Click here to Purchase full f ull Solution Manual at http://solutionmanuals.info http://solutionmanuals.info
CHAPTER 5 BEAMS AND FRAMES 5.1
I1 = 1.25 x 105 mm4 , I2 = 4.0 x 104 mm4 Q5
Q3
Q1
Q7 Q2I - 1
Q4
Q6
Q8
Q2
Q2I Node I
NE = 3, NL = 1 → F3 = -3000. DOF (degree of freedom) Boundary Conditions
Q1 = Q7 = 0 (ND = 2) i.e.,
DOF # 1 7
Specified displacement 0.0 0.0
Node 1 2 3 4
X-Coordinate (mm) 0. 150. 225. 350.
Solution using program BEAM
Deflection under load point, Q3 = - 0.133350 mm (down) Slope at at left left end, Q2 = - 0.001146 rad (i.e., clockwise rotation) Slope at right end Q8 = + 0.0015 rad
(
)
5.2
2
1
4
3
6
5
NE = 6 DOF 1 5 9 13
Specified displacement 0 0 0 0
The distributed load is handled as follows: W / unit length
WL/2
WL/2 W L2 / 12
W L2 / 12
1800 lb
3600 lb
1800 lb
900 ft lb
900 ft lb 5
6
Thus DOF 3 9 10 11 13 14
Load (inch, lb) - 5000 - 1800 -10800 - 3600 - 1800 +10800
7
7
Solution Using Program BEAM Node 2 4 6
Displ.(in) -13.76 x 10-3 4.45 x 10-3 - 3.45 x 10-3
DOF 1 5 9 13
Reaction 2,000 3,518 3,231 3,453
Rotation (rad) 3.3 x 10-5 -2.2 x 10-5 -0.3 x 10-5
(lb)
3518 lb
2000 lb
3231 lb
3453 lb
Free Body Diagram Shear Force and Bending Moment Diagrams may now be drawn from the above results. 5.3
1/2 Symmetry Model:
1
E1
2
E2
3
E3
4
Boundary Conditions: Q1 = 0, Q2 = 0 (fixed support) & Q8 = 0 12.83 I 1 = = 512 in 4 12 12.63 I 2 = = 216 in 4 12 12.43 I 3 = = 64 in 4 12 E = 4.5 x106 psi NE = 3
Click here to Purchase full Solution Manual at http://solutionmanuals.info
Loading:
We treat the self-weight as an uniform load:
lb 2 ω = 145 3 × ft 2 = 96.67 lb ft ft 3 Similar to Example 8.2, we obtain o btain
DOF# 1 2 3 4 5 6 7 8
Applied Load (in, lb) - 241.68 - 2416.80 - 422.93 604.32 - 241.65 1510.44 - 60.41 302.04
Note: Load applied at supports do not affect affect the displacements displacements and stresses stresses but do affect the reactions Solution (Using Program BEAM) Node 2 3 4
Displ.(in) - 2.96 x 10-2 - 8.41 x 10-2 - 9.70 x 10-2
DOF # 1 2 8
Rotation (rad) - 8.0 x 10-4 - 8.3 x 10-4 0.
Reaction 966.7 54,865. 8,570.
5.4 The bearing stiffnesses shown have to be added to the (1,1) and (7,7) locations of K , respectively. 1
k = = 20,000 N/mm
2
3
4
k = = 12,000 N/mm
Or add these these to the (1,1) and (7,1) locations of the banded banded stiffness S in the BEAM program. Specifically, in Program BEAM, after completion of element assembly and prior to considering boundary conditions, we insert: S(1,1) = S(1,1) + 20000 S(7,1) = S(7,1) + 12000 Further, we have ND = 0, since there are no specified displacement. displacement. Solution
Q3= - 0.22825 mm Q2= - 0.0012 rad Q8= - 0.0014 rad Q7= - 0.107 mm
5.5 k 1
EA =
k 2
=
I beam
L
=
30 × 10
6
0.08
12
30 × 10
6
0.08
×
20 =
×
0.64 in
=
=
200,000 lb in
120,000 lb in
4
Forces in the rods or springs: f s1 = 200000 × 1.186 × 10−2 = 2,372 lb f s = 120000 × 3.386 × 10− 2 = 4,063 lb 2
σ
1
σ
2
= =
2372 0.08 4063 0.08
= 29,650. psi = 50,788. psi
1
2
3
4
`
Q1 = 0
Q3 = 1.186
Q5 = 3.386
Q7 = 8.848 x 10-2 in.
Click here to Purchase full Solution Manual at http://solutionmanuals.info http://solutionmanuals.info 5.6
1
2
4
3
5
7
6
NE = 7 I1 = I3 = I5 = I7 = 4.123/12 = 576 in4 I2 = I4 = I6 = 4.123/12 - 4.63/12 = 504 in4 Without opening:
δ = PL3/3EI = 4.167 x 10-2 in With opening: Using BEAM program: Node 2 3 4 5 6 7 8
Displ.(in) -0.060 x 10-2 -0.536 x 10-2 -0.925 x 10-2 -1.935 x 10-2 -2.529 x 10-2 -3.826 x 10-2 -4.503 x 10-2
DOF # 1 2
Reaction 10,000 60,000
Rotation (rad) - 0.2 x 10-3 - 0.58 x 10-3 - 0.714 x 10-3 - 0.952 x 10-3 - 1.025 x 10-3 - 1.121 x 10-3 - 1.131 x 10-3
lb in-lb
5.7 Spindle
Some nodes have been introduced to avoid large variations in length between adjacent elements.
Node 10
Node 1 bearing
bearing
Node 1 2 3 4 5 6 7 8 9 10
I 1 =
(42) 4
π
64
X-Coord (mm) 0. 50. 125. 200. 275. 350. 380. 410. 425. 431.
−
(30) 4
π
64
= 112,984.5 mm 4
I 3 = I 3 = I 4 = I 5 = 267,036 mm 4 I 6 = 596,413.1 mm 4 I 7 = I 6 I 8 = 1,970,863.1 mm 4 I 9 = 267,036.mm 4
The bearing stiffnesses get added to the banded S before CALL BANSOL : S(3,1) = S(3,1) + 20000 S(12,1)= S(12,1) + 60000 S(11,1) = S(11, 1) + 8 x 108 The deformed spindle shape is: Node 1 2 3 4 5 6 7 8 9 10 Reactions f s = k r r x δ, Cs = k t θ
-3
Displ. x 10 mm 4.65 1.98 -2.08 -6.46 -11.46 -17.40 -20.30 -23.50 -25.20 -25.90 1000N 69,440 N-mm
40N 40N
1044N
Click here to Purchase full Solution Manual at http://solutionmanuals.info http://solutionmanuals.info 5.8 Frame Program
½ Symmetry Model DOF 3*I-1 DOF3*I
DOF3*I-2
2
3
1
Node 1 2 3
X-Coord (ft.) Y-Coord (ft.) 0 0 10 20 20 20
Element
Area (in )
2
1-2 2-3
15.0 7 .5
4
M_Inertia (in )
305. 125.
Connectivity: e 1 2
1
2
1 2
2 3
Boundary Conditions DOF # 1 2 3 7 9
Thus, ND = 5
Specified Displacement 0. 0. Fixed support 0. 0. Symmetry 0.
