... m
5kN
1. 2 m
1m
1m
1 kN
2 m2 m
1m
1m
2 kN
2 m2 m
1m
1m
2 kN
1 kN
Truss Analysis: Force Method, Part I by S. T. Mau
49
This particular cut separates the truss into two parts. We shall use the left part ... joint.
4
F
6
F
3
F
2
3
5
4
3
4
5
1
F
3
F
1
R
y1
R
x1
3
4
5
F
4
F
2
R
y5
5
R
x5
3
5
4
10 kN
3@2m=6m
1
3
2
4
5
6
7
1
2 3
5
6
7
8
10
11
2m
9
4
Truss Analysis: Force Method,...
... joint.
4
F
6
F
3
F
2
3
5
4
3
4
5
1
F
3
F
1
R
y1
R
x1
3
4
5
F
4
F
2
R
y5
5
R
x5
3
5
4
10 kN
3@2m=6m
1
3
2
4
5
6
7
1
2 3
5
6
7
8
10
11
2m
9
4
Truss Analysis: Force Method, Part I by S. T. Mau
50
Example ... F
4
(3/5) –F
1
= 0,
F
1
= –6 kN.
2
1
2
3
6 kN
4
5
6
R
y1
R
x1
1
3
4
5
R
x5
R
y5
F
5
3
4
5
6 kN
3
F
6
3
4
5
2
F
5
F
4
F
1
3
5
4
3
4
5
Truss Analysis: Force Method, Pa...
... Force Method, Part I by S. T. Mau
11 9
(11 ) (12 )
(13 ) (14 )
(15 ) (16 )
Problem 2. Frame problems.
5 m
5 m
10 kN
5 m
5 m
10 kN-m
5 m
5 m
10 kN
5 m
5 m
10 kN-m
5 m
5 m
10 kN
5 m
5 m
10 kN-m
Beam ... Beam
P
a/2EI
P
a/EI
2aaa
R
eactions
P
a/2EI
P
a/EI
11 Pa
2
/12 EI
5Pa
2
/12 EI
Shear(Rotation)Diagram
( Unit: Pa
2
/EI )
1
1/ 12
5 /12
1/ 6
M
oment(Deflection) Diagram
( Unit:...
... below.
Displaced configuration.
P
2P
2P
P
1
2
2
1
1
1/ L1/L
1
1/L
1/ L
1
1
2PL
2L
1
1
2L
2L
P
Beam and Frame Analysis: Force Method, Part II by S. T. Mau
14 0
The computing is carried out using the ...
∆
d
Load
Diagram
Moment
Diagram
(M)(m)(m)(m)(m)
a~b
EI
1
(
3
1
)
(2PL)(2L)(2L)
=
EI
PL
3
8
3
00
EI
1
(
3
1
)
(2PL)(2L)(2L)
=
EI
PL
3
8
3
b~c
EI2
1
(
3
1
)
(2PL)(2L)(...
... solutions.
P
1
2
3
θ
2
’
M
1
M
1
θ
11
θ
1
’
θ
3
’
M
1
θ
21
M
1
θ
31
M
1
M
2
θ
21
M
2
θ
22
M
2
θ
32
M
3
M
3
θ
31
M
3
θ
23
M
3
θ
33
θ
2
=0
θ
1
=0
θ
3
=0
P
Beam and Frame Analysis: Force Method, Part ...
∆
2
’
+ R
1
δ
21
+ R
2
δ
22
= 0
These two equations can be put in the following matrix form.
L
L
w
L
∆
1
’
∆
2
’
L
L
w
L
R
1
δ
11
R
1
δ
21
R...
... M
ab
M
ba
M
bc
M
cb
M
cd
M
dc
EAM 30
DM +15 +15
COM +7.50 +7.50
DM −4.69 −2. 81
COM −2.35 1. 41
DM +0. 71 +0.70
COM +0.36 0.35
DM −0.22 −0 .14
COM −0 .11 −0.07
DM +0.04 +0.03
COM +0.02 +0.02
DM −0. 01 −0. 01
COM 0.00 0.00
SUM ... distributed loads
−2.46
4.92
-14 .27
15 .73
-7.87
I
nflection point
Beam and Frame Analysis: Displacement Method, Part I by S. T. Mau
17 9
M
bc
=
2
1...
... condition is ∅.5 01 + ∅.005 = ∅.506. The hole has an MMC size limit of ∅. 511 and a positional
tolerance of ∅.005 at MMC. Since it’s an internal feature of size, its virtual condition is ∅. 511 − ∅.005 ... example, ∅.625 LC5 or 30 f7.
5-50 Chapter Five
Eliminate unnecessary trailing zeros.
25 .1 not 25 .10
12 not 12 .0
with not with
The exceptions are limit dimensions and bilateral (p...