1. No category

YANGON TECHNOLOGICAL UNIVERSITY
DEPARTMENT OF CIVIL ENGINEERING
CE 3013
ADVANCED THEORY OF STRUCTURES I
Sample Questions
(For Semester II)
August 2008
1
CE 3013 – Advanced Theory of Structures I
Sample Questions and Solutions for Semester II
Chapter 2: Force Method of Analysis
1. Use the force method to find the bending moment at the intermediate supports of the
continuous beam shown in the figure.
q / unit length
A
C
B
0.8 l
D
0.8 l
l
1
2
Coordinate
system
Solution: See in Worked-out examples No.2.
2. By using force method, find the reactions at the supports of the continuous beam shown
in the figure due to downward translation of A = l / 100. The beam has a constant flexural
rigidity EI.
B
A
l
C
A
C
B
2
1
l
Coordinate System
Solution: See in Example 2.2 of the Textbook.
3. By using force method, find the reactions at the supports of the continuous beam shown
in the figure due to downward translation of B = l / 100 together with a clockwise rotation
of C = 0.004 radian. The beam has a constant flexural rigidity EI.
B
A
l
C
A
l
C
B
2
1
Coordinate System
Solution: See in Example 2.2 of the Textbook.
4. Analyze the continuous beam shown in the figure due to a uniformly distributed load of
intensity q on all spans. The beam has a constant flexural rigidity EI.
Coordinate System
Solution: See in Example 2.3 of the Textbook.
2
5. Analyze the continuous beam shown in the figure due to a unit downward movement of
support A. The beam has a constant flexural rigidity EI.
Coordinate System
Solution: See in Example 2.3 of the Textbook.
6. Analyze the continuous beam shown in the figure due to a unit downward movement of
support B. The beam has a constant flexural rigidity EI.
Coordinate System
Solution: See in Example 2.3 of the Textbook.
7. Find the reactions at the supports of the continuous beam shown in the figure by using
force method. The beam has a constant flexural rigidity EI.
q per unit length
A
C
B
l
A
C
B
2
1
l
Coordinate System
Solution:
(1) The displacements at supports B and C due to the external load on the released structure,
q per unit length
where D1 = -
5ql 4
ql 3
, and D2 = .
24 EI
3EI
3
(2) The displacements at supports B and C due to the unit values of redundants, F1 = 1 and
F2 = 1, on the released structure,
where f11 =
2l
l3
l2
l2
.
, f21 =
, f12 =
, and f22 =
6 EI
4 EI
4 EI
3EI
(3) To find the redundants, F1 and F2, solve the following equation,
[f]{F} = {-D}
(1)
{F} = [f]-1{-D}
(2)
⎧F ⎫
where {F} = ⎨ 1 ⎬
⎩ F2 ⎭
3
2
f12 ⎤ ⎡l 6 EI l 4 EI ⎤
⎥
= ⎢
2l
⎥
f 22 ⎥⎦ ⎢l 2
3EI ⎦
⎣ 4 EI
⎧D ⎫
ql 3 ⎧5l ⎫
{-D}= - ⎨ 1 ⎬ =
⎨ ⎬
⎩ D2 ⎭ 24 EI ⎩ 8 ⎭
⎡f
[f] = ⎢ 11
⎣ f 21
and
[f]-1 =
12 EI
7l 3
⎡ 8 − 3l ⎤
⎢− 3l 2l 2 ⎥
⎣
⎦
⎧ F ⎫ ql ⎧16⎫
Then solving equation (2), {F} = ⎨ 1 ⎬ =
⎨ ⎬,
⎩ F2 ⎭ 14 ⎩ l ⎭
8
ql 2
.
where F1 = ql , and F2 =
7
14
(4) To find the reactions at supports A and C by the superposition of the effect of external
loads on the released structure and the effect of redundants,
17
1 ql 2 8 2
ql
(
RA =
+ ql ) =
28
2l 14 7
15
1 8 2 ql 2
RC =
ql
( ql )=
28
2l 7
14
The reactions at the supports of the continuous beam are shown in the following figure.
17
ql
28
15
ql
28
4
8. Find the reactions at the supports of the continuous beam shown in the figure by using
force method. Also draw the shear force diagram.
2q/unit length
A
3ql
2EI
1.5l
4ql
B
0.5l
3q/unit length
C
2EI
0.6l
1.8l
Solution:
To form {D}
ql 3
21Pl 2
21Pl 2
+
+
24 EI
384 EI
384 EI
3
2q (2l )
21(3ql )(2l ) 2
21(4ql )(2.4l ) 2
=
+
+
24(2 EI )
384(2 EI )
384(2 EI )
D1 =
0.333ql 3
0.33ql 3
0.63ql 3
+
+
EI
EI
EI
3
1.293ql
=
EI
=
15 Pl 2
ql 3
+
384 EI
24 EI
15(4ql )(2.4l ) 2
3q(1.5l ) 3
=
+
384(2 EI )
24 EI
D2 =
0.45ql 3
0.42ql 3
+
EI
EI
3
0.87ql
=
EI
=
{∆ - D} =
− ql 3 ⎧1.293⎫
⎨
⎬
EI ⎩ 0.87 ⎭
To form [f]
Ml
Ml
1 × 2l
1 × 2.4l
0.73l
=
+
=
+
3EI 3EI
3 × 2 EI
3 × 2 EI
EI
Ml
1 × 2.4l
0.2l
=
=
=
6 EI
6 × 2 EI
EI
Ml
1 × 2.4l
0.2l
=
=
=
6 EI
6 × 2 EI
EI
Ml
Ml
1 × 2.4l
1 × 1.5l
0.9l
=
=
+
=
+
3EI 3EI
3 × 2 EI
3EI
EI
f11 =
f21
f12
f22
EI
1.5l
D
5
[f] =
l ⎡0.73 0.2⎤
EI ⎢⎣ 0.2 0.9⎥⎦
EI ⎡ 0.9 − 0.2⎤
0.671l ⎢⎣− 0.2 0.73 ⎥⎦
[F] = [f]-1{∆-D}
EI ⎡ 0.9 − 0.2⎤ − ql 3 ⎧1.293⎫
=
x
⎨
⎬
0.671l ⎢⎣− 0.2 0.73 ⎥⎦
EI ⎩ 0.87 ⎭
[f]-1 =
=
− ql 2
0.617
⎧ 0.99 ⎫
⎨
⎬
⎩0.377⎭
⎧− 1.605ql 2 ⎫
=⎨
2⎬
⎩− 0.611ql ⎭
F1 = -1.605 ql2, F2 = -0.611 ql2
Support Reactions
RA
RB
RC
RD
= 1.9475 ql
= 8.4667 ql
= 3.2431 ql
= 1.8427 ql
2q/unit length
1.9475 ql
3ql
4ql
3q/unit length
8.4667 ql
3.2431 ql
1.8427 ql
Support Reactions
3.4142 ql
1.9475 ql
2.6573 ql
+
+
+
1.0525 ql
-
-
0.5858 ql
4.0525 ql
5.0525 ql
Shear Force Diagram
1.8427 ql
6
9. Find the reactions at the supports of the continuous beam shown in the figure by using
force method. Also draw the shear force diagram.
5k
A
2EI
6'
7k
3k/ft
B
EI
6'
10'
C
EI
7.5'
Solution:
To form {D}
D1 =
Pl 2
ql 3
5 × 12 2
3 × 10 3
295
+
=
+
=
16 EI 24 EI 16 × 2 EI
24 EI
2 EI
D2 =
ql 3
15Pl 2
3 × 10 3
15 × 7 × 10 2
4875
+
=
+
=
24 EI 384 EI
24 EI
384 EI
32 EI
1
{D}=
EI
⎧⎪ 295 ⎫⎪
1 ⎧4720⎫
2
⎨4875 ⎬ =
⎬
⎨
32 EI ⎩4875⎭
⎪⎩
⎪
32⎭
To form [f]
f11 =
Ml
Ml
12
10
16
=
+
=
+
3EI 3EI
3 × 2 EI
3EI
3EI
f21 =
Ml
10
=
6 EI
6 EI
f12 =
Ml
10
=
6 EI
6 EI
f22 =
Ml
Ml
10
10
20
=
+
=
+
3EI 3EI
3EI
3EI
3EI
[f] =
1
6 EI
[f]-1 =
6 EI
1180
⎡32 10 ⎤
⎢10 40⎥
⎣
⎦
⎡ 40 − 10⎤
⎢− 10 32 ⎥
⎣
⎦
D
2.5'
7
[F] = [f]-1{∆-D} =
6 EI
1180
⎡ 40 − 10⎤
− 1 ⎧4720⎫ ⎡− 22.25⎤
⎢− 10 32 ⎥ x 32 EI ⎨4875⎬ = ⎢ − 17.29 ⎥
⎭ ⎣
⎣
⎦
⎩
⎦
F1 = -22.25 k.ft, F2 = -17.29 k.ft
Support Reactions
RA
RB
RC
RD
= 0.65 k
= 19.85 k
= 17.97 k
= 3.521 k
5k
7k
3k/ft
0.65 k
19.85 k
17.97 k
3.521 k
Support Reactions
15.5 k
0.65 k
3.479 k
+
+
-
+
-
4.35 k
14.75 k
Shear Force Diagram
3.512 k
8
Chapter 3: Displacement Method of Analysis
1. Form [S] and {F} matrices for the frame shown in figure. Use the displacement method.
9P
2
3
1
l
4P
Constant EI
l
l
2l
Coordinate system
Solution: See in Worked-out examples No.3.
2. Draw the bending moment diagram for the frame shown in the figure, which has a
constant flexural rigidity EI. Use the displacement method.
Solution: See in Example 3.2 of the Textbook.
Coordinate System
3. By using the displacement method, draw the bending moment diagram for the frame
shown in the figure, which has a constant flexural rigidity EI. Deformations due to axial
forces are to be neglected.
Coordinate System
Solution: See in Example 3.3 of the Textbook.
9
4. Find the three reaction forces (vertical force, bending and twisting couples) at end A of
the horizontal grid shown in the figure due to a uniform vertical load of intensity q acting
on AC. All bars of the grid have the same cross section with the ratio of torsional and
flexural rigidities GJ / EI = 0.5.
Coordinate System
Solution: See in Example 3.4 of the Textbook.
5. The continuous beam shown in the figure of a constant flexural rigidity EI has two fixed
supports A and D and two roller supports B and C. Draw the bending moment diagram for
the beam when support A settles vertically a distance of ∆.
Coordinate System
Solution: See in Example 3.5 of the Textbook.
6. The continuous beam shown in the figure of a constant flexural rigidity EI has two fixed
supports A and D and two roller supports B and C. Draw the bending moment diagram for
the beam when the beam is rotated at B through an angle θ in a clockwise direction.
Coordinate System
Solution: See in Example 3.5 of the Textbook.
10
7. Form {F}, [S], {Ar} and [Au] matrices to find the three reaction components at joint 'B'
GJ
for the horizontal grid shown. Assume
= 0.5
EI
B
Downward load
2q/unit length
1.5l
A
C
Z
X
Y
E
Downward load
q/unit length
1.5l
D
l
2.5 l
Solution:
B3
B
A
B1
3
E
Z
B2
Y
2
X
1
D
C
To find {F}
q/unit length
2q/unit length
E
A
D
E
l
F1 =
2q × l q × 1.5l − 3.5ql
+
=
2
2
2
F2 =
− q × (1.5l ) 2 − 2.25ql 2
=
12
12
F3 =
+ 2q × (l ) 2 + ql 2
=
12
6
1.5 l
11
⎧ − 3.5ql ⎫
⎪
⎪ 2
⎪
⎪
⎪ − 2.25ql 2 ⎪
{F } = ⎨
⎬
⎪
⎪ 12
⎪
⎪ + ql 2
⎪
⎪ 6
⎭
⎩
To find [S]
S11
+
=
12 EI 12 EI
12 EI
12 EI
+
+
+
3
3
3
l
(2.5l )
(1.5l )
(1.5l ) 3
1
1
1 ⎫
⎧1
+
+
= + 12EI ⎨ 3 +
⎬
3
3
15.625l
3.375l
3.375l 3 ⎭
⎩l
19.88 EI
l3
=+
S21
= MED + MEB = + MED – MEB
6 EI
6 EI
−
=0
2
(1.5l )
(1.5l ) 2
= +
S31
= MEA + MEC = - MEA + MECw
=-
=
6 EI
6 EI
+
2
l
(2.5l ) 2
6 EI ⎛
1
⎜−1+
2 ⎜
l ⎝
(2.5) 2
E
A
1
⎞ − 5.04 EI
⎟⎟ =
l2
⎠
E
C
l
2.5 l
GJ 0.5EI
=
l
l
GJ 0.5EI 0.2 EI
=
=
2.5l
2.5l
l
E
D
4 EI
1.5l
∴S22 =
=
=
E
2.67 EI
l
4 EI
1.5l
0.5EI 0.2 EI 2.67 EI 2.67 EI
+
+
+
l
l
l
l
6.03EI
,
l
B
S32 = 0
12
1
A
E
C
E
1.5l
1.5 l
GJ 0.5EI
=
1.5l
1.5l
GJ 0.5EI 0.33EI
=
=
2.5l
2.5l
l
=
0.33EI
l
4 EI
l
A
E
E
C
4 EI 1.6 EI
=
2.5l
l
S33
=
0.33EI 0.33EI 4 EI 1.6 EI
+
+
+
l
l
l
l
=
6.27 EI
l
S23 = 0
S13
=
6 EI
6 EI 6 EI
− 2 = 2 (0.16 − 1)
2
l
l
(2.5l )
=
− 5.04 EI
l2
⎡ + 19.88EI
⎢
l3
⎢
⎢
[S] = ⎢
0
⎢
⎢
⎢ − 5.04 EI
⎢⎣
l2
+ 6.03EI
l
0
To find [Ar]
Ar1 = Ar2 = Ar3 = 0 ∴ {Ar} = {0}
⎤
symmetrical ⎥
⎥
⎥
⎥
⎥
⎥
6.27 EI ⎥
⎥⎦
l
13
To find [Au]
Au21 ==
1
E
Au11 =
1.5l
6 EI
− 2.67 EI
=
2
(1.5l )
l2
− 12 EI − 3.56 EI
=
(1.5l ) 3
l3
Au31 = 0
E
B
⎡ − 3.56 EI
⎢ l3
⎢ − 2.67 EI
[Au] = ⎢
2
⎢ l
⎢
0
⎢⎣
+ 6 EI + 2.67 EI
=
(1.5l ) 2
l2
Au12
=
Au32
=0
Au33
=
+ 2.67 EI
l2
2 EI
1.5l
0
GJ − 0.5EI − 0.33EI
=
=
l
1.5l
l
⎤
⎥
⎥
⎥
0
⎥
− 0.33EI ⎥
⎥⎦
l
0
14
8. Form {F}, [S], and [Au] matrices to find the three reaction components at joint 'B' for the
horizontal grid shown.
1 k/ ft
B
1
2
3
3 C
4
8'
4
1
10'
10 k
5
D
10'
2
A
12'
Solution:
To form {F}
b a 2 b ab 2
F1 = − P ( − 3 + 3 ) , a = 8', b = 10', l = 18', P = 10 k
l
l
l
2
10 8 × 10 8 × 10 2
+
F1 = -10 ( −
) = -5.83 k
18
18 3
18 3
10 × 8 2 × 10
Pa 2 b
= -19.75 k.ft
F2 = − 2 = −
l
18 2
Pab 2 ql 2 10 × 8 × 10 2 1 × 12 2
F3 =
=
= 12.69 k.ft
12
12
l2
18 2
ql 2 1 × 12 2
=
= 12 k.ft
F4 =
12
12
F5 = 0
⎧ − 5.83 ⎫
⎪− 19.75⎪
⎪⎪
⎪⎪
{F} = ⎨ 12.69 ⎬
⎪ 12 ⎪
⎪
⎪
⎪⎩ 0 ⎪⎭
To form [s]
D1 = 1
12EI
12EI
12 EI
12 EI
+
= 0.01406 EI
S11 = 3 + 3 =
3
l
l
18
10 3
6EI
6 EI
S21 = - 2 = - 2 = -0.0185 EI
18
l
6EI
6 EI
S31 = - 2 = - 2 = -0.0185 EI
l
18
Coordinate system
15
6EI
6 EI
= - 2 = -0.06 EI
2
l
10
6EI
6 EI
S51 = - 2 = - 2 = -0.06 EI
10
l
S41 = -
D2 = 1
6EI
6 EI
= - 2 = -0.0185 EI
2
l
18
4 EI
4EI
S22 =
=
= 0.222 EI
l
18
2 EI
2EI
=
= 0.111 EI
S32 =
l
18
S42 = S52 = 0
S12 = -
D3 = 1
6EI
6 EI
= - 2 = -0.0185 EI
2
l
18
2 EI
2EI
S23 =
=
= 0.111 EI
l
18
4EI
4EI
+
= 0.555 EI
S33 =
18
12
2EI
= 0.167 EI
S43 =
12
S53 = 0
S13 = -
D4 = 1
S14 = -
6EI
6 EI
= - 2 = -0.06 EI
2
l
10
S24 = 0
2EI
= 0.167 EI
S34 =
12
4EI
4EI
+
= 0.733 EI
S44 =
12
10
2EI
S54 =
= 0.2 EI
10
D5 = 1
6EI
6 EI
= - 2 = -0.06 EI
2
10
l
S25 = S35 = 0
2EI
= 0.2 EI
S45 =
10
4EI
= 0.4 EI
S55 =
10
S15 = -
16
⎡ 0.01406 − 0.0185 − 0.0185 0.06 0.06⎤
⎢− 0.0185
0.222
0.111
0
0 ⎥⎥
⎢
[S] = EI ⎢− 0.0185
0.111
0.555
0.167
0 ⎥
⎥
⎢
0
0.167
0.733 0.02⎥
⎢ − 0.06
⎢⎣ − 0.06
0
0
0.2
0.4 ⎥⎦
To form {Au}
6EI
6 EI
Au11 = - 2 = - 2
l
18
Au21 = Au31 = 0
6EI
6 EI
Au41 = - 2 = - 2
10
l
2EI
18
Au22 = Au32 = Au42 = 0
Au12 =
4EI
18
4EI
Au23 =
12
2EI
Au33 =
12
Au13 =
Au43 = 0
Au14 = 0
2EI
Au24 =
12
4EI
Au34 =
12
4EI
Au44 =
10
Au15 = Au25 = Au35 =0
2EI
Au45 =
10
2
4
⎡− 6
0
0 ⎤
18 18
⎢ 324
⎥
4
2
0
0 ⎥
⎢ 0
12
12
[Au] = EI ⎢
⎥
2
4
0
0
0 ⎥
⎢
12
12
⎢− 6
4
2 ⎥
0
0
⎢⎣ 100
10
10⎥⎦
17
9. Form {F}, [S], {Ar} and [Au] matrices to find the three reaction components at joint 'B'
for the horizontal grid shown.
18k
B
2
2
C
5
4k/ft
16'
3
10'
4
D
1
1
A
6'
6'
Solution:
To form {F}
F1 = 0
Pl
= -27 k.ft
F2 =
8
Pl ql 2
100
− 19
=
k.ft
F3 =
= 27 8
12
3
3
100
ql 2
k.ft
F4 =
=
12
3
ql
= -20 k
F5 = 2
⎧ 0 ⎫
⎪ − 27 ⎪
⎪ 19 ⎪
⎪
⎪
{F} = ⎨− 3 ⎬
⎪ 100 ⎪
⎪
⎪
⎪ 3 ⎪
⎩ − 20 ⎭
To form [S]
D1 = 1
4EI
=
S11 =
16
2EI
=
S21 =
16
S31 = S41 = 0
6EI
=
S51 =
16 2
EI
4
EI
8
3EI
128
Coordinate system
18
D2 = 1
2EI
S12 =
16
4EI
S22 =
16
2EI
S32 =
12
S42 = 0
6EI
S52 =
16 2
D3 = 1
S13 = 0
2EI
S23 =
12
4EI
S33 =
12
2EI
S43 =
10
6EI
S53 =
10 2
EI
8
4EI 7EI
+
=
12
12
EI
=
6
=
=
3EI
128
EI
6
4EI 11EI
+
=
10
15
EI
=
5
3EI
=
50
=
D4 = 1
S14 = S24 = 0
2EI EI
=
S34 =
10
5
4EI
2EI
=
S44 =
10
5
6EI
3EI
=
S54 =
50
10 2
D5 = 1
6EI
3EI
=
S15 =
2
128
16
6EI
3EI
=
S25 =
2
128
16
6EI
3EI
=
S35 =
2
50
10
6EI
3EI
=
S45 =
50
10 2
12EI
12EI
+
= 0.015 EI
S55 =
3
16
10 3
19
3
1
⎤
⎡ 1
0
0
8
128 ⎥
⎢ 4
7
3
1
0
⎥
⎢ 1
12
6
128 ⎥
⎢ 8
3 ⎥
1
11
1
[S] = EI ⎢ 0
6
15
5
50
⎥
⎢
3
1
2
0
⎢ 0
5
5
50 ⎥
⎢3
3
3
3
0.015⎥
50
50
⎦
⎣ 128 128
To form {Ar}
Ar1 = 0, Ar2 = -27, Ar3 = 27, Ar1 = -
⎧ 0 ⎫
⎪ − 27 ⎪
⎪
⎪
{Ar} = ⎨
27 ⎬
⎪
⎪
⎪− 100 ⎪
3⎭
⎩
To form [Au]
D1 = 1
EI
4 EI
=
16
4
Au21 = Au31 = Au41 = 0
Au11 =
D2 = 1
EI
2 EI
=
16
8
EI
4 EI
Au22 =
=
12
3
EI
2 EI
Au32 =
=
12
6
Au42 = 0
Au12 =
D3 = 1
Au13 = 0
EI
2 EI
Au23 =
=
12
6
EI
4 EI
Au33 =
=
12
3
4 EI
2 EI
Au43 =
=
10
5
D4 = 1
Au14 = Au24 = Au34 = 0
EI
2 EI
Au44 =
=
10
5
100
3
20
D5 = 1
6 EI
3EI
=
2
128
16
Au25 = Au35 = 0
6 EI
3EI
Au45 =
=
2
50
10
Au15 =
⎡1
⎢ 4
⎢0
[Au] = EI ⎢
⎢0
⎢
0
⎣⎢
1
1
1
8
3
6
0
0
1
6
1
3
2
5
0
0
0
1
5
3 ⎤
128⎥
0 ⎥
⎥
0 ⎥
3 ⎥
50 ⎦⎥
21
Chapter 4: Flexibility and Stiffness Matrices
1. The grid in figure is formed by four main girders and two cross girders of a bridge deck,
with a flexural rigidity in the ratio EIm : EIc = 2:1. The torsional rigidity is neglected.
Form [S], {F}, {Ar}, [Au] matrices for a cross girder due to a concentrated vertical load
"P" acting at joint 3.
1
3b
2
P
3b
3
3b
4
3b
3b
3b
hinge
Solution: See in Worked-out examples No.4.
2. Form [F] and [S] matrices for the simply supported beam shown in the figure.
Solution: See in Example 4.1 of the Textbook.
3. The grid shown in the figure is formed by four simply supported main girders and one
cross-girder of a bridge deck, with a flexural rigidity in the ratio EIm : EIc = 3 : 1. The
torsional rigidity is neglected. Draw the bending moment diagram for the cross-girder due
to a concentrated vertical load “P” acting at joint 1.
Solution: See in Example 4.2 of the Textbook.
22
4. The grid in figure is formed by four main girders and one cross girder of a bridge deck,
with a flexural rigidity in the ratio EIm : EIc = 2:1. The torsional rigidity is neglected.
Form [S], {F}, {Ar}, [Au] matrices for the cross girder due to a concentrated vertical load
"P" acting at the center of the cross girder.
2b
P
b
b
2b
4b
4b
Solution:
EIm : EIc = 2 : 1
Neglect torsional rigidity
To form {F}
{F} = {0 , −
P
P
, − , 0}
2
2
To Form [S]
D1 = 1
Main girder
- 23.9998 EI/l3
Cross-girder
EI
L3
1
3.6
- 1.6
-2.4
0.4
23
+ 1.60 EI C 24 EI m
+
(2b) 3
(4b) 3
EI
= 3C (0.2 + 1.125)
b
EI
= 1.325 3C
b
− 1.6 EI C
EI
S21 =
= - 0.2 3C
3
(2b)
b
2.4 EI C
EI
S31 =
= 0.3 3C
3
(2b)
b
− 0.4 EI C
EI
S41 =
= -0.05 3C
3
(2b)
b
S11 =
D2 = 1
Main girder
- 6 EI/l3
Cross-girder
EI
L3
3.6
8.4
-9.6
− 3.6 EI C
EI
= - 0.45 3C
3
(2b)
b
+ 9.6 EI C 6 × 3 EI C
+
S22 =
(2b) 3
(4b) 3
EI
= 1.48 3C
b
− 8.4 EI C
EI
S32 =
= - 1.05 3C
3
(2b)
b
+ 2.4 EI C
EI
S42 =
= 0.3 3C
3
(2b)
b
S12 =
-2.4
24
D3 = 1
Main girder
- 13.71428 EI/l3
Cross-girder
-4.15384
10.61538
3.69231
EI
L3
-10.15384
4.15384 EI C
EI
= 0.51923 3C
3
(2b)
b
− 10.61538 EI C
S23 =
(2b) 3
EI
= -1.32692 3C
b
13.71428 × 3EI C 10.15384 EI C
S33 =
+
(4b) 3
(2b) 3
EI
= 1.9121 3C
b
− 3.69231EI C
EI
S43 =
= -0.46154 3C
3
(2b)
b
S13 =
D4 = 1
Cross-girder
0.4
-2.4
EI
L3
3.6
-1.6
0.4 EI C
EI
= 0.05 3C
3
(2b)
b
2.4 EI C
EI
S24 =
= 0.3 3C
3
(2b)
b
− 3.6 EI C
EI
= -0.45 3C
S34 =
3
(2b)
b
S14 =
25
1.6 EI C 13.71428 × 3EI C
+
(2b) 3
(4b) 3
EI
= 0.84286 3C
b
S44 =
− 0.05 ⎤
⎡ 1.325 − 0.45 0.51923
⎢ − 0.2 1.48 − 1.32692
0.3 ⎥⎥
EI
[S] = 3C ⎢
1.9121
− 1.05
− 0.45 ⎥
b ⎢ 0.3
⎢
⎥
0.3
− 0.46154 − 0.84286⎦
⎣− 0.05
To form {Ar}
{Ar} = {0, -
Pb Pb
,
, 0}
4
4
To form [Au]
⎡− 1.6
[Au] = ⎢
⎣ 0.4
3.6
- 2.4
- 2.4
3.6
0.4⎤ EI C
- 1.6⎥⎦ (2b) 2
26
Chapter 5: Deflection of Various Structures Using Matrix Algebra
1. The rigid frame shown in the figure is made of a commonly used steel I sections which
has the following section properties: a = 1.34 x 10-4 l2, ar = aweb = 0.65 x 10-4 l2, I = 5.30 x
10-8 l4 and G = 0.4 E. Determine the contribution of bending deformation to the
displacement at the three coordinates shown in the figure.
Coordinate System
Solution: See in Example 5.1 of the Textbook.
2. The rigid frame shown in the figure is made of a commonly used steel I sections which
has the following section properties: a = 1.34 x 10-4 l2, ar = aweb = 0.65 x 10-4 l2, I = 5.30 x
10-8 l4 and G = 0.4 E. Determine the contribution of deformation due to axial force to the
displacement at the three coordinates shown in the figure.
Coordinate System
Solution: See in Example 5.1 of the Textbook.
3. The rigid frame shown in the figure is made of a commonly used steel I sections which
has the following section properties: a = 1.34 x 10-4 l2, ar = aweb = 0.65 x 10-4 l2, I = 5.30 x
10-8 l4 and G = 0.4 E. Determine the contribution of shear deformation to the
displacement at the three coordinates shown in the figure.
Coordinate System
Solution: See in Example 5.1 of the Textbook.
27
4. Determine the flexibility matrix of the frame shown in the figure corresponding to the
three coordinates. Consider the bending deformation only.
Coordinate System
Solution: See in Example 5.1 of the Textbook.
5. Determine the stiffness matrix of the frame shown in the figure by the assemblage of
members using the equation 5.38
Solution: See in Example 5.3 of the Textbook.
28
Chapter 6: Influence Lines for Beams, Frames and Grids
1. Calculate the influence ordinates for the end moment MCD in the bridge frame shown at
0.4l and 0.8l of each span. Use these influence ordinates to find the influence ordinates of
shear force Vn. The relative values of I are shown in the figure.
B
A
n
C
I=3
I=2
D
I =3
2b
2b
I =1
I =1
b
E
F
4b
5b
6b
Solution: See in Worked-out examples No.5.
2. Neglecting the torsional rigidity of the girders, form [F] and [S] matrices to find the
influence lines for the following actions.
(a)
bending moment at the center of girder AB
(b)
bending moment in the cross-girder at J
reaction RC at support C
(c)
Moment of inertia of main girder = 3I
Moment of inertia of cross girder = I
N
A
B
1
l/3
J
C
D
2
l/3
K
E
F
3
l
l
Solution: See in Worked-out examples No.6.
3. Determine the influence line for the end moment MBA in the bridge frame shown in the
figure. Use this influence line to find the influence ordinate of the bending moment MG at
the center of AB. The relative values of I are shown in the figure.
Solution: See in Example 6.1 of the Textbook.
29
4. Determine the influence line for the end moment MBA in the bridge frame shown in the
figure. Use this influence line to find the influence ordinate of the shear Vn at a point “n”
just to the left of B. The relative values of I are shown in the figure.
Solution: See in Example 6.1 of the Textbook.
5. Find the influence line for the bending moment at section n, just to the left of joint C, in
the grid shown in the figure. The main girders are encastre. The relative moment of
inertia is 4 for all main girders and 1 for the cross-girders. The ratio of the torsional
rigidity GJ to the flexural rigidity EI is 1 : 4 for all members.
Solution: See in Example 6.2 of the Textbook.
6. Neglecting the torsional rigidity of the girders, find the influence lines for the following
actions for the interconnected bridge system shown in the figure.
(a)
bending moment at the center of girder AB
(b)
bending moment at the center of girder CD
(a)
Solution: See in Example 6.3 of the Textbook.
(b)
30
7. Neglecting the torsional rigidity of the girders, find the influence lines for the following
actions for the interconnected bridge system shown in the figure.
(a)
bending moment in the cross-girder at J
(b)
reaction RC at support C
(b)
(b)
Solution: See in Example 6.3 of the Textbook.
8. Find the influence line for the reaction at B and the forces in the members labeled Z1 and
Z2 in the truss shown in the figure. The unit load can act at the nodes of the lower chord
only. All the members are assumed to have the same value of L/AE, L being the length
and A the cross-sectional area of the members.
Solution: See in Example 6.4 of the Textbook.
9. Determine the influence line for the end moment MCD in the bridge frame shown in the
figure. Use this influence line to find the influence ordinate of the bending moment Mn
and of the shear Vn at a point “n” to the right of C at 0.3l, 0.6l and 0.8l of each span. The
relative values of I are shown in the figure.
A
B
n
C
I=3
I=3
I =3
3b
I =1
4b
4b
I =1
E
F
4b
D
6b
31
Solution:
Introduce a unit rotation (Counterclockwise) at end 'C' of 'CD'.
1
D
C
− 3E (3I )
EI
⎛ 3EI ⎞
=
= 1.5
⎟
6b
b
⎝ l ⎠ CD
-⎜
KAB
3 3I
=- ×
= 0.5625
4 4b
KBE
=
KBC
=
ΣK
= 1.5625
I
= 0.25
4b
3I
= 0.75
4b
0.5625
= 0.36
1.5625
0.25
=
= 0.16
1.5625
0.75
=
= 0.48
1.5625
DFBA =
DFBE
DFBC
End
DF's
FEM's
Multiplier
EI
100b
BA
0.36
0
BE
0.16
0
-15
-7
-1
-16
BC
0.48
0
+41
-20
+3
-1
+23
-1
-8
CB
0.545
0
+82
-10
+5
CF
0.182
0
+27
CD
0.273
-150
+41
+2
+3
+77
+29
-106
Ordinates of I.L for End Moment Mn ( b / 10 )
Inf: coefficients
ηMs
I
ηMCD
2
Inf: ordinates
-
Member AB
0.3 l
0.6 l
0.8 l
0
0
0
MemberBC
0.3 l 0.6 l
0.8 l
0
0
0
Member CD
0.3 l 0.6 l 0.8 l
9
12
6
0.194 0.273
0.205 -0.57
-0.97
-0.79
-3.79 -3.56
-2.04
0.194 0.273
0.205 -0.57
-0.97
-0.79
5.21
3.96
8.44
32
ηv
= ηvs -
0
1
(ηMCD + ηMDC)
l
= ηvs -
1
ηMCD
6b
Ordinates of I.L for shear Vn
Member AB
Inf: coeff
0.3 l
0.6 l
0.8 l
ηVs
I
ηMCD
6b
Inf:
ordinates
-
MemberBC
0.3 l
0.6 l
0.8 l
Member CD
0.3 l 0.6 l 0.8 l
0
0
0
0
0
0
-0.3
0.4
0.2
-0.006
-0.009
-0.007
0.019
0.032
0.026
0.126
0.119
0.068
-0.006
-0.009
-0.007
0.019
0.032
0.026
-0.174
0.519
0.268
33
Chapter 1: Effects of Axial Forces
1. Find the end-moments for the members in the frame shown in the figure, taking into
account the beam-column effect. The relative values of I are shown in the figure.
Figure: (a) Frame properties and loading, (b) Coordinate system.
Solution: See in Example 1.2 of the Textbook.