矩阵位移法题目及答案

1．作图示刚架的
F、S F、M图，已知各杆截面均为矩形，柱截面宽
N
0.4m,高0.4m, 大跨梁截面宽0.35m,高0.85m，小跨梁截面宽0.35m,高0.6m，各杆E=3.0×104 MPa。

10分
2、计算图示桁架各杆的轴力。

已知A=2400mm2,E=2.0×105MPa。

5分
3．作图示连续梁的
F、M图，已知各梁截面面积A=6.52m,惯性矩
S
I=5.504m，各杆E=3.45×104MPa。

5分
答案1
*******************************************************************
* *
* 1 composite beam 2012.10.17 *
* *
*******************************************************************
3E10 16 13 9 1
1 2 0.2975 17.912E-3
2 3 0.2975 17.912E-3
3 4 0.21 6.3E-3
1 5 0.16 2.133E-3
3 6 0.16 2.133E-3
4 7 0.16 2.133E-3
5 6 0.2975 17.912E-3
6 7 0.21 6.3E-3
5 8 0.1
6 2.133E-3
6 9 0.16 2.133E-3
7 10 0.16 2.133E-3
8 9 0.2975 17.912E-3
9 10 0.21 6.3E-3
8 11 0.16 2.133E-3
9 12 0.16 2.133E-3
10 13 0.16 2.133E-3 0 10.9
3.8 10.9
7.6 10.9
11.4 10.9
0 7.7
7.6 7.7
11.4 7.7
0 4.5
7.6 4.5
11.4 4.5
0 0
7.6 0
11.4 0
111 0
112 0
113 0
121 0
122 0
123 0
131 0
132 0
133 0
4
1 100E3 0 0
2 0 0 -15E3
3 0 0 -15E3
5 100E3 0 0
7
12 2 -26E3 3.8
13 2 -26E3 2.7
7 4 -36E3 7.6
8 4 -36E3 3.8
12 4 -36E3 7.6
13 4 -36E3 3.8
14 3 20E3 4.5
第一题结果
******************************************************************* * * * 1 composite beam 2012.10.17 * * * *******************************************************************
The Input Data
The General Information
E NM NJ NS NLC
3.000E+10 16 13 9 1
The Information of Members
member start end A I
1 1
2 2.975000E-01 1.791200E-02
2 2
3 2.975000E-01 1.791200E-02
3 3
4 2.100000E-01 6.300000E-03
4 1
5 1.600000E-01 2.133000E-03
5 3
6 1.600000E-01 2.133000E-03
6 4
7 1.600000E-01 2.133000E-03
7 5 6 2.975000E-01 1.791200E-02
8 6 7 2.100000E-01 6.300000E-03
9 5 8 1.600000E-01 2.133000E-03
10 6 9 1.600000E-01 2.133000E-03
11 7 10 1.600000E-01 2.133000E-03
12 8 9 2.975000E-01 1.791200E-02
13 9 10 2.100000E-01 6.300000E-03
14 8 11 1.600000E-01 2.133000E-03
15 9 12 1.600000E-01 2.133000E-03
16 10 13 1.600000E-01 2.133000E-03
The Joint Coordinates
joint X Y
1 .000000 10.900000
2 3.800000 10.900000
3 7.600000 10.900000
4 11.400000 10.900000
5 .000000 7.700000
6 7.600000 7.700000
7 11.400000 7.700000
8 .000000 4.500000
9 7.600000 4.500000
10 11.400000 4.500000
11 .000000 .000000
12 7.600000 .000000
13 11.400000 .000000
The Information of Supports
IS VS
111 .000000
112 .000000
113 .000000
121 .000000
122 .000000
123 .000000
131 .000000
132 .000000
133 .000000
( NA= 357 )
( NW= 1167 )
Loading Case 1
The Loadings at Joints
NLJ= 4
ILJ PX PY PM
1 100000.0000 .0000 .00000
2 .0000 .0000 -15000.00000
3 .0000 .0000 -15000.00000 5 100000.0000 .0000 .00000
The Loadings at Members
NLM= 7
ILM ITL PV DST
12 2 -26000.0000 3.800000
13 2 -26000.0000 2.700000
7 4 -36000.0000 7.600000
8 4 -36000.0000 3.800000
12 4 -36000.0000 7.600000
13 4 -36000.0000 3.800000
14 3 20000.0000 4.500000
The Results of Calculation
The Joint Displacements
joint u v phi
1 1.845349E-0
2 -1.982711E-04 -1.263100E-04
2 1.841771E-02 -3.424398E-04 -8.180773E-06
3 1.838193E-02 -5.043591E-0
4 -1.356524E-04
4 1.836317E-02 -3.892198E-04 -1.683554E-04
5 1.608566E-02 -2.069957E-04 -9.278065E-04
6 1.601139E-02 -5.147233E-04 6.593305E-05
7 1.599555E-02 -3.701310E-04 -4.819689E-04
8 1.132049E-02 -1.535800E-04 -1.283845E-03
9 1.131820E-02 -3.849935E-04 3.869225E-05
10 1.130585E-02 -2.796765E-04 -9.193725E-04
11 7.105234E-18 -1.638186E-17 -1.781240E-17
12 9.610834E-18 -4.106598E-17 -2.156936E-17
13 7.783932E-18 -2.983216E-17 -1.882119E-17
The Terminal Forces
member N(st) Q(st) M(st) N(en) Q(en) M(en)
1 84035.890 -13086.980 -41569.990 -84035.890 13086.980 -8160.553
2 84035.890 -13086.980 -6839.447 -84035.890 13086.980 -42891.100
3 31099.080 -28633.300 -52776.720 -31099.080 28633.300 -56029.810
4 -13086.980 15964.110 41569.990 13086.980 -15964.110 9515.139
5 -15546.320 52936.810 80667.820 15546.320 -52936.810 88729.980
6 28633.300 31099.080 56029.810 -28633.300 -31099.080 43487.230
7 87221.910 93210.620 -62622.290 -87221.910 180389.400 -268657.000
8 26256.560 29751.550 -2861.126 -26256.560 107048.500 -144003.000
9 80123.640 28742.190 53107.150 -80123.640 -28742.190 38867.850
10 194594.600 113902.200 182788.200-194594.600-113902.200 181698.700
11 135681.800 57355.640 100515.700-135681.800 -57355.640 83022.300
12 2689.851 83695.000 -146729.300 -2689.851 215905.000 -355668.700
13 20483.680 160.171 -42824.000 -20483.680 162639.800 -245087.300
14 163818.600 26052.340 107861.500-163818.600-116052.300 211874.000
15 410659.800 96108.340 216794.000-410659.800 -96108.340 215693.500
16 298321.600 77839.320 162065.000-298321.600 -77839.320 188211.900
( NA= 357 )单位（N m）
( NW= 1195 )
第二题答案
******************************************************************* * * * 2 composite beam 2012.10.17 * * * ******************************************************************* 2E11 14 9 4 1
1 2 2.4E-3 1E-10
2 3 2.4E-3 1E-10
3 4 2.4E-3 1E-10
4 5 2.4E-3 1E-10
1 8 2.4E-3 1E-10
1 6 2.4E-3 1E-10
2 6 2.4E-
3 1E-10
3 6 2.4E-3 1E-10
3 7 2.4E-3 1E-10
4 7 2.4E-3 1E-10
5 7 2.4E-3 1E-10
5 9 2.4E-3 1E-10
6 8 2.4E-3 1E-10
7 9 2.4E-3 1E-10
0 6
2 6
4 6
6 6
8 6
2 3
6 3
0 0
8 0
81 0
82 0
91 0
92 0
5
1 0 -50E3 0
2 0 -50E
3 0
3 0 -50E3 0
4 0 -50E3 0
5 -10E3 -50E3 0
第二题结果
******************************************************************* * * * 2 composite beam 2012.10.17 * * * *******************************************************************
The Input Data
The General Information
E NM NJ NS NLC
2.000E+11 14 9 4 1
The Information of Members
member start end A I
1 1
2 2.400000E-0
3 1.000000E-10
2 2
3 2.400000E-03 1.000000E-10
3 3
4 2.400000E-03 1.000000E-10
4 4
5 2.400000E-03 1.000000E-10
5 1 8 2.400000E-03 1.000000E-10
6 1 6 2.400000E-03 1.000000E-10
7 2 6 2.400000E-03 1.000000E-10
8 3 6 2.400000E-03 1.000000E-10
9 3 7 2.400000E-03 1.000000E-10
10 4 7 2.400000E-03 1.000000E-10
11 5 7 2.400000E-03 1.000000E-10
12 5 9 2.400000E-03 1.000000E-10
13 6 8 2.400000E-03 1.000000E-10
14 7 9 2.400000E-03 1.000000E-10
The Joint Coordinates
joint X Y
1 .000000 6.000000
2 2.000000 6.000000
3 4.000000 6.000000
4 6.000000 6.000000
5 8.000000 6.000000
6 2.000000 3.000000
7 6.000000 3.000000
8 .000000 .000000
9 8.000000 .000000
The Information of Supports
IS VS
81 .000000
82 .000000
91 .000000
92 .000000
( NA= 270 )
( NW= 907 )
Loading Case 1
The Loadings at Joints
NLJ= 5
ILJ PX PY PM
1 .0000 -50000.0000 .00000
2 .0000 -50000.0000 .00000
3 .0000 -50000.0000 .00000
4 .0000 -50000.0000 .00000
5 -10000.0000 -50000.0000 .00000
The Loadings at Members
NLM= 0
The Results of Calculation
The Joint Displacements
joint u v phi
1 -1.052370E-04 -9.375000E-04 -7.026900E-05
2 -1.746814E-04 -1.193735E-0
3 1.087907E-04
3 -2.441259E-0
4 -8.137530E-04 -3.230397E-05
4 -3.552370E-04 -1.302888E-03 -1.022944E-04
5 -4.663480E-04 -9.375000E-04 1.412285E-04
6 3.860367E-04 -8.812350E-04 1.226822E-04
7 -7.938921E-04 -9.903881E-04 -6.211760E-05
8 -3.833334E-18 -1.325000E-17 -2.257494E-04
9 2.833334E-18 -1.175000E-17 3.510750E-04
The Terminal Forces
member N(st) Q(st) M(st) N(en) Q(en) M(en)
1 16666.660 .009 .007 -16666.660 -.009 .011
2 16666.660 -.009 -.008 -16666.660 .009 -.011
3 26666.660 .011 .011 -26666.660 -.011 .010
4 26666.660 -.010 -.012 -26666.660 .010 -.007
5 75000.000 -.001 -.003 -75000.000 .001 -.004
6 -30046.240 -.002 -.004 30046.240 .002 -.002
7 49999.980 -.002 -.003 -49999.980 .002 -.003
8 39060.150 -.002 -.005 -39060.150 .002 -.003
9 21032.390 .002 .004 -21032.390 -.002 .003
10 49999.980 .002 .002 -49999.980 -.002 .003
11 -30046.240 .002 .005 30046.240 -.002 .002
12 75000.000 .001 .003 -75000.000 -.001 .004
13 69106.410 .003 .008 -69106.410 -.003 .004
14 51078.650 -.004 -.009 -51078.650 .004 -.004
( NA= 270 )
( NW= 907 )
第三题答案
*******************************************************************
* *
* 3 composite beam 2012.10.17 *
* *
******************************************************************* 3.45E10 4 5 6 1
1 2 6.5 5.5
2 3 6.5 5.5
3 4 6.5 5.5
4 5 6.5 5.5
0 0
40 0
60 0
80 0
120 0
11 0
12 0
13 0
22 0
42 0
52 0
1
3 0 -320E3 -100E3
4
1 4 -10.5E3 40
2 4 -10.5E
3 20
3 4 -10.5E3 20
4 4 -10.5E3 40
第三题结果
******************************************************************* * * * 3 composite beam 2012.10.17 * * * *******************************************************************
The Input Data
The General Information
E NM NJ NS NLC
3.450E+10 4 5 6 1
The Information of Members
member start end A I
1 1
2 6.500000E+00 5.500000E+00
2 2
3 6.500000E+00 5.500000E+00
3 3
4 6.500000E+00 5.500000E+00
4 4
5 6.500000E+00 5.500000E+00
The Joint Coordinates
joint X Y
1 .000000 .000000
2 40.000000 .000000
3 60.000000 .000000
4 80.000000 .000000
5 120.000000 .000000
The Information of Supports
IS VS
11 .000000
12 .000000
13 .000000
22 .000000
42 .000000
52 .000000
( NA= 66 )
( NW= 299 )
Loading Case 1
The Loadings at Joints
NLJ= 1
ILJ PX PY PM
3 .0000 -320000.0000 -100000.00000
The Loadings at Members
NLM= 4
ILM ITL PV DST
1 4 -10500.0000 40.000000
2 4 -10500.0000 20.000000
3 4 -10500.0000 20.000000
4 4 -10500.0000 40.000000
The Results of Calculation
The Joint Displacements
joint u v phi
1 0.000000E+00 3.713943E-18 4.951923E-17
2 0.000000E+00 -2.916827E-17 -5.219418E-05
3 0.000000E+00 -1.405865E-03 1.038816E-06
4 0.000000E+00 -3.431731E-17 4.276883E-05
5 0.000000E+00 6.771635E-18 5.239688E-05
The Terminal Forces
member N(st) Q(st) M(st) N(en) Q(en) M(en)
1 .000 172860.600 904807.700 .000 247139.400-2390385.000
2 .000 359543.300 2390385.000 .000-149543.300 2700481.000
3 .000-170456.700-2800481.000 .000 380456.700-2708654.000
4 .000 277716.300 2708654.000 .000 142283.700 .000
( NA= 66 )
( NW= 315 )。