Problem 7.1 Determine the soil pressure distribution under the footing. Elevation. Plan. M 180 e 1.5 ft P 120. (a) B= L= 8 ft L e 1.5 ft 1.

Problem 7.1 Determine the soil pressure distribution under the footing. Elevation Plan M 180 e 1.5 ft P 10 (a) B= L= 8 ft L e 1.5 ft 1.33 ft 6 1

q q P 6 (P e) 180 6 (180) 4.9 kip/ft B L B L 8(8) 8 3 P 6 P e 180 6 (180) 0.7 kip/ft B L B L 8(8) 8 1 3 (b) L=10 ft B =5 ft L e 1.5 ft 1.66 ft 6 P 6 (P e) 180 6 (180) q 5.76 kip/ft B L B L 5(10) (5)(10) P 6 P e 180 6 (180) q1 1.4 kip/ft B L B L 5(10) (5)(10) Problem 7. Determine the soil pressure distribution under the footing. Use a factor of 1. for DL and 1.6 for LL. Elevation Plan P 1. P 1.6 P 1.(00) 1.6(50) 640 kn u D L M 1. M 1.6 M 0 1.6(64) 64 kn-m u D L

M 64 L P 640 6 u e 0.1 m 0.5 m u q q P 6 (P e) 640 6 (64) 85.3 kn/m B L B L 3(3) 3 u u u1 3 P 6 P e 640 6 (64) 56.88 kn/m B L B L 3(3) 3 u u u 3 Problem 7.3 The effective soil pressure is 4 kip/ft. Determine the maximum allowable value for L. Elevation Plan M 48 e.3478 ft P 138 L P 6 M 138 6 (48) Assume e 4 3 3 6 L L L L 4L 138L 88 0 L 6.7 ft 3 3 req req

Problem 7.4 The allowable soil pressure isq P D = 1000 kn and P L = 1400 kn. allowable 50 kn/m, γsoil 3 18 kn/m, γ 3 conc 4 kn/m, Plan P D+L 1000 1400 400 kn Elevation q q γ t γ (h-t) 50 4(.4) 18(1.4) 9.6 kn/m effective allowable conc soil P u 1. PD 1.6 PL 1.(1000) 1.6(1400) 3440 kn Pu qu L1 L L1 a V u ( x) L q u ( ) M ( x) u L qu L1 a 4

(a) A square footing (L 1 = L = L) P D+L q effective 9.6 kn/m L 400 L 3.3 m use L 3.5 m 9.6 q P 3440 36 kn/m L 3.5(3.5) u u L a Vu max L q u ( ) (3.5)(36)(1.65.5) 1483 kn LquL a (3.5)(36) M u (1.65.5) 1038 kn-m max (b)a rectangular footing with L =.5 m. P D+L qeffective 9.6 kn/m L (L ) 1 400 L1 4.18 m use L1 4. m.5(9.6) 5

q u P 3440 L L.5(4.) u 1 37.6 kn/m L a 1 Vu max L q u ( ) (.5)(37.6)(.1.5) 1536 kn L qu L1 a (.5)(37.6) M u (.1.5) 1440 kn-m max Problem 7.5 A 350 mm x 350 mm column is to be supported on a shallow foundation. Determine the dimensions (either square or rectangular) for the following conditions. The effective soil pressure is q 180 kn/m. effective (a) The centerline of the column coincides with the centerline of the footing. Plan Elevation q P 6 (P e) q L L L L 1 1 effective 40 6 (550) 180.5L.5L 1 1 180L 0L 576 0 L.5 m 1 1 1 6

(b) The center line of the column is 0.75meter from the property line. Plan Elevation q P 6 (P e) q L L L L 1 1 effective L 550.5L.5L 1 6 (.75)(550) 40 180 1 1 1 1 1 1 180 L 880 L 65.5 0 L 3.97 m use L 4 m (c) The center line of the column is 0.5 meter from the centroid of the footing. 7

Plan Elevation q P 6 (P e) q L L L L 1 1 effective 550 6 (515) 180.5L.5L 1 1 180 L 0 L 136 0 L 3.3 m 1 1 1 Problem 7.6 A combined footing supports two square columns: Column A is 14 inches x 14 inches and carries a dead load of 140 kips and a live load of 0 kips. Column B is 16 inches x 16 inches and carries a dead load of 60 kips and a live load of 300 kips. The effective soil pressure is q 4.5 k/ft. Assume the soil pressure distribution is uniform. Determine the footing e 8

dimensions for the following geometric configurations. Establish the shear and moment diagrams corresponding to the factored loading, P u =1. P D +1.6 P L. 9

Problem 7.7 Column A is 350 mm x 350 mm and carries a dead load of 1300 kn and a live load of 450 kn. Column B is 450 mm x 450 mm and carries a dead load of 1400 kn and a live load of 800 kn. The combined footing shown below is used to support these columns. Determine the soil pressure distribution and the shear and bending moment distributions along the longitudinal direction corresponding to the factored loading, P u =1. P D +1.6 P L. Elevation 0

1 Plan

Problem 7.8 Dimension a strap footing for the situation shown. The exterior column A is 14 inches x 14 inches and carries a dead load of 160 kips and a live load of 130 kips; the interior column B is 18 inches x 18 inches and carries a dead load of 00 kips and a live load of 187.5 kips; the distance between the center lines of the columns is 18 ft. Assume the strap is placed such that it does not bear directly on the soil. Take the effective soil pressure as qe 4.5 k/ft. Draw shear and moment diagrams using a factored load of P u =1. P D +1.6 P L. 4

Plan Elevation 5

Problem 7.9 Column A is 350 mm x 350 mm and carries a dead load of 1300 kn and a live load of 450 kn. Column B is 450 mm x 450 mm and carries a dead load of 1400 kn and a live load of 800 kn. A strap footing is used to support the columns and the center line of Column A is 0.5 meter from the property line. Assume the strap is placed such that it does not bear directly on the soil. Determine the soil pressure distribution and the shear and bending moment distributions along the longitudinal direction corresponding to the factored loading, P u =1. P D +1.6 P L. 6

Elevation Plan 7

Problem 7.10 An exterior 18 in x18 in column with a total vertical service load of P 1 = 180 kips and an interior 0 in x0 in column with a total vertical service load of P = 40 kips are to be supported at each column by a pad footing connected by a strap beam. Assume the strap is placed such that it does not bear directly on the soil. (a) Determine the dimensions L 1 and L for the pad footings that will result in a uniform effective soil pressure not exceeding 3 k/ft under each pad footing. Use ¼ ft increments. (b) Determine the soil pressure profile under the footings determined in part (a) when an additional loading, consisting of an uplift force of 80 kips at the exterior column and an uplift force of 5 kips at the interior column, is applied. 9

Problem 8.1 For the concrete retaing wall shown, determine the factors of safety against sliding and overturning and the base pressure distribution. 1 1 1 Pa γsoil H k a (18)(3.5) ( ) 49 kn 3 H 3.5 M P a ( ) 49( ) 57.16 kn-m overturning 3 3 W (4)(0.75)(3)(1) 54 kn 1 3 W (4)( )(1.5)(1) 45 kn W (4)()(0.5)(1) 4 kn 3 N N W 13 kn max i F N.5(13) 61.5 kn F 61.5 sliding Pa 49 M W 1(1.65) W ( )(1.5) W 3(1) 149. kn-m resisting 3 M resisting 149. F.S..6 overturning M 57.16 max F.S. 1.5 overturning M M M 9 kn-m net resisting overturning Page 1 of Chapter I 1

M net x.748 N L e x=.5 m N 6e 13 6(.5) q1 1 1 107.6 kn/m L L N 6e 13 6(.5) q 1 1 15.3 kn/m L L Problem 8. For the concrete retaing wall shown, determine the factors of safety against sliding and overturning and the base pressure distribution. Use the Rankine theory for soil pressure computations. Page of Chapter I

1 1 1 Pa γsoil H k a (18)(3.5) ( ) 49 kn 3 H 3.5 M P a ( ) 49( ) 57.16 kn-m overturning 3 3 W (4)(0.75)(3)(1) 54 kn 1 3 W (4)( )(1.5)(1) 45 kn W (4)(3)(0.5) 36 kn 3 W (18)(3)(1) 4 kn 4 N W 189 kn max i F N.5(13) 94.5 kn max F.S. 1.98 sliding M resisting F 94.5 P 49 a 314.3 kn-m M resisting 314.3 F.S. 5.49 overturning M 57.16 overturning M M M 57 kn-m net resisting overturning M net x 1.36 m N L e x 1.5 1.36.14 m N 6e q1 1 45.4 kn/m L L N 6e q 1 8.1 kn/m L L Page 3 of Chapter I 3

Problem 8.3 For the concrete retaining wall shown, determine the factors of safety against sliding and overturning and the base pressure distribution. Use the Rankine theory for soil pressure computations Page 4 of Chapter I 4

1 1 1 3 Pa ka γsh ( )(.1)(14) 3.9 kip W =.15 () (1) 1 3 4 3.6 kip 1 W =.15 ( )(4 )(1) 3.6 kip W =.15 () (13) 3.9 kip W =.1 (4) (1) 5.76 kip The normal and horizontal forces are N W W W W 16.86 kip max 1 3 4 F N.5 (16.86) = 8.43 kips Next we compute the factors of safety. F 8.43 max F.S. =.15 sliding Pa 3.9 H 14 MB P overturning a( ) 3.9( ) 18.4 k-ft 3 3 M W (8)+W (.67)+W (6.5)+W (11) Bresisting 1 3 4 3.6 (8)+3.6 (.67)+3.9(6.5)+5.76(11) 17.1 k-ft MB resisting 17.1 F.S. 6.97 overturning M 18.4 Boverturning Page 5 of Chapter I 5

M M M 108.86 k-ft Bnet Bresisting Boverturning x M 108.86 N 16.86 Bnet 6.45ft L 13 e x 6.45.05 ft Using the above values, the peak pressures are N 6e 16.86 6(.05) q (1 ± ) (1 ) q 1.38 kip/ft q 1.7 kip/ft L L 13 13 1 Problem 8.4 Page 6 of Chapter I 6

μγ b b c 1 F.S. sliding ( ) ( ) (1 ) ka γs H b.5 (.15) b 1 1.5 ( ) (1 ) b 8.33 (1/ 3) (.1) 14 b γc b 1 b 1 b 1 F.S. 1 overturning ka γs H b b μγ b b c 1 F.S. sliding ( ) ( ) (1 ) ka γs H b (.15) b 1 1 1 F.S. 1 1.75 overturning (1/3) (.1) 14 b b For b 8.33 (.15) 8.33 1 1 1 F.S. 1.95 1.75 overturning (1/3) (.1) 14 8.33 8.33 Problem 8.5 Assume: γ soil= 0.1 k/ft 3, γ concrete = 0.15 k/ft 3, µ=.5 and Φ = 30 0 Page 7 of Chapter I 7

1 1 1 Pa γsoil H k a (.1)() ( ) 9.68 kip 3 1 Ps ka ws H = ( ) (.) () 1.47 3 1 ' 1 Pp kp γsh (3)(.1)(4).88 kip F P + P P 8.7 kip horizontal a s p H H M P a ( ) P s( ) 9.68 ( ) 1.47( ) 87.15 overturning 3 3 W.15 (1) (0.5) 3.075 1 0.5 W.15 (.5) ( ).77 W.15 (b11) (1.5).475 3 W.1 (6) (0.5) 14.76 4 W.1 (.5) (3.5) 1.05 5 N W F N 11 max i F 11 max F.S. = 1.33 sliding Fhorizontal 8.7 Page 8 of Chapter I 8

M W (4.5)+W (3.83)+W (5.5)+W (8)+W (1.75)+P (1.33) Bresisting 1 3 4 5 p 3.075(4.5)+.768(3.83)+.475(5.5)+14.76(8)+1.05(1.75)+.88(1.33) 154 kip-ft MB resisting 154 F.S. 1.76 overturning M 87.15 Boverturning M M M 66.85 kip-ft Bnet Bresisting Boverturning x M 66.85 N Bnet 3ft L 11 L e x 3 3.5 ft 1.83 6 q (3a) ()() 3(3) 1 N q 1 =4.89 kip/ft Problem 8.6 Allowable soil pressure = 5.0 ksf, γ soil 3 =.1 k/ft and γ =.15 k/ft concrete 3 Page 9 of Chapter I 9

1 1 1 3 H 4 MB P overturning a( ) 11.5( ) 9.16 kip-ft 3 3 Pa ka γsh ( )(.1)(4) 11.5 kip Page 10 of Chapter I 10

W =.15 (1) (1) 1 3 4 3.15 kip 1 W =.15 ( )(.5 )(1).78 kip W =.15 (3) (13) 5.85 kip W =.1 (8) (1) N W 30 kip max i 0.1 kip F N.5 (30) = 15 kips F 15 max F.S. = 1.3 sliding Pa 11.5 M W (4.5)+W (3.83)+W (6.5)+W (9) Bresisting 1 3 4 3.15(4.5)+.78(3.83)+5.85(6.5)+0.1 (9) 36 kip-ft MB resisting 36 F.S..56 overturning M 9.16 Boverturning M M M 143.84 kip-ft Bnet Bresisting Boverturning x M 143.84 N 30 Bnet 4.79ft L 13 e x 4.79 1.7 ft N 6e 30 6(1.7) q (1 ± ) (1 ) q 4.1 kip/ft q 0.5 kip/ft L L 13 13 1 Page 11 of Chapter I 11

1 p ka γsh ( )(.1)(4) 0.96 kip/ft 3 Problem 8.7 Suggest values for b1 and b. Take the safety factors for sliding and over turning to be equal to. 3 3 Assume: γ =.1 k/ft, γ =.15 k/ft, µ =.57, and Φ = 30 0 soil concrete Page 1 of Chapter I 1

1 1 1 Pa γsoil H k a (.1)(16) ( ) 5.1 kip 3 1 Ps ka ws H = ( ) (.1) (16).53 3 1 ' 1 Pp kp γsh (3)(.1)(4).88 kip F P + P P.77 horizontal a s p H H 16 16 M P a ( ) P s( ) 5.1 ( ).53( ) 31.55 overturning 3 3 W.15 (3) (14) 6.3 1 W.15 () (b +b +3).3(b +b +3) 1 1 W.1 (b ) (0).4b 3 W.1 (b ) ().4b 4 1 1 N W 6.3+.3(b +b +3)+.4b.4b 7..7(b +b ) i 1 1 1 F N.57 (7..7(b +b )) 4.11.54(b +b ) max 1 1 F 4.11.54(b +b ) max 1 F.S. = b sliding 1+b 0.935 Fhorizontal.77 b 0.935 b1 Page 13 of Chapter I 13

M W (1.5 b )+W (.5b +.5b +1.5)+W (b +.5b +3)+W (.5)+P (1.33) Bresisting 1 1 1 3 1 4 p 6.3(1.5 b ).3(b +b +3)(.5b +.5b +1.5) 16.8(b +.5b +3).4b (.5).88(1.33) 106.4 3.9b 1 1 1 1 1 1 M 106.4 3.9b Bresisting 1 F.S. b overturning 1 0 b.935ft MB 31.55 overturning Problem 8.8 Determine the minimum value of w at which soil failure occurs (i.e., the soil pressure exceeds the allowable soil pressure). 3 3 Assume: q 5 k/ft, γ =.1 k/ft, γ =.15 k/ft, µ =.57 and Φ = 30 0 allowable soil concrete Page 14 of Chapter I 14

1 1 1 Pa γsoil H k a (.1)() ( ) 9.68 kip 3 1 Ps ka ws H = ( ) ( ws ) () 7.33 ws 3 1 ' 1 Pp kp γ s(h ) (3)(.1)(4).88 kip F P + P P 9.68 7.33 w.88 6.88 7.33 w horizontal a s p s s H H M P a ( ) P s( ) 9.68( ) 7.33 ws (11) 71 80.6 ws overturning 3 3 W.15 () (0) 6 kip 1 W.15 () (14) 4. kip W.1 (7) (0) 16.8 kip 3 W.1 (5) () 1. kip 4 N W 8. kip max i F N.57 (8.) = 16 kip N 6 e 8. 6 e q (1 ) 5 (1 ) 5 e 3.46 ft L L 14 14 M M net net x N 8. L M 8. net e x 7 3.46 M 99.8 net M W (6)+W (7)+W (10.5)+W (.5)+P (1.33) Bresisting 1 3 4 p 6(6) 4.(7) 16.8(10.5) 1.(.5).88(1.33) 48.6 M M M 99.8 48.6 (71 80.6 w ) net resisting overturning w s.96kip/ft s Page 15 of Chapter I 15

Problem 8.9 (a) The answer is case (a). Problem 8.10 (a) Determine the factor of safety with respect to overturning and sliding. (b) Identify the tension areas in the stem, toe, and heel and show the reinforcing pattern. (c) Determine the location of the line of action of the resultant at the base of the footing Page 16 of Chapter I 16

1 1 Pa γsoil H k a (.1)(1) (.4).88 kip H 1 M P a ( )+(15).88( ) 30 41.5 kip-ft overturning 3 3 W (1.5)(13)(3)(.15).95 kip 1 W ()(1)(.15) 3.6 kip W (7)()(010)(.1) 7 kip 3 N W 13.55 kip F N.58(13.55) 7.84 kip max max F.S..7 sliding F 7.84 P.88 a M resisting 93.53 kip-ft M resisting 93.53 F.S..5 overturning M 41.5 overturning Page 17 of Chapter I 17

M M M 5 kip-ft net resisting overturning M net 5 x 3.84 N 13.55 L L e x 6 3.84.16 ft ft 6 Page 18 of Chapter I 18

Problem 9.1 Take E=9,000 ksi and I= 00 4 in. 4 4 3 17wL 17()(30) (1) B,0 33.78 in 43EI 43(9000)(00) 3 3 3 8L 8(30) (1) δbb 0.794 in 81EI 81(9000)(00) 1 in B Page 1 of Chapter 9

R δ 1 in B,0 B BB R 1 33.78 1 B,0 B 41.8 kip δbb 0.794 Problem 9. Take E=00 GPa and 6 4 I= 80(10) mm. R δ 1 mm B,0 B BB 1 mm B Page of Chapter 9

4 4 9 wl 15(6) (10) B,0 6 151.875 mm 8EI 8(00)(80)10 Table 3.1 3 3 9 L (6) (10) δbb 4.5 mm 6 3EI 3(00)(80)10 R 1 151.875 1 B,0 B 31.08 kn δbb 4.5 Problem 9.3 kv= 60 k/in, E=9,000 ksi, I= 00 4 in Page 3 of Chapter 9

4 4 3 17wL 17(1.0)(0) (1) D,0.11 in 384EI 384(9000)(00) 3 3 3 L (0) (1) δdd 0.0993 in 4EI 4(9000)(00) F CD.11 1 1 δdd 0.0993 k 60 D,0 V 18.19 kip Problem 9.4 L = 5 m, E = 00 GPa, (a) 6 4 I= 170(10) mm Page 4 of Chapter 9

B,0 110.49 mm δ 3 3 9 L (1) (10) 1.059 mm 48EI 48(00)(170)10 BB 6 (b) R B,0 B 110.4 kn δbb Page 5 of Chapter 9

δ 3 3 9 L (1) (10) 1.059 mm 48EI 48(00)(170)10 BB 6 R 8.33 B B 7.86 kn δbb 1.059 (c) δ 3 3 9 L (1) (10) 1.059 mm 48EI 48(00)(170)10 BB 6 0 B RB 18.89 kn RB 18.89 kn δbb 1.059 (d) Page 6 of Chapter 9

9 P 40(10) B,0 6 5.946 5.946 30.5 mm EI (00)(170)(10) 3 3 9 L (1) (10) δbb 1.059 mm 6 48EI 48(00)(170)(10) F B 30.5 1 1 δbb 1.059 k 40 B,0 V 8.15 kn (e) Page 7 of Chapter 9

4 4 9 57wL 57(0)(1) (10) D,0 6 113.16 mm 6144EI 6144(00)(170)(10) 3 3 9 3L 3(1) (10) δdd 0.596 mm 6 56EI 56(00)(170)(10) F D 113.16 1 1 δdd 0.596 k 40 D,0 V 17.97 kn Problem 9.5 4 I 400 in, E 9,000 ksi, L 54 ft, w.1 kip/ft, B 1. in Page 8 of Chapter 9

(i) The distributed load shown R δ 0 B,0 B BB The deflection terms are given in Chapter 3 4 4wL B,0 δ BB 79 EI 14.6 in 3 4L.386 in 43EI Page 9 of Chapter 9

R 14.6 B,0 B 37.8 kip δbb 0.386 (ii) The support settlement at B R δ 1. in B BB 1. 1. RB 3.1 kip RB 3.1 kip δbb 0.386 Problem 9.6 A 100 mm, L 9 m, P 40 kn and E 00 GPa. C Page 10 of Chapter 9

Page 11 of Chapter 9

Problem 9.7 x x Io y 4h ( ) I L L cos θ (a) Determine the horizontal reaction at B due to the concentrated load. Page 1 of Chapter 9

Page 13 of Chapter 9

Page 14 of Chapter 9

(b) Utilize the results of part (a) to obtain an analytical expression for the horizontal reaction due to a distributed loading, wx. ( ) (c) Specialize (b) for a uniform loading, w( x) w0. Page 15 of Chapter 9

(d) Suppose the horizontal support at B is replaced by a member extending from A to B. Repeat part (a). Page 16 of Chapter 9

Problem 9.8 Consider the semi-circular arch shown below. Determine the distribution of the axial and shear forces and the bending moment. The cross section properties are constant. Page 17 of Chapter 9

Page 18 of Chapter 9

Problem 9.9 6 4 Take A 0,000 mm I 400(10) mm and E 00 GPa Page 19 of Chapter 9

Case (a): M 0(no bending moment) max Deflected shape Case (b): Page 0 of Chapter 9

Mmax 1101 kn-m Deflected shape Case (c): Page 1 of Chapter 9

Mmax 59 kn-m Deflected shape Problem 9.10 Consider the following values for the area of the tension rod AC: 4 A 30 in I 1000 in E 9,000 ksi. 4 in, 8 in, 16 in. 4 A 30 in I 1000 in E 9,000 ksi Page of Chapter 9

Bending moment Axial force Bending moment Page 3 of Chapter 9

Axial force Bending moment Axial force Problem 9.11 Page 4 of Chapter 9

4 A 40 in I 100 in E 9,000 ksi Case(a): Case (b): Mmax 0 (no bending moment) Page 5 of Chapter 9

Mmax 130 kip-ft Case (c): Mmax 43.3 kip-ft Case (d): Page 6 of Chapter 9

Mmax 10.9 kip-ft Problem 9.1 Determine the horizontal reaction at support D. (See Figure 9.38) H D 1 P 1 kip Page 7 of Chapter 9

(See example 9.15) I I I I r 5 L 16 h 0 r 4 1 g rg rc c wl 1 (3) 1 D rg 5 H.33 kip 1h 1(0) 1 1 ( ) 3r 3 4 c H H H 1.33 1.33 kip D D1 D Problem 9.13 Page 8 of Chapter 9

Consider h = m, 4 m, 6 m. x1 0 x1 9 M 0( x1 ) 101.5 x1 7.5 x1 δm( x1 ) 6 9h 0 x 9 M 0( x) 101.5 x δm( x) 6 x 9h Page 9 of Chapter 9

9 9 1 1 x1 1 x B,0 0 1 1 1 EI EI x x x x x 9h EI 9h s 0 0 Δ M δm ds (101.5 7.5 )( 6) d (101.5 )( 6) d 9 9 1 1 x1 1 x1 BB 1 EI EI x x 9h EI 9h s 0 0 δ δm ds ( 6) d ( 6) d Then H B δ B,0 BB 6 6 0.0676(10) for h m 0.0063(10) for h m 6 6 EIΔ B,0 0.0847(10) for h 4m EIδ BB 0.00357(10) for h 4m 6 6 0.1067(10) for h 6m 0.0049(10) for h 6m 5.7 kn for h m HB 3.7 kn for h 4m 1.78 kn for h 6m Page 30 of Chapter 9

Problem 9.14 Determine the peak positive and negative moments as a function of h. Consider h = 10 ft, 0 ft, 30 ft. See equation (9.51) and Figure 9.49 Page 31 of Chapter 9

M BA wl 1 1.(50) 1 50 1 1 rg 1 1 h h 1 1 ( ) 1 r 5 50 c wl 1 ME 1-375 1-8 3 1 r g h 1+ 3(1 ) r 50 c 08 kip-ft for h 10ft M max MBA 178.6 kip-ft for h 0ft 156. kip-ft for h 30ft 166.6 kip-ft for h 10ft M max ME 196.4 kip-ft for h 0ft 18.7 kip-ft for h 30ft Problem 9.15 Using a Computer software system, determine the bending moment distribution and deflected shape due to the loading shown. Take 4 I1 1000 in, I 4 000 in, E=9,000 ksi and A=0 in all members Page 3 of Chapter 9

Problem 9.16 Take E=00 GPa, 6 4 I 400(10) mm, A=100000 mm. Use a Computer software system. mm and AC 100 mm, 400 mm, 4800 Page 33 of Chapter 9

Page 34 of Chapter 9 (a)

Page 35 of Chapter 9

Page 36 of Chapter 9 (b)

Page 37 of Chapter 9

Page 38 of Chapter 9

Problem 9.17 Both cases (a) and (b) will are rigid frames and will have large horizontal displacements and will behave the same. In case (b) the added member will not carry any load. In case (c) the diagonal member changes the behavior. The frame will behave like a truss. All the members only carry axial forces and lateral movement will be very small. Case (a): Case (b): Case (c): Page 39 of Chapter 9

Problem 9.18 E= 9,000 ksi, A= 1 in all members F 0 δf member L, in F 0 δf F δf L δf L F 0 + X 1 δf ab 14.14(1) -3.53 -.707 43.47 84.81 4.6 bc 14.14(1) -17.67 -.707 119.75 84.81-9.51 cd 10(1) 1.5.5 750 30 6.73 da 10(1) 1.5.5 750 30 6.73 bd 10(1) 0 1 0 10-11.53 F δf L 4043 δf L 349.6 Page 40 of Chapter 9

L 4043 1,0 F0 δf 0.139 AE 9000 L 349.6 δ11 δf 0.01 AE 9000 0.139 1,0 1,0 δ11 X1 0 X1 11.53 δ11 0.01 Knowing the value of X 1, we determine the member forces and reactions by using superposition. Member forces are listed below. Problem 9.19 Assume the vertical reaction at d as the force redundant. -6 0 0 E= 00 GPa, A= 660 mm all members, α= 1x10 / C, T 10 C Page 41 of Chapter 9

member L, mm δf e temp = α ΔTL δf e temp δf L X1 δf ab 5657 -.606 0.67884 -.4073 077.4-6.9 bc 5657 -.714 0.67884 -.4819 883.9-8.1 cd 3000.48 0 0 549.5 4.8 da 4000.48 0 0 73.7 4.8 bd 4000 1 0 0 4000 11.4 e δf 0.889 1,0 temp e δf 0.889 temp L 1043.5 δ11 δf 0.0776 AE 660(00) 0.889 L δf 1043.5 1,0 1,0 δ11 X1 0 X1 11.4 δ11 0.0776 Knowing the value of 1 X, we determine the member forces and reactions by using superposition. Member forces are listed below. Page 4 of Chapter 9

Problem 9.0 E=9,000 ksi and A= 1 in all members. member L, in F 0 δf δf L F0 δf L F 0 + X 1 δf ab 1(1) 0 -.8 9.16 0 7.6 bc 9(1) -10 -.6 38.88 648-4.3 cd 1(1) -19.33 -.8 9.16 6.8-11.7 da 9(1) 0 -.6 3.88 0 5.7 ac 15(1) 16.66 1 180 998.8 7. bd 15(1) 0 1 180 0-9.5 δf 616 F δfl 58736 0 Page 43 of Chapter 9

L 5873.6 1,0 F0 δf 0.05 AE 900(1) L 616 δ11 δf 0.01 AE 900(1).05 1,0 1,0 δ11 X1 0 X1 9.5 δ 11.01 Knowing the value of X 1, we determine the member forces and reactions by using superposition. Member forces are listed below. Problem 9.1 A A A A 10 1 3 4 in, A 5 = 5 in, α= -6 0 6.5x10 / F, ΔT= 0 60, E= 9,000 ksi Page 44 of Chapter 9

member L, in L/AE e temp = α ΔTL F 0 δf etemp δf F0 δfl AE δf L F 0 + X 1 δf AE ab 8.84(1) 0.001193 0.13497 1-1.8 -.49 -.057.00386-19.1 bc 8.84(1) 0.001193 0.13497-6 -1.8 -.49.018.00386-37.1 cd 0(1) 0.00087 0.0936 4.16.5 0.0086.00516 47.3 da 0(1) 0.00087 0.0936 4.16.5 0.0086.00516 47.3 bd 1(1) 0.000993 0.05616 5 3 0.0148.00893 56.8 F0 δfl 1,0 etemp δf.019 0.485 0.466 AE δf L δ11 0.07 AE 0.466 1,0 1,0 δ11 X1 0 X1 17.6 δ11 0.07 Knowing the value of X 1, we determine the member forces and reactions by using superposition. Member forces are listed below. Page 45 of Chapter 9

Problem 9. Assume the force in member ac and the reaction at support f as force redundants. A=1000 mm and E=00 GPa for all the members Page 46 of Chapter 9

F 0 δf 1 δf member L, mm F 0 δf 1 δf F 0 δf1 L F 0 δf L (δf 1) L (δf ) L δf 1 δf L ab 4000-0 -.5 -.8 40000 64000 1000 560 1000 bc 3000 15 -.375 -.6-16875 -7000 4 1080 675 cd 3000 30 0 0 0 0 0 0 0 de 4000 0 0 0 0 0 0 0 0 ef 3000 15.375 0 16875 0 4 0 0 fa 3000 30 0 -.6 0-54000 0 1080 0 bf 5000-5.65 1 7815-15000 1953 5000 315 ac 5000 0 0 1 0 0 0 5000 0 cf 4000 0.5 -.8 40000-64000 1000 560-1000 ce 5000-5 -.65 0-7815 0 1953 0 0 F δfl 80000 0 1 0 F δf L 06000 (δf 1) L 6750 (δf ) L 1780 1 δf δf L 3800 Page 47 of Chapter 9

F δfl 80000 EA 00(1000) 0 1 0 1,0.4 mm,0 F δf L 06000 EA 00(1000) -1.03 mm (δf 1) L 6750 δ11 0.03375 EA 00(1000) (δf ) L 1780 δ 0.0864 EA 00(1000) δfδf L 3800 EA 00(1000) 1 δ1 δ1 0.019 δf 1 δf δ X δ X 0 1,0 11 1 1 δ X δ X 0,0 1 1 X X 1 1. kn 16.58 kn Knowing the value of 1 X and X, we determine the member forces and reactions by using superposition. Page 48 of Chapter 9

member F 0 δf 1 δf F 0 + X 1 δf 1 + X δf ab -0 -.5 -.8 -.67 bc 15 -.375 -.6 13.0 cd 30 0 0 30 de 0 0 0 0 ef 15.375 0 7 fa 30 0 -.6 0 bf -5.65 1-1.67 ac 0 0 1 16.58 cf 0.5 -.8-3.86 ce -5 -.65 0-11.75 Member forces are listed below. Page 49 of Chapter 9

Problem 10.1 Take E = 00 GPa and A 000 mm 1 (a) A1 A AE (000)00 kn 94.8 L1 3 (1000) mm AE 1 (1000)00 kn 66.67 L 3(1000) mm AE (000)00 kn 66.67 L 6(1000) mm 3 AE AE F cos45 u sin45 v 66.67u 66.67v (1) L1 L1 AE F v 66.67v 1 () L AE AE F - cos30 u sin30 v 57.73u 33.33v (3) L3 L3 1

F 0 F cos45 F cos30 45 0 x (1) (3) F 0 F sin45 F F sin30 90 0 y (1) () (3) 0.707 F 0.86 F 45 (1) (3) 0.707 F F 0.5 F 90 (1) () (3) v v v v v 0.707 66.67u 66.67 0.86 57.73u 33.33 45 0.707 66.67u 66.67 66.67 0.5 57.73u 33.33 90 96.78 u 18.47 v 45 18.7 u 130.47 v 90 u v 0.34 mm 0.645 mm Next, we substitute for u and v to determine member forces F 66.67u 66.67v 65.8 kn (1) F 66.67v 43 kn () F 57.73u 33.33v 1.75 kn (3)

(b) A 1 A AE (000)00 kn 94.8 L1 3 (1000) mm AE 1 (4000)00 kn 66.67 L 3(1000) mm AE (000)00 kn 66.67 L 6(1000) mm 3 AE AE F cos45 u sin45 v 66.67u 66.67v (1) L1 L1 AE F v 66.67v 1 () L AE AE F - cos30 u sin30 v 57.73u 33.33v (3) L3 L3 F 0 0.707 F 0.86 F 45 x (1) (3) F 0 0.707 F F 0.5 F 90 y (1) () (3) 96.78 u 18.47 v 45 18.7 u 330.47 v 90 u v 0.417 mm 0.5 mm Next, we substitute for u and v to determine member forces F 66.67u 66.67v 44.47 kn (1) F 66.67v 66.67 kn () F 57.73u 33.33v 15.74 kn (3) 3

(c) Results based on computer based analysis listed below 1 For A1 A F(1) 65.6 kn F() 4.79 kn F (3) 1.6 kn u v 0.34 mm 0.64 mm 90.00 FX 6.56E+01 o 45.00 3 1 FX 4.79E+01 FX 1.619E+00 For A 1 A F(1) 44.37 kn F() 66.48 kn F (3) -15.73 kn u v 0.416 mm 0.49 mm 4

90.00 o 45.00 FX 4.438E+01 3 1 FX 6.648E+01 FX -1.573E+01 Problem 10. Take A=0.1 in, A =0.4 in, and E=9,000 ksi. 1 cosθ 0.6 sinθ 0.8 5

AE AE (0.1)9000 kip AE 1 (0.4)9000 kip 9.67 96.67 L L 5(1) in L 10(1) in 1 3 (a) The loading shown AE AE F cosθ u sinθ v 5.8 u 7.74 v (1) L1 L1 AE F v 96.67 v 1 () L AE AE F - cosθ u sinθ v 5.8 u 7.74 v (3) L3 L3 F 0 0.6 F 0.6 F 6 0 x (1) (3) F 0 0.8 F F 0.8 F 10 0 y (1) () (3) 6.96 u 6 109 v 10 u v 0.86 in 0.09 in Next, we substitute for u and v to determine the member forces F 5.8 u 7.74 v 5.7 kip (1) F 96.67 v 8.89 kip () F 5.8 u 7.74 v 4.3 kip (3) (b) Support #1 moves as follows: 1 1 u inch and v inch 8 6

AE AE F cosθ u sinθ v 5.8 u 7.74 v (1) L1 L1 AE 1 1 F ( v ) 96.67 ( v ) L 1 () AE AE F - cosθ u sinθ v 5.8 u 7.74 v (3) L3 L3 F 0 F F u 0 x (1) (3) F 0 0.8 (F F ) F 0 y (1) (3) () 1 0.8 (15.48 v) 96.67 ( v ) 0 v 0.443 in F F 3.4 kip F (1) (3) () 5.5 kip For the following beams and frames defined in Problems 10.3-10.18, determine the member end moments using the slope-deflection equations. Problem 10.3 Assume E=9,000 ksi, I 4 00 in, L 30 ft, C.6 in and w 1. kip/ft. 7

EI EI k k k k L L Let member AB member BC L 0.6 1 L 60(1) 100 B A ρab 0 ρ C B BC EI 9000(00) 1 k 671.3 kip-ft L (30) (144) The slope-deflection equations take the form: AB BA B B M 4k θ 90 M 4k θ 90 M k θ - 3( ) k θ - 3ρ L C B BC B B BC M k θ - 3( ) k θ - 3ρ L C B CB B B BC Enforce moment equilibrium at node B: M M 0 BA BC B B BC B BC 8kθ 90 k θ - 3ρ 0 1kθ 6k ρ 90 1 90 6(671.3)( ) θ 100 B 0.01076 rad counterclockwise 1(671.3) 8

The corresponding end moments are: AB BA M 4(671.3) 0.01076 90 118.89 kip-ft M 4(671.3) (0.01076) 90 3.1 kip-ft 1 MBC (671.3) (0.01076) - 3(- ) 3.5 kip-ft 100 1 MCB (671.3) 0.01076-3(- ) 17.80 kip-ft 100 Problem 10.4 Assume E= 00 GPa, I L 6 4 80(10) mm and 6 m. EI EI (a) Let kmember AB k kmember BC k L L F (6) F (6) MAB 66 kn-m MBA 66 kn-m 1 1 F 45(6) F 45(6) MBC 33.75 kn-m MCB 33.75 kn-m 8 8 The slope-deflection equations take the form: M k θ 66 AB BA B M 4k θ 66 B M 4k θ 33.75 9 BC M k θ 33.75 CB B B Enforce moment equilibrium at node B: MBA MBC 0 4k θb 66 4k θb 33.75 0 k θb 4.03

The corresponding end moments are: M k θ 66 74.06 kn-m AB BA B M 4k θ 66-49.9 kn-m B M 4k θ 33.75 49.9 kn-m BC CB B M k θ 33.75 5.7 kn-m B E(I ) EI (b) Let kmember AB k kmember BC k L L The slope-deflection equations take the form: M 4k θ 66 AB BA B M 8k θ 66 B M 4k θ 33.75 BC M k θ 33.75 CB B B Enforce moment equilibrium at node B: MBA MBC 0 8k θb 66 4k θb 33.75 0 k θb.68 The corresponding end moments are: M 4k θ 66 76.7 kn-m AB BA B M 8k θ 66 44.5 kn-m B 10

M 4k θ 33.75 44.5 kn-m BC CB B M k θ 33.75 8.4 kn-m B E(I ) EI (c) Let kmember AB k kmember BC k (L ) L F (1) F (1) MAB 64 kn-m MBA 64 kn-m 1 1 F 45(6) F 45(6) MBC 33.75 kn-m MCB 33.75 kn-m 8 8 The slope-deflection equations take the form: M k θ 64 AB M 4k θ 64 BA B B M 4k θ 33.75 BC M k θ 33.75 CB B B Enforce moment equilibrium at node B: MBA MBC 0 4k θb 64 4k θb 33.75 0 k θb 8.78 The corresponding end moments are: M k θ 64 31.5 kn-m AB BA B M 4k θ 64-148.9 kn-m B M 4k θ 33.75 148.9 kn-m BC CB B M k θ 33.75 3.8 kn-m B 11

Problem 10.5 E=9,000 ksi and I 300 in 4, L=0 ft EI k k k L Let member AB member BC F 1.5(0) F 1.5(0) MAB 50 kip-ft MBA 50 kip-ft 1 1 F 10(0) F 10(0) MBC 5 kip-ft MCB 5 kip-ft 8 8 The slope-deflection equations take the form: M k θ 50 AB BA B M 4k θ 50 B M k (θ θ ) 5 BC B C M k (θ θ ) 5 CB B C Enforce moment equilibrium at nodes B and C: M 5 0 k (θ θ ) 5 5 0 θ θ CB B C C B M M 0 4k θ 50 k (θ θ ) 5 0 8k θ kθ 5 BA BC B B C B C k θ 5 5 B k θc 7 14 The corresponding end moments are: M k θ 50 57.1 kip-ft AB BA B M 4k θ 50 35.7 kip-ft B M k (θ θ ) 5 35.7 kip-ft BC B C M k (θ θ ) 5 5 kip-ft CB B C 1

Problem 10.6 E= 00 GPa, I. 6 4 80(10) mm, P=45kN, h =3 m and L 9 m The slope-deflection equations take the form: M 0 M AB BAmodified M 0 CB 3EI L 3EI MBCmodified θb L Enforce moment equilibrium at node B: θ B M M 135 BAmodified 6EI L B BCmodified θ 135 EI θ B.5 L The corresponding end moments are: M BA =67.5 kn-m counterclockwise M BC =67.5 kn-m counterclockwise 13

Problem 10.7 E=9,000 ksi, I 4 00 in, L 18 ft, and w 1. k/ft. Because of symmetry θa θd and θb θc EI Let k L EI EI kmember BC 0.5 k kmember AB kmember CD k L L The slope-deflection equations take the form: M 0 AB M M 3k θ BAmodified BA modified B F M k θ θ M k θ θ 19.6 k θ 19.6 BC B C BC B B B M M CB BC Enforce moment equilibrium at node B: 14

M M 0 BAmodified B B BC 4kθ 19.6 0 kθ 3.4 The corresponding end moments are: M 3k θ 97. M 97. kip-ft clockwise BA B BA MBC k θb 19.6 97. kip-ft counterclockwise Because of symmetry M M 97. kip-ft clockwise CB CD BC M M 97. kip-ft counterclockwise BA Problem 10.8 Assume E= 00 GPa and I 80(10) mm 6 4 EI EI k k 1.5k k k 1 18 Let member AB member CD member BC F 30(1) F 30(1) MAB 144 kn-m MBA 16 kn-m 30 0 F 45(18) F 45(18) MBC 101.5 kn-m MCB 101.5 kn-m 8 8 F 30(1) F 30(1) MCD 16 kn-m MDC 144 kn-m 0 30 Because of symmetry θb θc, MBC MCB 15

The slope-deflection equations take the form: M M (1.5k) θ 144 AB DC B M M 4(1.5k) θ 16 BA CD B MBC MCB k (θb θ C) 101.5 k θb 101.5 Enforce moment equilibrium at either node B or C: MBA MBC 0 4(1.5k) θb 16 k θb 101.5 0 k θb 14.34 The corresponding end moments are: M M (1.5k) θ 144 187 kn-m AB DC B M M 4(1.5k) θ 16 19.9 kn-m BA CD B M M k θ 101.5 19.9 kn-m BC CB B Problem 10.9 E=9,000 ksi, I 4 400 in. k ( EI ) 3 k k EI k k ( EI ) 5 k 0 30 18 3 Let member AB member BC member CD 16

The slope-deflection equations take the form: M 0 17 AB 3 M BA modified 3( k) θb 75 BC B C CB B C M k θ θ 90 M k θ θ 90 5 M CD ( k) θc 3 36 5 M DC ( k) θc 3 36 Enforce moment equilibrium at nodes B and C: M M 0 BA CB BC M M 0 CD 3 3( k) θ B 75 k θb θc 90 0 5 k θ B θc 90 ( k) θc 3 36 0 8.5k θ k θ 15 B B k θ 10.667k θ 54 The corresponding end moments are: M 0 AB 3 M BA 3( k) θb 75 88.9 kip-ft BC B C CB B C B C k θ 3.08 k θ 5.64 M k θ θ 90 88.9 kip-ft M k θ θ 90 73.6 kip-ft 5 M CD ( k) θc 3 36 73.6 kip-ft 5 M DC ( k) θc 3 36 17. kip-ft C C

Problem 10.10 Assume E= 00 GPa, and I 6 4 100(10) mm. EI k k k L Let member AB member BC F 90(6)(3) F 90(6) (3) MAB 60 kn-m M BA 10 kn-m 9 9 30(9) F 10(0) CB F MBC 0.5 kn-m M 0.5 kn-m 1 8 The slope-deflection equations take the form: M k θ 60 AB BA B M 4k θ 10 B M k (θ θ ) 0.5 BC B C M k (θ θ ) 0.5 CB B C Enforce moment equilibrium at nodes B and C: M 135 0 k (θ θ ) 0.5 135 0 k (θ θ ) 33.75 CB B C B C M M 0 4k θ 10 k (θ θ ) 0.5 0 8k θ kθ 8.5 BA BC B B C B C k θ 16.6 k θ 5.18 B C The corresponding end moments are: M k θ 60 16.8 kn-m AB BA B M 4k θ 10 186.4 kn-m B 18

M k (θ θ ) 0.5 186.4 kn-m BC B C M k (θ θ ) 0.5 135 kn-m CB B C Problem 10.11 Assume E 9,000ksi and I 4 100 in. E(4I) EI EI k k ) k k k 0 5 10 Let AB BC BD modified The modified slope-deflection equations take the form: M M 3(k) θ BA modified BC modified B M 3(k) θ BD modified B Enforce moment equilibrium at node B: M M M 1 0 BA modified BC modified BD modified 3(k) θ 3(k) θ 3(k) θ 1 B B B k θb 0.8 The corresponding end moments are: M M 3(k) θ 4.8 kip-ft BA modified BC modified B M 3(k) θ.4 kip-ft BD modified B 19

Problem 10.1 Assume E 00 GPa, I (a) Ib Ic (b) Ib 1.5I c 6 4 c 10(10) mm, L 8 m, h 4 m and P=50 kn. 0

(a) I I 10(10) mm b c 6 4 Ic 10(10) mm 6 4 θ θ B B A M M AD AB (b) Ic 6 4 10(10) mm, I 1.5I 180(10) mm b c 6 4

Problem 10.13 4 Assume E 9,000ksi andi 00 in. For no axial deformation, bending moment is zero throughout the structure. Problem 10.14 Assume E 00 GPa, and I 6 4 80(10) mm. 3

Problem 10.15 4 I 600 in A 6 in E 9, 000 k/in 5

Problem 10.16 6 4 Assume E 00 GPa, and I 10(10) mm. Neglecting axial deformation. HC HD.5, θa θc EI E(I) k AC k k BC k 6 4 M 4k (θ 3 ) CA M 4k (θ 3 ) AC A A M k (θ θ ) 6k θ AB A C A M M 50.6 kn-m AB AC BA M M 50.6 kn-m CA BD M M 84.4 kn-m DB 6

Problem 10.17 Assume E 9,000ksi and I 4 00 in. For no axial deformation, bending moment is zero throughout the structure. Problem 10.18 Assume E 00 GPa, and I 6 4 80(10) mm. 7

Problem 10.19 Assume E 9,000ksi and I 4 00 in. 9

Deflection profile Moment Diagram For the following beams and frames defined in Problems 10.0-10.34, determine the member end moments using moment distribution. Problem 10.0 E = 9,000 ksi, I = 300 4 in 30

M M F AB F BA (45) 135 kip-ft 30 (45) 0.5 kip-ft 0 M M F BC F CB 10(60) 75 kip-ft 8 75 kip-ft M M F CD F DC 1.0(45) 168.75 kip-ft 1 168.75 kip-ft DF DF 0 DC AB I I I 7I ( ) ( ) L 45 60 180 I 4 At joints B or C DF 45 BA DFCD 7I 7 180 4 3 DFCB DFBC 1-7 7 Case(a): 31

Case (b): F F 6EIδB 6(9000)(300)(.5) M AB set. M BA set. 7.45 kip-ft 3 L (45) (1) F F 6(9000)(300)(.5) M BC set. M CB set. 4.19 kip-ft 3 (60) (1) 3

Case (c): Moment diagram-loading kip-ft Moment diagram settlement kip-ft Moment diagram -( loading + settlement) kip-ft 33

Problem 10.1 Check your results with Computer analysis. Assume E 00 GPa, and I 6 4 75(10) mm. 34

Shear diagram Moment diagram Problem 10. Assume EI constant. (a) 35

(b) 36

Problem 10.3 Determine the bending moment distribution. Assume EI is constant. 37

Problem 10.4 Determine the bending moment distribution. Assume I 1 1.4I. 38

Problem 10.5 Determine the bending moment distribution. 39

Problem 10.6 Solve for the bending moments. δ B.4 in,e=9,000 ksi and I 4 40 in. 40

Problem 10.7 Determine the bending moment distribution and the deflected shape. E=9,000 ksi (a) Take I1 I 1000 in 4 41

Moment diagram Deflected shape (b) Take I 1.5 I 1. Use Computer analysis. 4

Moment diagram Deflected shape Case (a) has larger moment and deflection compare to case(b). 43

Problem 10.8 Determine the axial, shear, and bending moment distributions. Take I g=i c Axial Force 44

Shear diagram Moment Diagram (kn-m) Problem 10.9 Determine the member forces and the reactions. a) Consider only the uniform load shown b) Consider only the support settlement of joint D ( δ D =.5 in ) /0 F c) Consider only the temperature increase of ΔT 80 for member BC. E = 9,000 ksi, I I 100 in, I 400 in, α = 4 4 AB CD BC -6 0 6.5x10 / F 45

(a) (b) 46

Moment diagram (kn-m) (b) Symmetrical loading- no sway Moment diagram (kn-m) 48

Problem 10.31 Solve for the bending moments. Moment diagram (kn-m) 50

Moment diagram Problem 10.3 Determine the bending moment distribution. 51

Problem 10.33 Solve for the bending moments. Moment diagram (kip-ft) Symmetrical loading- no sway 5

Moment diagram Problem 10.34 For the frame shown, determine the end moments and the reactions. Assume E I 6 4 40(10) mm. 00 GPa, and M 0 M F AB F BA 4(6) 8 108 kn-m I 3 4I 3 4I 3 I 5I ( ) ( ) ( ) L 4 6 4 6 4 3 4 Joint B I DF BA DFCB 0.4 DF DB. 5I 4 53

Problem 10.35 Determine analytic expression for the rotation and end moments at B. TakeI in for all members, and α =1.0,.0, 5.0. Is there a upper limit for the end moment, 4 1000 in, A=0 M BD? 54

M BD is independent of α. MBD 40 kip-ft 55

EI k k k 0 AB BC kbd αi E αk 0 Δ 0 M 3(k) θ BA modified M 3(k) θ BC modified M 3(αk) θ 3(αk) BD modified B B B M BA modified 3(k) θb 180 k θb 60 joint B M 0-3(αk) θ 3(αk) 6(k) θ 10 0 B B 80 60α M BD modified 3(α)(-60) 3(αk) 40 k α EI 9000(1000) 1 k 10069.4 k-ft 0 0 (1) 56

3.33 in for α 1.0 (80 60α)(0)1 Δ L.38 in for α.0 αk 1.81 in for α 5.0 Problem 10.36 Compare the end moments and horizontal displacement at A for the rigid frames shown below. Check your results for parts (c) and (d) with a computer based analysis. Take E 00 GPa, and I 6 4 10(10) mm. A=10000 Case (a) mm for all members. 57

Moment diagram kn-m Case (b) Moment diagram kn-m Case (c) 58

Moment diagram kn-m Case (d) 59

60 Moment diagram kn-m

http://www.springer.com/978-1-4614-361-6