Structural Analysis: Solution manual

553 Pages·137.422 MB·English

by  Hibbeler R.C.

© 2012 Pearson Education,Inc.,Upper Saddle River,NJ.All rights reserved.This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced,in any form or by any means,without permission in writing from the publisher.
16–1. Determine the structure stiffness matrix K for 8 2
the frame.Assume 1 and 3 are fixed.Take E = 200 GPa, 12 kN/m
I = 300 106 mm4,A = 10 103 mm2for each member. 9 3
1 2 1 2 1
7
1 2
1
2 m
4 m
2
10 kN
5
2 m
6
3
4
Member Stiffness Matrices.
The orgin of the global coordinate system will be set at joint 1 .
For member ƒ 1 ƒ and ƒ 2 ƒ,L = 4m
AE 0.01 200(109)
= 3 4 = 500(106) N/m
L 4
12EI 12 200(109) 300(10-6)
= 3 43 4 = 11.25(106) N/m
L3 43
6EI 6 200(109) 300(10-6)
= 3 43 4 = 22.5(106) N
L2 42
4EI 4 200(109) 300(10-6) #
= 3 43 4 = 60(106) N m
L 4
2EI 2 200(109) 300(10-6) #
= 3 43 4 = 30(106) N m
L 4
4 - 0 0 - 0
For member ƒ 1 ƒ,l = = 1and l = = 0.Thus,
x 4 y 4
7 8 9 1 2 3
500 0 0 -500 0 0 7
0 11.25 22.5 0 -11.25 22.5 8
0 22.5 60 0 -22.5 30 9
k1 = -500 0 0 500 0 0 1 (106)
0 -11.25 -22.5 0 11.25 -22.5 2
0 22.5 30 0 -22.5 60 3
F V
4 - 4 -4 - 0
For member ƒ 2 ƒ,l = = 0and l = = -1.Thus,
x 4 y 4
1 2 3 4 5 6
11.25 0 22.5 -11.25 0 22.5 1
0 500 0 0 -500 0 2
22.5 0 60 -22.5 0 30 3
k = (106)
2 -11.25 0 -22.5 11.25 0 -22.5 4
0 -500 0 0 500 0 5
22.5 0 30 -22.5 0 60 6
F V
535
© 2012 Pearson Education,Inc.,Upper Saddle River,NJ.All rights reserved.This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced,in any form or by any means,without permission in writing from the publisher.
16–1. Continued
Structure Stiffness Matrix.It is a 9 * 9 matrix since the highest code number is 9.Thus,
1 2 3 4 5 6 7 8 9
511.25 0 22.5 -11.25 0 22.5 -500 0 0 1
0 511.25 -22.5 0 -500 0 0 -11.25 -22.25 2
22.5 -22.5 120 -22.5 0 30 0 22.5 30 3
-11.25 0 -22.5 11.25 0 -22.5 0 0 0 4
Ans.
K = 0 -500 0 0 500 0 0 0 0 5 (106)
22.5 0 30 -22.5 0 60 0 0 0 6
-500 0 0 0 0 0 500 0 0 7
I 0 -11.25 22.5 0 0 0 0 11.25 22.5 Y 8
0 -22.5 30 0 0 0 0 25.5 60 9
16–2. Determine the support reactions at the fixed 8 2
supports 1 and 3 .Take E = 200 GPa,I = 300 106 mm4, 12 kN/m
1 2
A = 10 103 mm2for each member. 9 3
1 2 1
7
1 2
1
2 m
4 m
2
10 kN
5
2 m
6
3
4
Known Nodal Loads and Deflections.The nodal load acting on the
unconstrained degree of freedom (code number 1,2 and 3) are shown in
Fig.aand b.
0 4
0 5
-5(103) 1
0 6
Q = -24(103) 2 and D =
k k 0 7
11(103) 3
0 8
0 9
C S F V
Loads-Displacement Relation.Applying Q = KD,
-5(103) 511.25 0 22.5 -11.25 0 22.5 -500 0 0 D
1
-24(103) 0 511.25 -22.5 0 -500 0 0 -11.25 -22.5 D
2
11(103) 22.5 -22.5 120 -22.5 0 30 0 22.5 30 D
3
Q -11.25 0 -22.5 11.25 0 -22.5 0 0 0 0
4
Q = 0 -500 0 0 500 0 0 0 0 (106) 0
5
Q 22.5 0 30 -22.5 0 60 0 0 0 0
6
Q -500 0 0 0 0 0 500 0 0 0
7
Q 0 -11.25 22.5 0 0 0 0 11.25 22.5 0
I 8 Y I Y I Y
Q 0 -22.5 30 0 0 0 0 22.5 60 0
9
536
© 2012 Pearson Education,Inc.,Upper Saddle River,NJ.All rights reserved.This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced,in any form or by any means,without permission in writing from the publisher.
16–2. Continued
From the matrix partition,Q = K D + K D ,
k 11 u 12 k
-5(103) = (511.25D + 22.5D )(106) (1)
1 3
-24(103) = (511.25D - 22.5D )(106) (2)
2 3
11(103) = (22.5D - 22.5D + 120D )(106) (3)
1 2 3
Solving Eqs.(1) to (3),
D = -13.57(10-6) m D = -43.15(10-6)m D = 86.12(10-6)rad
1 2 3
Using these results and applying Q = K D + K D ,
u 21 u 22 k
Q = -11.25(106)(-13.57)(10-6) + (-22.5)(106)(86.12)(10-6) = -1.785 kN
4
Q = -500(106)(-43.15)(10-6) = 21.58 kN
5
#
Q = 22.5(106)(-13.57)(10-6) + 30(106)(86.12)(10-6) = 2.278 kN m
6
Q = -500(106)(-13.57)(10-6) = 6.785 kN
7
Q = -11.25(106)(-43.15)(10-6) + 22.5(106)(86.12)(10-6) = 2.423 kN
8
#
Q = -22.5(106)(-43.15)(10-6) + 30(106)(86.12)(10-6) = 3.555 kN m
9
Superposition these results to those of FEM shown in Fig.a,
R = -1.785 + 5 = 3.214 kN = 3.21 kN : Ans.
4
R = 21.58 + 0 = 21.58 kN = 21.6 kN c Ans.
5
# #
R = 2.278 - 5 = -2.722 kN m = 2.72 kN m b Ans.
6
R = 6.785 + 0 = 6.785 kN = 6.79 kN : Ans.
7
R = 2.423 + 24 = 26.42 kN = 26.4 kN c Ans.
8
# #
R = 3.555 + 16 = 19.55 kN m = 19.6 kN m d Ans.
9
537
© 2012 Pearson Education,Inc.,Upper Saddle River,NJ.All rights reserved.This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced,in any form or by any means,without permission in writing from the publisher.
16–3. Determine the structure stiffness matrix K for 5 m
tEhe= fr2a0m0 eM.PAas,suIm=e 3300. i1s0 6pi nmnmed4, aAnd= 121. i1s0 3fix medm.2Tafkoer 89 300 kN (cid:2) m 2 3
1 2 1 2 1
each member.
7 1
1 2
4 m
2
6
4
3
5
For member 1
5 - 0
l = = 1 l = 0
x 5 y
AE (0.021)(200)(106) 12EI (12)(200)(106)(300)(10-6)
= = 840000 = = 5760
L 5 L3 53
6EI 6(200)(106)(300)(10-6) 2EI 2(200)(106)(300)(10-6)
= = 14400 = = 24000
L2 52 L 5
4EI 4(200)(106)(300)(10-6)
= = 48000
L 5
840000 0 0 -840000 0 0
0 5760 14400 0 -5760 14400
0 14400 48000 0 -14400 24000
k =
1 -840000 0 0 840000 0 0
0 -5760 -14400 0 5760 -14400
0 14400 24000 0 -14400 48000
F V
For member 2
0 - (-4)
l = 0 l = = 1
x y 4
AE (0.021)(200)(106) 12EI (12)(200)(106)(300)(10-6)
= = 1050000 = = 11250
L 5 L3 43
6EI 6(200)(106)(300)(10-6) 2EI 2(200)(106)(300)(10-6)
= = 22500 = = 30000
L2 42 L 4
4EI 4(200)(106)(300)(10-6)
= = 60000
L 4
11250 0 -22500 -11250 0 -22500
0 1050000 0 0 -1050000 0
-22500 0 60000 22500 0 30000
k =
2 -11250 0 22500 11250 0 22500
0 -1050000 0 0 1050000 0
-22500 0 30000 22500 0 60000
F V
538
© 2012 Pearson Education,Inc.,Upper Saddle River,NJ.All rights reserved.This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced,in any form or by any means,without permission in writing from the publisher.
16–3. Continued
Structure Stiffness Matrix.
851250 0 22500 22500 -11250 0 -840000 0 0
0 1055760 -14400 0 0 -1050000 0 -5760 -14400
22500 -14400 108000 30000 -22500 0 0 144000 24000
22500 0 30000 60000 -22500 0 0 0 0
K = -11250 0 -22500 -22500 11250 0 0 0 0 Ans.
0 - 1050000 0 0 0 1050000 0 0 0
-840000 0 0 0 0 0 840000 0 0
0 -5760 14400 0 0 0 0 5760 14400
I Y
0 -14400 24000 0 0 0 0 14400 48000
*16–4. Determine the support reactions at 1 . and 3 . 5 m
Take E = 200 MPa, I = 30011062 mm4, A = 2111032 mm2 8 300 kN (cid:2) m 2
for each member. 9 3
1
7 1
1 2
4 m
2
6
4
3
5
0
0
0
0
D = 0 Q =
k k 300
0
0
0
E U D T
0 851250 0 22500 22500 -11250 0 -840000 0 0 D
1
0 0 1055760 -14400 0 0 -1050000 0 -5760 -14400 D
2
300 22500 -14400 108000 30000 -22500 0 0 14400 24000 D
3
0 22500 0 30000 60000 -22500 0 0 0 0 D
4
Q = -11250 0 -22500 -22500 11250 0 0 0 0 0
5
Q 0 -1050000 0 0 0 1050000 0 0 0 0
6
Q -840000 0 0 0 0 0 840000 0 0 0
7
Q 0 -5760 14400 0 0 0 0 5760 14400 0
I 8Y I Y I Y
Q 0 -14400 24000 0 0 0 0 1440 48000 0
9
539
© 2012 Pearson Education,Inc.,Upper Saddle River,NJ.All rights reserved.This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced,in any form or by any means,without permission in writing from the publisher.
16–4. Continued
Partition matrix
0 851250 0 22500 22500 D 0
1
0 0 1055760 -14400 0 D 0
= 2 +
300 22500 -14400 108000 30000 D 0
3
0 22500 0 30000 60000 D 0
4
0D = 8T51250DD + 22500D + 22500D T D T D T
1 3 4
0 = 1055760D - 14400D
2 3
300 = 22500D - 14400D + 108000D + 30000D
1 2 3 4
0 = 22500D + 30000D + 60000D
1 3 4
Solving.
D = -0.00004322 m D = 0.00004417 m D = 0.00323787 rad
1 2 3
D = -0.00160273 rad
4
Q -11250 0 -22500 -22500 0
5 -0.00004322
Q 0 -1050000 0 0 0
6 0.00004417
Q = -840000 0 0 0 + 0
7 0.00323787
Q 0 -5760 14400 0 0
8 -0.00160273
Q 0 -14400 24000 0 0
9
E U E U D T E U
Q = -36.3 kN Ans.
5
Q = -46.4 kN Ans.
6
Q = 36.3 kN Ans.
7
Q = 46.4 kN Ans.
8
#
Q = 77.1 kN m Ans.
9
Check equilibrium
aaFx = 0; 36.30 - 36.30 = 0 (Check)
+caFy = 0; 46.37 - 46.37 = 0 (Check)
a+ aM1 = 0; 300 + 77.07 - 36.30(4) - 46.37(5) = 0 (Check)
540
© 2012 Pearson Education,Inc.,Upper Saddle River,NJ.All rights reserved.This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced,in any form or by any means,without permission in writing from the publisher.
16–5. Determine the structure stiffness matrix Kfor the frame. 9 60 kN 2
Take E = 200 GPa, I = 350 106 mm4, A = 15 103 mm2
1 2 1 2
5 3
for each member.Joints at 1 and 3 are pins.
1
8
1 2
1
2 m 2 m
2
4 m
7
4
3
6
Member Stiffness Matrices.The origin of the global coordinate system will be set at joint 1 .
For member ƒ 1 ƒ and ƒ 2 ƒ,L = 4m.
AE 0.015[200(109)]
= = 750(106) N m
L 4 >
12EI 12[200(109)][350(10-6)]
= = 13.125(106) N m
L3 43 >
6EI 4[200(109)][350(10-6)]
= = 26.25(106 ) N
L2 42
4EI 4[200(109)][350(10-6)] #
= = 70(106) N m
L 4
2EI 2[200(109)][350(10-6)] #
= = 35(106) N m
L 4
4 - 0 0 - 0
For member ƒ 1 ƒ,l = = 1and l = = 0.Thus,
x 4 y 4
8 9 5 1 2 3
750 0 0 -750 0 0 8
0 13.125 26.25 0 -13.125 26.25 9
0 26.25 70 0 -26.25 35 5
k = (106)
1 -750 0 0 750 0 0 1
0 -13.125 -26.25 0 13.125 -26.25 2
0 26.25 35 0 -26.25 70 3
F V
4 - 4 -4 - 0
For member ƒ 2 ƒ,l = = 0,and l = = -1.Thus,
x 4 y 4
1 2 3 6 7 4
13.125 0 26.25 -13.125 0 26.25 1
0 750 0 0 -750 0 2
26.25 0 70 -26.25 0 35 3
k = (106)
2 -13.125 0 -26.25 13.125 0 -26.25 6
0 -750 0 0 750 0 7
26.25 0 35 -26.25 0 70 4
F V
541
© 2012 Pearson Education,Inc.,Upper Saddle River,NJ.All rights reserved.This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced,in any form or by any means,without permission in writing from the publisher.
16–5. Continued
Structure Stiffness Matrix.It is a 9 * 9 matrix since the highest code number is 9.Thus
1 2 3 4 5 6 7 8 9
763.125 0 26.25 26.25 0 -13.125 0 -750 0 1
0 763.125 -26.25 0 -26.25 0 -750 0 -13.125 2
26.25 -26.25 140 35 35 -26.25 0 0 26.25 3
26.25 0 35 70 0 -26.25 0 0 0 4
K = (106) Ans.
0 -26.25 35 0 70 0 0 0 26.25 5
-13.125 0 -26.25 -26.25 0 13.125 0 0 0 6
0 -750 0 0 0 0 750 0 0 7
I -750 0 0 0 0 0 0 750 0 Y 8
0 -13.125 26.25 0 26.25 0 0 0 13.125 9
16–6. Determine the support reactions at pins 1 and 3 . 9 60 kN 2
Take E = 200 GPa, I = 350 106 mm4, A = 15 103 mm2
1 2 1 2 5 3
for each member.
1
8
1 2
1
2 m 2 m
2
4 m
7
4
3
6
Known Nodal Loads and Deflections.The nodal load acting on
the unconstrained degree of freedom (code numbers 1,2,3,4,and 5)
are shown in Fig.aand Fig.b.
0 1
0 6
-41.25(103) 2
0 7
Q = 45(103) 3 and D =
k k 0 8
0 4
0 9
0 5
E U D T
Loads-Displacement Relation.Applying Q = KD,
0 763.125 0 26.25 26.25 0 -13.125 0 -750 0 D
1
-41.25(103) 0 763.125 -26.25 0 -26.25 0 -750 0 -13.125 D
2
45(103) 26.25 -26.25 140 35 35 -26.25 0 0 26.25 D
3
0 26.25 0 35 70 0 -26.25 0 0 0 D
4
0 = 0 -26.25 35 0 70 0 0 0 26.25 (106) D
5
Q -13.125 0 -26.25 -26.25 0 13.125 0 0 0 0
6
Q 0 -750 0 0 0 0 750 0 0 0
7
Q -750 0 0 0 0 0 0 750 0 0
I 8 Y I Y I Y
Q 0 -13.125 26.25 0 26.25 0 0 0 13.125 0
9
542
© 2012 Pearson Education,Inc.,Upper Saddle River,NJ.All rights reserved.This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced,in any form or by any means,without permission in writing from the publisher.
16–6. Continued
From the matrix partition,Q = K D + K D ,
k 11 u 12 k
0 = (763.125D + 26.25D + 26.25D )(106) (1)
1 3 4
-41.25(103) = (763.125D - 26.25D - 26.25D )(106) (2)
2 3 5
45(103) = (26.25D - 26.25D + 140D + 35D + 35D )(106) (3)
1 2 3 4 5
0 = (26.25D + 35D + 70D )(106) (4)
1 3 4
0 = (-26.25D + 35D + 70D )(106) (5)
2 3 5
Solving Eqs.(1) to (5)
D = -7.3802(10-6) D = -47.3802(10-6) D = 423.5714(10-6)
1 2 3
D = -209.0181(10-6) D = -229.5533(10-6)
4 5
Using these results and applying Q = K D + K D ,
u 21 u 22 k
Q = (-13.125)(106) - 7.3802(10-6) - 26.25(106)423.5714(10-6) - 26.25(106) - 209.0181(10-6) + 0 = -5.535 kN
6
Q = -750(106) - 47.3802(10-6) + 0 = 35.535 kN
7
Q = -750(106) - 7.3802(10-6) + 0 = 5.535 kN
8
Q = -13.125(106) - 47.3802(10-6) + 26.25(106) + 423.5714(10-6) + 26.25(106) - 229.5533(10-6) + 0 = 5.715 kN
9
Superposition these results to those of FEM shown in Fig.a,
R = -5.535 kN + 0 = 5.54 kN Ans.
6
R = 35.535 + 0 = 35.5 kN Ans.
7
R = 5.535 + 0 = 5.54 kN Ans.
8
R = 5.715 + 18.75 = 24.5 kN Ans.
9
543