7 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
8 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
5.38
Figure P5–38
NODE X- Y- Z- X- Y- Z-
number translation translation translation rotation rotation rotation (deg)
1 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
4 1.4282E–01 –2.1948E–03 0.0000E+00 0.0000E+00 0.0000E+00 2.2790E–01
5 1.4265E–01 –4.7155E–04 0.0000E+00 0.0000E+00 0.0000E+00 –5.1623E–01
6 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
7 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
BEAM ELEMENT FORCES AND MOMENTS
ELEMENT CASE AXIAL SHEAR SHEAR TORSION BENDING BENDING
NO. (MODE) FORCE FORCE FORCE MOMENT MOMENT MOMENT
R1 R2 R3 M1 M2 M3
1 1 –2.907E+04 –9.878E+05 0.000E+00 0.000E+00 0.000E+00 –1.538E+06
–2.907E+04 2.122E+05 0.000E+00 0.000E+00 0.000E+00 –3.605E+04
2 1 –2.122E+05 –2.907E+04 0.000E+00 0.000E+00 0.000E+00 –3.605E+04
–2.122E+05 7.593E+04 0.000E+00 0.000E+00 0.000E+00 –2.001E+05
3 1 –1.024E+05 4.084E+04 0.000E+00 0.000E+00 0.000E+00 2.022E+05
5.39
(a) Truss model
NUMBER OF ELEMENTS (NELE) = 15
NUMBER OF NODES (KNODE) = 8
NODE POINTS
285
ELEMENTS
K NODE(I,K) E(K) A(K)
1 1 3 2.0000E+11 2.0000E–04
2 1 4 2.0000E+11 2.0000E–04
3 2 3 2.0000E+11 2.0000E–04
14 6 8 2.0000E+11 2.0000E–04
15 7 8 2.0000E+11 2.0000E–04
NUMBER OF NONZERO UPPER CO-DIAGONALS (MUD)-11
DISPLACEMENTS X Y Z
NODE NUMBER 1 0.0000E+00 0.0000E+00 0.0000E+00
NODE NUMBER 2 0.0000E+00 0.0000E+00 0.0000E+00
NODE NUMBER 3 0.7935E–02 0.3730E–02 0.0000E+00
STRESSES IN ELEMENTS (IN CURRENT UNITS)
ELEMENT NUMBER STRESS
1 = 0.24864E+09
2 = 0.14809E+09
3 = – 0.16441E+09
4 = – 0.23886E+09
5 = – 0.30998E+08
6 = 0.10311E+09
(b) Rigid frame model
NUMBER OF ELEMENTS = 15
NUMBER OF NODES = 8
286
NODE POINTS
K IFIK XC(K) YC(K) ZC(K) FORCE(1,K) FORCE(2,K) FORCE (3,K)
1 1 1 0 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
2 1 1 0 4.000000 0.000000 0.000000 0.000000 0.000000 0.000000
3 0 0 0 0.000000 3.000000 0.000000 20000.000000 0.000000 0.000000
ELEMENTS
K NODE(I, K) E(K) G(K) A(K) XI(K)
1 1 3 2.0000000E+11 0.0000000E+00 2.0000000E–04 4.0000000E–04
2 1 4 2.0000000E+11 0.0000000E+00 2.0000000E–04 4.0000000E–04
3 2 3 2.0000000E+11 0.0000000E+00 2.0000000E–04 4.0000000E–04
4 2 4 2.0000000E+11 0.0000000E+00 2.0000000E–04 4.0000000E–04
NODE DISPLACEMENTS Z-ROTATION
X Y THETA
1 0.00000E+00 0.00000E+00 – 0.14936E–02
2 0.00000E+00 0.00000E+00 – 0.14200E–02
3 0.54772E–02 0.32294E–02 – 0.17329E–02
5 0.11556E–01 0.40230E–02 – 0.20908E–02
7 0.18021E–01 0.42395E–02 – 0.21639E–02
8 0.17667E–01 – 0.42509E–02 – 0.21228E–02
ELEMENTS
K NODE X-FORCE Y-FORCE Z-MOMENT X-FORCE Y-FORCE Z-MOMENT
(I,K)
1 1 3 – 0.4306E+05 0.2267E+05 0.4038E+05 0.4306E+05 – 0.2267E+05 0.2762E+05
2 1 4 – 0.1745E+05 – 0.1746E+05 – 0.4038E+05 0.1745E+05 0.1746E+05 – 0.4692E+05
9 4 6 0.1087E+05 0.1228E+05 0.2893E+05 – 0.1087E+05 – 0.1228E+05 0.7899E+04
10 5 6 0.3563E+04 – 0.4091E+04 – 0.8150E+04 – 0.3563E+04 0.4091E+04 – 0.8214E+04
11 5 7 – 0.2887E+04 0.2978E+04 0.6416E+04 0.2887E+04 – 0.2978E+04 0.2517E+04
12 5 8 0.6004E+03 – 0. 1903E+04 – 0.4245E+04 – 0.6004E+03 0.1903E+04 – 0.5270E+04
(c) Use program PFRAME to model a truss
(Use PFRAME to model a Truss, i.e., MAKE I 0).
NUMBER OF ELEMENTS = 15
NUMBER OF NODES = 8
NODE POINTS
K IFIX XC(K) YC(K) ZC(K) FORCE(1,K)
1 1 1 0 0.000000 0.000000 0.000000 0.000000
2 1 1 0 4.000000 0.000000 0.000000 0.000000
5 0 0 0 0.000000 6.000000 0.000000 20000.000000
6 0 0 0 4.000000 6.000000 0.000000 0.000000
ELEMENTS
K NODE(I,K) E(K) G(K) A(K) XI(K)
5 3 4 2.0000000E+11 0.0000000E+00 2.0000000E–04 1.0000000E–06
6 3 5 2.0000000E+11 0.0000000E+00 2.0000000E–04 1.0000000E–06
11 5 7 2.0000000E+11 0.0000000E+00 2.0000000E–04 1.0000000E–06
12 5 8 2.0000000E+11 0.0000000E+00 2.0000000E–04 1.0000000E–06
13 6 7 2.0000000E+11 0.0000000E+00 2.0000000E–04 1.0000000E–06
NODE DISPLACEMENTS Z-ROTATION
X Y THETA
1 0.00000E+00 0.00000E+00 – 0.21033E–02
2 0.00000E+00 0.00000E+00 – 0.19714E–02
4 0.73056E–02 – 0.35821E–02 – 0.23447E–02
6 0.16772E–01 – 0.48468E–02 – 0.29231E–02
8 0.25670E–01 –0.50248E–02 – 0.27544E–02
ELEMENTS
K NODE X-FORCE Y-FORCE Z-MOMENT X-FORCE Y-FORCE Z-MOMENT
(I,K) at 1st node at 2nd node
1 1 3 – 0.4970E+05 0.1044E+03 0.1762E+03 0.4970E+05 – 0.1044X+03 0.1372E+03
288
7 3 6 – 0.1548E+05 – 0.2166E+02 – 0.3305E+02 0.1548E+05 0.2166E+02 – 0.7524E+02
8 4 5 0.2170E+05 0.2894E+00 0.2253E+02 – 0.2170E+05 – 0.2894E+00 – 0.2108E+02
9 4 6 0.1686E+05 0.1391E+03 0.2472E+03 – 0.1686E+05 – 0.1391E+03 0.1701E+03
10 5 6 0.5641E+04 – 0.5654E+02 – 0.1114E+03 – 0.5641E+04 0.5654E+02 – 0.1148E+03
11 5 7 – 0.5143E+04 0.1572E+02 0.1480E+02 0.5143E+04 – 0.1572E+02 0.3237E+02
Comparison of TRUSS, PFRAME and modeling a truss using PFRAME
DISPLACEMENTS
u5 v5 u7 v7
TRUSS 0.01738 0.005276 0.02603 0.005662
Note: Global displacements in meters
FORCES
ELEMENT 1 ELEMENT 2 ELEMENT 3
f1x f1y f1x f1y f2x f2y
Truss –49728 0 –29618 0 32882 0
Note 1: From equilibrium, only forces for one node of an element are shown
5.40
For the two-story, two-bay rigid frame shown, determine (1) the nodal displacement
components and (2) the shear force and bending moments in each member. Let E = 200 GPa, I =
2 10–4 m4 for each horizontal member and I = 1.5 10–4 m4 for each vertical member.
Figure P5–40
Displacements/Rotations (degrees) of nodes
NODE X– Y– Z– X– Y– Z–
number translation translation translation rotation rotation rotation
1 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
5 0.0000E+00 –3.1986E–06 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
6 7.2482E-04 –1.4007E–06 0.0000E+00 0.0000E+00 0.0000E+00 –1.6616E–02
7 –7.7467E–07 –2.8014E–06 0.0000E+00 0.0000E+00 0.0000E+00 –6.6411E–02
8 –3.8733E–07 –9.2660E–03 0.0000E+00 0.0000E+00 0.0000E+00 1.6572E–02
9 0.0000E+00 –6.3972E–06 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
10 3.8733E–07 –9.2660E–03 0.0000E+00 0.0000E+00 0.0000E+00 –1.6572E–02
1 **** BEAM ELEMENT FORCES AND MOMENTS
ELEMENT CASE AXIAL SHEAR SHEAR TORSION BENDING BENDING
NO. (MODE) FORCE FORCE FORCE MOMENT MOMENT MOMENT
R1 R2 R3 M1 M2 M3
1 1 –1.121E+05 8.348E+03 0.000E+00 0.000E+00 0.000E+00 1.391E+04
–1.121E+05 8.348E+03 0.000E+00 0.000E+00 0.000E+00 –6.955E+03
2 1 –2.559E+05 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
–5.484E+04 2.384E+04 0.000E+00 0.000E+00 0.000E+00 –5.963E+03
8 1 –1.303E+05 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
–1.303E+05 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
9 1 –5.484E+04 –2.384E+04 0.000E+00 0.000E+00 0.000E+00 –5.364E+04
–5.484E+04 –2.384E+04 0.000E+00 0.000E+00 0.000E+00 5.963E+04
5.41 For the two-story, three-bay rigid frame shown, determine (1) the nodal displacements and (2)
moment diagrams for each member. Let E = 200 GPa, I = 1.29 10–4 m4 for the beams and I
= 0.462 10–4 m4 for the columns. The properties for I correspond to a W 610 155 and a
W 410 114 wide-flange section, respectively, in metric units.
Figure P5–41
1**** BEAM ELEMENT FORCES AND MOMENTS
ELEMENT CASE AXIAL SHEAR SHEAR TORSION BENDING BENDING
NO. (MODE) FORCE FORCE FORCE MOMENT MOMENT MOMENT
R1 R2 R3 M1 M2 M3
–1.347E+04 –1.750E+04 0.000E+00 0.000E+00 0.000E+00 –4.685E+03
5 1 1.347E+04 –1.750E+04 0.000E+00 0.000E+00 0.000E+00 –4.685E+03
1.347E+04 –1.750E+04 0.000E+00 0.000E+00 0.000E+00 4.782E+04
6 1 2.471E+03 –2.000E+04 0.000E+00 0.000E+00 0.000E+00 –2.184E+03
2.471E+03 –2.000E+04 0.000E+00 0.000E+00 0.000E+00 5.782E+04
7 1 –2.470E+03 –2.000E+04 0.000E+00 0.000E+00 0.000E+00 –2.184E+03
–2.686E+03 –3.024E+03 0.000E+00 0.000E+00 0.000E+00 5.469E+03
13 1 2.686E+03 –3.022E+03 0.000E+00 0.000E+00 0.000E+00 5.469E+03
2.686E+03 –3.022E+03 0.000E+00 0.000E+00 0.000E+00 1.151E+04
14 1 1.063E+03 –9.477E+03 0.000E+00 0.000E+00 0.000E+00 2.269E+03
1.063E+03 –9.477E+03 0.000E+00 0.000E+00 0.000E+00 2.122E+04
291
5.42 For the rigid frame shown, determine (1) the nodal displacements and rotations and (2) the
member shear forces and bending moments. Let E = 200 GPa, I = 0.795 10–4 m4 for the
292
5.43 For the rigid frame shown, determine (1) the nodal displacements and rotations and (2) the
Figure P5–43
8 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
9 1.4331E–03 –4.4935E–05 0.0000E+00 0.0000E+00 0.0000E+00 –1.2573E–03
10 2.6228E–03 –8.8701E–05 0.0000E+00 0.0000E+00 0.0000E+00 –1.3909E–03
11 3.5926E-03 –1.4025E-04 0.0000E+00 0.0000E+00 0.0000E+00 –1.5614E–03
12 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
13 1.3394E–03 –1.3694E–04 0.0000E+00 0.0000E+00 0.0000E+00 –1.3297E–03
14 2.4458E–03 –2.0556E–04 0.0000E+00 0.0000E+00 0.0000E+00 –1.3014E–03
15 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
16 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
1 **** BEAM ELEMENT FORCES AND MOMENTS
ELEMENT CASE AXIAL SHEAR SHEAR TORSION BENDING BENDING
NO. (MODE) FORCE FORCE FORCE MOMENT MOMENT MOMENT
R1 R2 R3 M1 M2 M3
Columns
1 1 6.363E+03 –1.139E+04 0.000E+00 0.000E+00 0.000E+00 –1.444E+05
–1.952E+03 –8.662E+03 0.000E+00 0.000E+00 0.000E+00 1.828E+04
7 1 –1.902E+03 –5.757E+03 0.000E+00 0.000E+00 0.000E+00 – 4.814E+04
–1.902E+03 –5.757E+03 0.000E+00 0.000E+00 0.000E+00 3.822E+04
8 1 –2.240E+03 –3.983E+03 0.000E+00 0.000E+00 0.000E+00 –3.620E+04
–2.240E+03 –3.983E+03 0.000E+00 0.000E+00 0.000E+00 2.355E+04
5.44 A structure is fabricated by welding together three lengths of I-shaped members as shown in
Figure P5-44. The yield strength of the members is 36 ksi, E = 29e6 psi, and Poisson’s ratio
is 0.3. The members all have cross-section properties corresponding to a W 18 76. That is,
A = 22.3 in.2, depth of section is d = 18.21 in., Ix = 1330 in.4, Sx =
146 in.3, Iy = 152 in.4, and Sy = 27.6 in.3. Determine whether a load of Q = 10,000 lb
294
5.45
Tapered beam using 1 element
L
x
2
Tapered beam using 2 elements
I1–2
4
L
= 125 in.4 ; I2–3 =
3
4
L
= 175 in.4
I1–2
8
L
= 112.5 in.4 ; I2–3
3
8
L
= 137.5 in.4 ; I3–4
5
8
L
= 162.5 in.4
7
L
I1–2
16
L
= 106.25 I2–3
3
16
L
= 118.75
I3–4
5
16
L
= 131.25 in.4 I4–5
7
16
L
= 143.75 in.4
I5–6
9
16
L
= 156.25 in.4 I6–7
11
16
L
= 168.75 in.4
296
I7–8
13
16
L
= 181.25 in.4 I8–9
15
16
L
= 193.75 in.4
Analytical solution
2
2
0
Pl
n EI
21
2
n l n n
l n n l l
A =
2
2
0
Pl
n EI
[ln (1 + n) – (1 + n)]
B =
3
2
0
Pl
n EI
11
(1 + )+ 1
2
n
ln n
nn
x = 0, n = 1, P = 500, l= 100, E = 30 106, I0 = 100
2
2 6 4
(500) (100 in.)
(1) (30 10 ) (100 in. )
= – 2.1781 10–3
3
500 (100 in.) 1
v =
2
26
500 (100 in.) 100 in.
(1)
(1) (30 10 ) (100 in.)
– 1.3448 10–1
PROBLEM 2 USING 1 ELEMENT—
NUMBER OF ELEMENTS = 1
NUMBER OF NODES = 2
NODE POINTS
K IFIX XC(K) YC(K) ZC(K) FORCE(1,K) FORCE(2,K) FORCE(3,K)
ELEMENTS
K NODE(I,K) E(K) G(K) A(K) XI(K)
1 1 2 3,0000000E+07 1.0000000E+00 1.0000000E–01 1.5000000E+02
NODE DISPLACEMENTS Z-ROTATION
X Y THETA
ELEMENTS
K NODE(I,K) X-FORCE Y-FORCE Z-MOMENT X-FORCE Y-FORCE Z-MOMENT
1 1 2 0.0000E+00 0.5000E+03 0.2434E–03 0.0000E+00 – 0.5000E+03 0.5000E+05
PROBLEM 2 USING 2 ELEMENTS—
297
NUMBER OF ELEMENTS = 2
NUMBER OF NODES = 3
NODE POINTS
K IFIX XC(K) YC(K) ZC(K) FORCE(1,K) FORCE(2,K) FORCE(3,K)
1 0 0 0 0.000000 0.000000 0.000000 0.000000 500.000000 0.000000
ELEMENTS
K NODE(1,K) E(K) G(K) A(K) XI(K)
NODE DISPLACEMENTS Z-ROTATION
X Y THETA
1 0.00000E+00 0.33333E–01 – 0.52381E–03
ELEMENTS
K NODE X-FORCE Y-FORCE Z-MOMENT X-FORCE Y-FORCE Z-MOMENT
For 1 element (I1 evaluated at x = L/2)
7
For 2 elements
I1 = 100
11
4
= 275 in.4
25
4
For 4 elements
15
8
298
57
43
8
For 8 elements
I1 = 100
23
16
= 143.75 in.4
37
16
65
16
I5 = 100
79
16
= 493.75 in.4
I6 = 100
93
16
= 581.25 in.4
107
16
The analytical solution is
3
6
1 (500) (100)
17.55 (30 10 ) (100)
FEM
NUMBER OF ELEMENTS = 1
NUMBER OF NODES = 2
NODE POINTS
K IFIK XC(K) YC(K) ZC(K) FORCE(1,K) FORCE(2,K) FORCE(3,K)
ELEMENTS
ymax
Analytical 0.0095 in.
1 element 0.0123 in.
2 elements 0.0103 in.
4 elements 0.0097 in.
8 elements 0.0096 in.