77 4 k-ft @
Ans.
:
:
M DF ¼ 148 8 k-ft
’
Ans.
M FD ¼ 208 3 k-ft
’
Ans.
:
:
’
M CD ¼ 13 2 k-ft :
Ans.
M DC ¼ 329 6 k-ft ’ M EF ¼ 77 4 k-ft
or
M FE ¼ 208 3 k-ft
or
:
329 6 k-ft @ :
Ans. Ans.
:
:
208 3 k-ft @ :
Ans.
Back substitution of the numerical values of member end moments into the equilibrium equations yields M CA þ M CD þ M CE ¼ 66 5 þ 13 2 79 7 ¼ 0
Checks
M DB þ M DC þ M DF ¼ 180 8 329 6 þ 148 8 ¼ 0
Checks
M EC þ M EF ¼ 77 4 þ 77 4 ¼ 0
Checks
M FD þ M FE ¼ 208 3 208 3 ¼ 0
Checks
:
:
:
:
:
:
:
:
:
:
M CE þ M EC þ M DF þ M FD ¼ 79 7 77 4 þ 148 8 þ 208 3 ¼ 200
Checks
M AC þ M CA þ M BD þ M DB ¼ 147 8 þ 66 5 þ 204 9 þ 180 8 ¼ 600
Checks
:
:
:
:
:
:
:
:
Member End Shears and Axial Forces See Fig. 16.19(e). Support Reactions See Fig. 16.19(f). Equilibrium Check The equilibrium equations check.
Ans.
702
CHAPTER 16
Slope-Deflection Method
SUMMARY In this chapter, we have studied a classical formulation of the displacement (sti¤ness) method, called the slope-deflection method, for the analysis of beams and frames. The method is based on the slope-deflection equation: M nf ¼
2 EI L
ð2yn þ y f 3cÞ þ FEMnf
(16.9)
which relates the moments at the ends of a member to the rotations and displacements of its ends and the external loads applied to the member. The procedure for analysis essentially involves (1) identifying the unknown joint displacements (degrees of freedom) of the structure; (2) for each member, writing slope-deflection equations relating member end moments to the unknown joint displacements; (3) establishing the equations of equilibrium of the structure in terms of member end moments; (4) substituting the slope-deflection equations into the equilibrium equations and solving the resulting system of equations to determine the unknown joint displacements; and (5) computing member end moments by substituting the values of joint displacements back into the slope-deflection equations. Once member end moments have been evaluated, member end shears and axial forces, and support reactions, can be determined through equilibrium considerations.
PROBLEMS Section 16.3 16.1 through 16.5 Determine the reactions and draw the
16.6 Solve Problem 16.2 for the loading shown in Fig. P16.2 and a settlement of 12 in. at support B .
shear and bending moment diagrams for the beams shown in Figs. P16.1–P16.5 by using the slope-deflection method.
20 k 3 k/ft
1.5 k/ft A
15 ft E =
FIG. P16.1
C
B
FIG. P16.2,
15 ft 29,000 ksi
P16.6
20 ft I =
1,650 in. 4
Problems
16.7 Solve Problem 16.4 for the loading shown in Fig. P16.4 and the support settlements of 50 mm at B and 25 mm at C .
703
16.8 through 16.14 Determine the reactions and draw the
shear and bending moment diagrams for the beams shown in Figs. P16.8–P16.14 by using the slope-deflection method.
1.5 k/ft A
D B
25 ft
C
20 ft EI = constant
25 ft
FIG. P16.8 FIG. P16.3
25 kN/m A
C B
8m E =
FIG. P16.4,
8m 70 GPa
I =
1,300 (10 6) mm4
P16.7
FIG. P16.9,
P16.15
3 k/ft A
C B
25 ft 2 I
15 ft I
E = 29,000 ksi I = 2,500 in. 4
FIG. P16.5
FIG. P16.10
35 k
1 k/ft
2 k/ft E
A
B
10 ft
C
10 ft
D
10 ft EI =
FIG. P16.11
constant
20 ft
704
CHAPTER 16
Slope-Deflection Method
120 kN A
120 kN C
150 kN E G
B
6m
D
4m
6m
F
4m 2 I
I
4m
4m I
E = I =
FIG. P16.12,
200 Gpa 500 (106) mm4
P16.16
FIG. P16.17,
P16.21
FIG. P16.18,
P16.22
FIG. P16.13
3 k/ft C
E D
FIG. P16.14
10 ft
16.15 Solve Problem 16.9 for the loading shown in Fig. P16.9 and a settlement of 25 mm at support C .
I B
16.16 Solve Problem 16.12 for the loading shown in Fig. P16.12 and support settlements of 10 mm at A ; 65 mm at C ; 40 mm at E ; and 25 mm at G .
15 k 5 ft
A
20 ft
Section 16.4
5 ft
16.17 through 16.20 Determine the member end moments
2 I
and reactions for the frames shown in Figs. P16.17–P16.20 by using the slope-deflection method.
E =
FIG. P16.19
constant
Problems
30 kN/m C
705
2 k/ft C
25 k
D
B
8m 20 ft A
B
10 m EI =
A
constant
FIG. P16.20
15 ft
16.21 Solve Problem 16.17 for the loading shown in Fig. P16.17 and a settlement of 50 mm at support D .
EI =
constant
FIG. P16.24
16.22 Solve Problem 16.18 for the loading shown in Fig. P16.18 and a settlement of 14 in. at support A . 16.23 Determine the member end moments and reactions
for the frame in Fig. P16.23 for the loading shown and the support settlements of 1 in. at A and 112 in. at D. Use the slope-deflection method.
FIG. P16.25
3 k/ft 40 k
C
D
15 ft FIG. P16.23 A
Section 16.5
30 ft
16.24 through 16.31 Determine the member end moments
and reactions for the frames shown in Figs. P16.24–P16.31 by using the slope-deflection method.
B
EI =
FIG. P16.26
constant
