This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **EML 4507 Fall 2010 Quiz 4B-V1 A beam is clamped at the left end and rests on a
spring at the right end. The right support is such
that the beam cannot rotate at that end. Thus,
the only active DOF is v2. A force F = 3,000 N
acts downward at the right end as shown. The
structure is modeled using two elements: one
beam element and one spring element. The
spring stiffness k = 1000 N/m. The beam
properties are L = 1 m, E] = 1000 N—mz. (a) Write the element stiffness matrices of both
elements and clearly show the row addresses. (b) Assemble the two element matrix equations
and apply boundary conditions to obtain the
global matrix equation. (c) Solve for unknown displacement 122. (d) Draw the BM and SF diagrams for the
beam. 76) Solution
(a) The stiffness matrix of the beam element is:
12 6 —12 6 '01
6 4 —6 2 91
— 12 —6 1 2 —6 122
6 2 —6 4 02 [km] = 1000 The stiffness matrix of the spring element is:
' 1 —1 v
[162)] =1000 2
—1 1 v3 (b) Assembling we obtain the global
" equations as 12 6 —12 6 0 v1 R1
6 4 —6 2 o 191 C1
1000 ~12 —6 12+1 —6 —1 v2 = —3000
6 2 —6 4 0 02 o
0 0 ~1 0 +1 v3 R3 (0) Deleting all zero DOFs we obtain the
global equation as 13,0001)2 = —3,000 :> v2=—0.231 m (d) Use the equation 12 6L ~12 6L {,1 4x,”1
k E] 6L 4L2 —6L 2L2 01 —M1
[Md—F42 —6L 12 —6L 112 _ +Vy2 6L 2L2 —6L 4L2 62 +M2 Substituting for E1, L and {41(1)} —12 2, 772 —Vy1 ~6 1,386 —M1
1000 {v2 = —0.23 1} = = 12 —2, 772 +Vy2 —6 1,386 +M2 SF 1
—2,772 N 2:17.210 . .3000 N
g N5358é New
‘585 NT“ T225110 1
i
1
l
1
l
1 EML 4507 Fall 2010 Quiz 4B-V2 A beam is clamped at the left end and rests on a
spring at the right end. The right support is such
that the beam cannot rotate at that end. Thus,
the only active DOF is 122. A force F = 1,200 N
acts downward at the right end as shown. The
structure is modeled using two elements: one
beam element and one spring element. The
spring stiffness k = 500 N/m. The beam
properties are L = l m, E] = 500 N—mz. (a) Write the element stiffness matrices of both elements and clearly show the row addresses. (b) Assemble the two element matrix equations and apply boundary conditions to obtain the global matrix equation.
(0) Solve for unknown displacement v2. (d) Draw the BM and SF diagrams for the
beam. [email protected] Solution (a) The stiffness matrix of the beam ' element is: 12 6 —12 6 '01
6 4 ~6 2 61
—12 —6 12 —6 112
6 2 —6 4 92 [1(0)] = 500 The stiffness matrix of the spring element is: 1 ——1 v
[k‘2)]=500[1 1L: (b) Assembling we obtain the global
equations as 12 6 —12 6 0 v1 R1
6 4 —6 2 ’0 91 C1
500 —12 —6 12+1 —6 —1 v2 = —1200
6 2 —6 4 0 92 o
0 0 —1 0 +1 v3 R3 (c) Deleting all zero DOFs we obtain the
global equation as 6,5001}2 = ~1200 2) v2=-0.185 m ((1) Use the equation 12 6L —12 6L {,1 _Vy1
A E] 6L 4L2 —6L 2L2 01 _ —M1
[HQ—F42 —6L 12 —6L 112 _ +Vy2 6L 2L2 ~6L 4L2 62 +M2 Substituting for E], L and {gm} —12 1,110 —Vyl
—6 555 —M1
500 {v2 = —O.185} = =
12 —1,110 +Vy2
—6 555 +M2
SF
_ -1,110N
' +555 N—In
BM
~555 N—m
ES: Al“) N
WC ’1‘
mo N y ESSN’Y“
E i5 1 EML 4507 Fall 2010 A beam is clamped at the left end and rests on a
spring at the right end. The right support is such
that the beam cannot rotate at that end. Thus,
the only active DOF is v2. A force F = 2,400 N acts-downward at the right end as shown. The ‘ structure is modeled using two elements: one beam element and one spring element. The
spring stiffness k = 1000 N/m. The beam properties are L = 1 m, E] = 1000 N-m2. (a) Write the element stiffness matrices of both
elements and clearly show the row addresses. (b) Assemble the two element matrix equations
and apply boundary conditions to obtain the
global matrix equation. (c) Solve for unknown displacement v2. (d) Draw the BM and SF diagrams for the
beam. 7 Solution (a) The stiffness matrix of the beam
_ element is: ' 12 6 —12 6 121 6 4 —6 2 61
—12 —6 12 —6 v2 6 2 —6 4 6 [km] = 1000 The stiffness matrix of the spring element is: 1 —1v
[1<<2>]=1000[_1 1}): (b) Assembling we obtain the global
equations as 12 6 —12 6 0 v1 R1
6 4 ~6 2 0 .91 C1
1000 —12 —6 12+1 —6 —1 v2 = —2400
6 2 ~6 4 o 62 0
0 0 —1 0 +1 v3 R3 (0) Deleting all zero DOFs we obtain the
global equation as 13,0001)2 = —2400 :> v2=—0.185 m ((1) Use the equation 12 6L —12 6L 1,1 44,1
k E] 6L 4L2 —6L 21;2 01 —M1
[HQ—E42 —6L 12 —6L 112 _ +Vy2 6L 2L2 —6L 4L2 62 +M2 Substituting for E1, L and {qm} —12 2,215 —Vyl
—6 1,108 4141
1000 {v2 = —0.185} = =
12 —2,215 +Vy2
—6 1,108 +M2
SF
_ -2,215 N +1,108 N—m —1,108N—m
15"I '
9-9— 9.qu N
g = 1108 N“
HOS—NH“ /Yj
\"b\r“\l\m i5 ...

View
Full Document