701_aKassimali Structural Analysis 4th US&SI Txtbk

8/18/2019 701_aKassimali Structural Analysis 4th US&SI Txtbk

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frame to an arbitrary horizontal displacement D and draw a qualitativedeflected shape of the frame, which is consistent with its support con-ditions as well as with our assumption that the members of the frame

are inextensible. To draw the deflected shape, which is shown in Fig.16.16(b), we first imagine that the members BD and CD are dis-connected at joint D. Since member AC is assumed to be inextensible,

joint C can move only in an arc about point A. Furthermore, since thetranslation of joint C is assumed to be small, we can consider the arc tobe a straight line perpendicular to member AC .

Thus, in order to move joint C horizontally by a distance D, wemust displace it in a direction perpendicular to member AC by a dis-tance CC 0 (Fig. 16.16(b)), so that the horizontal component of CC 0

equals D. Note that although joint C is free to rotate, its rotation is ig-nored at this stage of the analysis, and the elastic curve AC 0 of memberAC is drawn with the tangent at C 0 parallel to the undeformed directionof the member. The member CD remains horizontal and translates as arigid body into the position C 0D1 with the displacement DD1 equal toCC 0, as shown in the figure. Since the horizontal member CD is as-sumed to be inextensible and the translation of joint D is assumed to besmall, the end D of this member can be moved from its deformed posi-

tion D1 only in the vertical direction. Similarly, since member BD is also

FIG. 16.16 (contd.)

686 CHAPTER 16 Slope-D ef lect ion Met hod


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assumed to be inextensible, its end D can be moved only in the directionperpendicular to the member. Therefore, to obtain the deformed posi-tion of joint D, we move the end D of member CD from its deformed

position D1 in the vertical direction and the end D of member BD in thedirection perpendicular to BD, until the two ends meet at point D 0,where they are reconnected to obtain the displaced position D 0 of

joint D. By assuming that joint D does not rotate, we draw the elasticcurves C 0D 0 and BD 0, respectively, of members CD and BD, to com-plete the deflected shape of the entire frame.

The chord rotation of a member can be obtained by dividing therelative displacement between the two ends of the member in the direc-tion perpendicular to the member, by the member’s length. Thus we cansee from Fig. 16.16(b) that the chord rotations of the three members of the frame are given by

cAC ¼ CC 0

L1cBD ¼

DD 0

L2cCD ¼

D1D0

L (16.26)

in which the chord rotations of members AC and BD are considered tobe negative because they are clockwise (Fig. 16.16(c)). The three chord

rotations can be expressed in terms of the joint displacement D

by con-sidering 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 horizon-tal. Thus, from the displacement diagram of joint C (triangle CC 0C 2), wecan see that

CC 0 ¼ D

cos b 1(16.27)

Next, let us consider the displacement diagram of joint D (triangleDD1D

0). It has been shown previously that DD1 is equal in magnitudeand parallel to CC 0. Therefore,

DD2 ¼ DD1 cos b 1 ¼ D

Since DD 0 is perpendicular to member BD, it makes an angle b 2 withthe horizontal. Thus, from the displacement diagram of joint D,

DD 0 ¼ DD2cos b 2

¼ Dcos b 2

(16.28)

and

D1D0 ¼ DD1 sin b 1 þ DD

0 sin b 2 ¼ D

cos b 1sin b 1 þ

D

cos b 2sin b 2

or

D1D0

¼ Dðtan b 1 þ tan b 2Þ (16.29)

SECTION 16.5 Analysis of Frames with Sidesway 687


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By substituting Eqs. (16.27) through (16.29) into Eq. (16.26), we obtainthe chord rotations of the three members in terms of D:

cAC ¼ D

L1 cos b 1 (16.30a)

cBD ¼ D

L2 cos b 2(16.30b)

cCD ¼ D

Lðtan b 1 þ tan b 2Þ (16.30c)

The foregoing expressions of chord rotations can be used to write

the slope-deflection equations, thereby relating member end momentsto the three unknown joint displacements, yC ; yD, and D. As in the caseof the rectangular frames considered previously, the three equilibriumequations necessary for the solution of the unknown joint displacementscan be established by summing the moments acting on joints C and Dand by summing the horizontal forces acting on the entire frame. How-ever, for frames with inclined legs, it is usually more convenient toestablish the third equilibrium equation by summing the moments of all the forces and couples acting on the entire frame about a momentcenter 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 themoment center O can be determined by using the conditions (see Fig.16.16(d))

a1 cos b 1 ¼ a2 cos b 2 (16.31a)

a1 sin b 1 þ a2 sin b 2 ¼ L (16.31b)

By solving Eqs. (16.31a) and (16.31b) simultaneously for a1 and a2, weobtain

a1 ¼ L

cos b 1ðtan b 1 þ tan b 2Þ (16.32a)

a2 ¼ L

cos b 2ðtan b 1 þ tan b 2Þ (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 multistoryframes subjected to sidesway, as illustrated by Example 16.12. How-ever, because of the considerable amount of computational e¤ort in-volved, the analysis of such structures today is performed on computers

using the matrix formulation of the displacement method presented inChapter 18.



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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 yC ;yD, and D (see Fig. 16.17(b)).

continued

FIG. 16.17



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FIG. 16.17 (contd.)

continued



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Fixed-End Moments By using the fixed-end moment expressions given inside the back cover of the book, we obtain

FEMCD ¼

40ð3Þð4Þ2

ð7Þ2 ¼ 39:

2 kN m

’

or þ39:

2 kN m

FEMDC ¼ 40ð3Þ2ð4Þ

ð7Þ2 ¼ 29:4 kN m @ or 29:4 kN m

FEMAC ¼ FEMCA ¼ FEMBD ¼ FEMDB ¼ 0

Chord Rotations From Fig. 16.17(b), we can see that

cAC ¼ D

7

cBD ¼ D

5

cCD ¼ 0

Slope-Deflection Equations

M AC ¼ 2EI

7 yC 3

D

7

¼ 0:286EI yC þ 0:122EI D (1)

M CA ¼ 2EI

7 2yC 3

D

7

¼ 0:571EI yC þ 0:122EI D (2)

M BD ¼ 2EI

5 yD 3

D

5

¼ 0:4EI yD þ 0:24EI D (3)

M DB ¼ 2EI

5 2yD 3

D

5

¼ 0:8EI yD þ 0:24EI D (4)

M CD ¼ 2EI

7 ð2yC þ yDÞ þ 39:2 ¼ 0:571EI yC þ 0:286EI yD þ 39:2 (5)

M DC ¼ 2EI

7

ðyC þ 2yDÞ 29:4 ¼ 0:286EI yC þ 0:571EI yD 29:4

(6)

Equilibrium Equations By considering the moment equilibrium of joints C and D, we obtain the equilibriumequations

M CA þ M CD ¼ 0 (7)

M DB þ M DC ¼ 0 (8)

To establish the third equilibrium equation, we apply the force equilibrium equationP

F X ¼ 0 to the free body of theentire 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 inFig. 16.17(c). To express the column end shears in terms of column end moments, we draw the free-body diagramsof 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



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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 CAÞ 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 slope-deflection 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 m2

EI yD ¼ 34:24 kN m2

EI D ¼ 25:177 kN m3

Member End Moments By substituting the numerical values of EI yC ; EI yD, and EI D into the slope-deflectionequations (Eqs. (1) through (6)), we obtain

M AC ¼ 14:6 kN m or 14:6 kN m @ Ans.

M CA ¼ 26 kN m or 26 kN m @ Ans.

M BD ¼ 7:7 kN m ’

Ans.

M DB ¼ 21:3 kN m ’

Ans.

M CD ¼ 26 kN m ’

Ans.

M DC ¼ 21:3 kN m or 21:3 kN m @ Ans.

To check that the solution of the simultaneous equations (Eqs. (10) through (12)) has been carried out correctly, wesubstitute 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 CAÞ þ 7ðM BD þ M DB Þ ¼ 5ð14:6 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 shownin Fig. 16.17(e).

Member Axial Forces With end shears known, member axial forces can now be evaluated by considering theequilibrium of joints C and D. The axial forces thus obtained are shown in Fig. 16.17(e).

Support Reactions See Fig. 16.17(f). Ans.

Equilibrium Check The equilibrium equations check.



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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 yC ; yD, and D.

Fixed-End Moments Since no external loads are applied to the members, the fixed-end moments are zero.

continued

FIG. 16.18



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FIG. 16.18 (contd.)continued



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Chord Rotations From Fig. 16.18(b), we can see that

cAC ¼ CC 0

20 ¼

5

4 D

20 ¼ 0:0625D

cBD ¼ DD 0

16 ¼

D

16 ¼ 0:0625D

cCD ¼ C 0C 1

20 ¼

3

4

D

20 ¼ 0:0375D

continued

FIG. 16.18 (contd.)



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Slope-Deflection Equations

M AC ¼

2EI

20 ½yC 3ð0:

0625DÞ ¼ 0:

1EI yC þ 0:

0188EI D (1)

M CA ¼ 2EI

20 ½2yC 3ð0:0625DÞ ¼ 0:2EI yC þ 0:0188EI D (2)

M BD ¼ 2EI

16 ½yD 3ð0:0625DÞ ¼ 0:125EI yD þ 0:0234EI D (3)

M DB ¼ 2EI

16 ½2yD 3ð0:0625DÞ ¼ 0:25EI yD þ 0:0234EI D (4)

M CD ¼ 2EI

20 ½2yC þ yD 3ð0:0375DÞ ¼ 0:2EI yC þ 0:1EI yD 0:0113EI D (5)

M DC ¼ 2EI

20 ½2yD þ yC 3ð0:0375DÞ ¼ 0:2EI yD þ 0:1EI yC 0:0113EI D (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 freebody 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 BDð42:67Þ 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 CA20 and S BD ¼ 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 equilibriumequations (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-ft2

EI yD ¼ 125:912 k-ft2

EI D ¼ 5;233:6 k-ft3

continued



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Member End Moments By substituting the numerical values of EI yC ; EI yD, and EI D into the slope-deflectionequations (Eqs. (1) through (6)), we obtain

M AC ¼ 91:7 k-ft ’

Ans.

M CA ¼ 85:1 k-ft ’

Ans.

M BD ¼ 106:7 k-ft ’

Ans.

M DB ¼ 91 k-ft ’

Ans.

M CD ¼ 85:1 k-ft or 85:1 k-ft @ Ans.

M DC ¼ 91 k-ft or 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).



Example 16.12

Determine the member end moments, the support reactions, and the horizontal deflection of joint F of the two-storyframe 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 jointsC 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, theframe has six degrees of freedom—that is, yC ; yD; yE ; yF ;D1, and D2.

Fixed-End Moments The nonzero fixed-end moments are

FEMCD ¼ FEMEF ¼ 200 k-ft

FEMDC ¼ FEMFE ¼ 200 k-ft

continued



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Chord Rotations See Fig. 16.19(b).

cAC ¼ cBD ¼

D1

20

cCE ¼ cDF ¼ D2

20

cCD ¼ cEF ¼ 0

Slope-Deflection Equations Using I column ¼ I and I girder ¼ 2I , 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



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FIG. 16.19 (contd.)

continued



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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 lowerends of the columns of each story of the frame and apply the equation of horizontal equilibrium ð

PF X ¼ 0Þ 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 PF X ¼ 0 to the top story of the frame (Fig. 16.19(c)), we obtain

S CE þ S DF ¼ 10

Similarly, by applyingP

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 M AC þ M CA

20 S BD ¼

M BD þ M DB 20

S CE ¼ M CE þ M EC

20 S DF ¼

M DF þ M FD20

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 equilibriumequations (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



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By solving Eqs. (19) through (24) by the Gauss-Jordan elimination method (Appendix B), we determine

EI yC ¼ 812:988 k-ft2

EI yD ¼ 241:556 k-ft2

EI yE ¼ 789:612 k-ft2

EI yF ¼ 353:248 k-ft2

EI D1 ¼ 15;272:728 k-ft3 or D1 ¼ 0:0758 ft ¼ 0:91 in: !

EI D2 ¼ 10;787:878 k-ft3 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

D1 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 obtainM 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 or 79:7 k-ft @ Ans.

M EC ¼ 77:4 k-ft 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 or 329:6 k-ft @ Ans.

M EF ¼ 77:4 k-ft ’

Ans.

M FE ¼ 208:3 k-ft or 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).





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SUMMARY

In this chapter, we have studied a classical formulation of the displace-

ment (sti¤ness) method, called the slope-deflection method, for the anal-ysis of beams and frames. The method is based on the slope-deflectionequation:

M nf ¼ 2EI

L ð2yn þ y f 3cÞ þ FEMnf (16.9)

which relates the moments at the ends of a member to the rotations anddisplacements of its ends and the external loads applied to the member.

The procedure for analysis essentially involves (1) identifying theunknown joint displacements (degrees of freedom) of the structure; (2)for each member, writing slope-deflection equations relating memberend moments to the unknown joint displacements; (3) establishing theequations of equilibrium of the structure in terms of member end mo-ments; (4) substituting the slope-deflection equations into the equili-brium equations and solving the resulting system of equations to de-termine the unknown joint displacements; and (5) computing memberend moments by substituting the values of joint displacements back intothe slope-deflection equations. Once member end moments have beenevaluated, 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 theshear and bending moment diagrams for the beams shownin Figs. P16.1–P16.5 by using the slope-deflection method.

FIG. P16.1

16.6 Solve Problem 16.2 for the loading shown in Fig.P16.2 and a settlement of 12 in. at support B .

15 ft

E = 29,000 ksi I = 1,650 in.4

15 ft 20 ft

B A C

20 k

1.5 k/ft

3 k/ft

FIG. P16.2, P16.6


http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/


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16.7 Solve Problem 16.4 for the loading shown in Fig. P16.4and the support settlements of 50 mm at B and 25 mm at C .

FIG. P16.3

8 m 8 m

B

C A

25 kN/m

E = 70 GPa I = 1,300 (106) mm4

FIG. P16.4, P16.7

A C

B25 ft

2 I

15 ft

I

3 k/ft

E = 29,000 ksi

I = 2,500 in.4

FIG. P16.5

16.8 through 16.14 Determine the reactions and draw theshear and bending moment diagrams for the beams shownin Figs. P16.8–P16.14 by using the slope-deflection method.

1.5 k/ft

25 ft20 ft

EI = constant

25 ft

B C D A

FIG. P16.8

FIG. P16.9, P16.15

FIG. P16.10

10 ft 10 ft 10 ft 20 ft

EI = constant

E B C D A

35 k 2 k/ft1 k/ft

FIG. P16.11

Problems 703


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6 m 4 m 6 m 4 m 4 m 4 m

I I 2 I

A C

B D F

E G

120 kN 120 kN 150 kN

E = 200 Gpa

I = 500 (106) mm4

FIG. P16.12, P16.16

FIG. P16.13

FIG. P16.14

16.15 Solve Problem 16.9 for the loading shown in Fig.P16.9 and a settlement of 25 mm at support C .

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 .

Section 16.4

16.17 through 16.20 Determine the member end momentsand reactions for the frames shown in Figs. P16.17–P16.20by using the slope-deflection method.

FIG. P16.17, P16.21

FIG. P16.18, P16.22

3 k/ft

D

C E

B

A

10 ft

5 ft15 k

I

20 ft 5 ft

2 I

E = constant

FIG. P16.19



20/20

30 kN/m

C D

A B

10 m

EI = constant

8 m

FIG. P16.20

16.21 Solve Problem 16.17 for the loading shown in Fig.P16.17 and a settlement of 50 mm at support D.

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 reactionsfor the frame in Fig. P16.23 for the loading shown and thesupport settlements of 1 in. at A and 112 in. at D. Use theslope-deflection method.

FIG. P16.23

Section 16.5

16.24 through 16.31 Determine the member end momentsand reactions for the frames shown in Figs. P16.24–P16.31

by using the slope-deflection method.

2 k/ft

25 k

20 ft

B

C

A

15 ft

EI = constant

FIG. P16.24

FIG. P16.25

30 ft

EI = constant

3 k/ft

40 k C D

B A

15 ft

FIG. P16.26

Problems 705

701_aKassimali Structural Analysis 4th US&SI Txtbk

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