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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Example No. 1 The steel jointed building frame show in the figure carry the lateral load as shown. Analyse the frame and draw the bending moment diagram using the cantilever method. 10 kN
1.5 m
10 kN
2.0 m
2.0 m
2.5 m
3.0 m
Solution Step # 1 : Locate point of inflection X = 3.5 m
10 kN A
C
E
G 1.5 m
10 kN
2.0 m
2.0 m
2.5 m
3.0 m
Step # 2 : Locate the centre of gravity X=
- Cantilever Method
=
A(0) A(2) A( 4.5) A(7.5) 3.5m 4A
Page 1
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Step # 3 : Determine the axial force in each column 10 kN
P
P1
P2
P3
Since any column stress, , is proportional to its distance from the neutral axis, NA, we can relate the column stress by proportional triangles. P3 P2
P1 P P P1 , P1 = 0.429P = 0.429x0.833=0.357kN 3.5 1.5 P P2 , P2 = 0.286P = 0.238kN 3.5 1 P P3 , P3 = 1.143P = 0.952kN 3.5 4
Calculate the value of “P” using equation of equilibrium Mh
+=0
10 (0.75) + P1 (2) - P2 (4.5) - P3 (7.5) = 0 10 (0.75) + (0.429P)(2) – (0.286P)(4.5) – (1.143P)(7.5) = 0 P = 0.833 kN
- Cantilever Method
Page 2
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Step # 4 :
Determine Shear forces of each part of frame and determine member end moments
At Point “a” Vb 10 kN
Mb
Hb
+=0
Hh (0.75) = 0.833(1) Hh = 1.111 kN #
Hh
FY+ = 0
FX+ = 0
-P + VAB = 0
10 – Hb - Hh = 0
Vb = P
Hb = 10 - Hh
Vb = 0.833 kN #
Hb = 8.889 kN #
P = 0.833 kN
Vb
Ma
Ma
Hb
Ma
+=0
+=0
(down)
Ma = Vb(1)
Ma=0.75(1.111)
Ma = (0.833)(1)
Ma=0.833kNm(column)
Ma = 0.833 kNm (beam)
At Point “c” Vd
Vb = 0.833 kN
Md Hd
Hb = 8.889 kN
FY+ = 0
+=0
Hi(0.75)-0.833(2.25)
Vd-0.357-0.833=0 8.889-3.094-Hd
-0.357(1.25)=0
Vd = 1.19kN #
=0 Hd = 5.795kN #
Hi = 3.094kN #
Hi
FX+ = 0
P1 =0.357kN
Mc
Vd Mc
Mi
Mc=3.094(0.75)
Hd Md
+=0
Hi
Mc = 2.321kNm (column)
+=0
Mc=1.19(1.25)kNm
P1
Mc = 1.488kNm (beam) - Cantilever Method
Page 3
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
At Point “e” Vf
Vd = 1.19 kN
Mf Hf
Hd = 5.79 kN
FY+ = 0
+=0
FX+ = 0
Hj(0.75)-1.19(2.75) Vf-1.19-0.238=0 -Hf+5.79-4.84 Vf = 1.428kN #
-0.238(1.5)=0
Hf = 0.96 kN #
Hj = 4.84kN #
Hj
P2 =0.238kN
=0
Me
Vf
Mj
+=0
Me=4.84(0.75) Me
Hf Mf
Me = 3.63kNm (column)
Hj
+=0
P2
Me=1.428(1.5) Me = 2.142kNm (beam)
At Point “g” Vf Hf
Vf
FX+ = 0 Hk=Hf Hk
Hk = 0.96kN #
Hf Mf
P3 = 0.952 kN
Mg
+=0
Mg=1.428(1.5) Mg = 2.142kNm (beam) Mf
+=0
Mg=0.75(0.96) Mg=0.72kNm(column)
- Cantilever Method
Page 4
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Step # 5 : Draw bending moment diagram (Beam ACEG) 1.488kNm
2.14kNm
0.833kNm
0.833kNm
1.488kNm
2.14kNm
1.5 m
2.0 m
2.0 m
2.5 m
3.0 m
1.5 m
2.0 m
2.0 m
- Cantilever Method
2.5 m
3.0 m
Page 5
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Example No. 2 The building frame in figure below is subjected to horizontal forces of 100 kN, 80 kN and 50 kN at node D, H and L, respectively, and all columns has a dimension of 250 mm x 250 mm. Using the cantilever method; i. ii. iii.
establish the location of the centroidal axis of the frame calculate the axial forces for columns EI, FJ, GK and HL draw the bending moment diagram for beam ABCD
A
B
C
D
100 kN 3.2 m
E
F
G
H
80 kN 3.2 m
I
J
K
L
50 kN
4.0 m M
N
4.0 m
- Cantilever Method
O
6.0 m
P
8.0 m
Page 6
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Solution i.
Establish the location of the centroidal axis of the frame
X=
=
A(0) A(4) A(10) A(18) 8m 4A
8m
4.0 m
ii.
4.0 m
2.0 m
8.0 m
calculate the axial forces for columns EI, FJ, GK and HL From the distribution diagram shown calculate the value of P1, P2 and P3 in term of P P P1
P2 P3
P P1 , P1 = 0.5 P 8 4 P P2 8m P , P2 = 0.25 8 2
100 kN
P P3 , P3 = 1.25P 8 10
3.2 m
80 kN 1.6 m 4.0 m
4.0 m
2.0 m
8.0 m
- Cantilever Method
P
P1
P2
P3
Page 7
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
By taking Moment at mid point of column EI, calculate the value of P MEI
+=0
P2 (10) + P3 (18) – P1 (4) – 80 (1.6) – 100 (4.8) = 0 (0.25P)(10) + (1.25P)(18) – (0.5P)(4) – (80)(1.6) – (100)(4.8) =0 2.5P + 22.5P – 2P – 128 – 480 = 0 P = 26.435 kN Therefore the value of axial forces for columns EI, FJ, GK and HL are as follows; EI = P = 26.435 kN FJ = P1 = 0.5P = (0.5)( 26.435) = 13.218 kN GK = P2 = 0.25P = (0.25)(26.435) = 6.609 kN HL = P3 = 1.25P = (1.25)(26.435) = 33.044 kN
iii.
draw the bending moment diagram for beam ABCD
- Cantilever Method
Page 8
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
8m 100 kN 1.6 m 4.0 m
P
4.0 m
P1
2.0 m
8.0 m
P2
P3
By taking Moment at mid point of column AE, calculate the value of P MAE
+=0
P2 (10) + P3 (18) – P1 (4) – 100 (1.6) = 0 (0.25P)(10) + 1.25P(18) – (0.5P)(4) – (100)(1.6) =0 2.5P + 22.5P – 2P – 160 = 0 P = 6.957 kN Therefore the value of axial forces for columns AE, BF, CG and DH are as follows; AE = P = 6.957 kN BF = P1 = 0.5P = (0.5)( 6.957) = 3.479 kN CG = P2 = 0.25P = (0.25)( 6.957) = 1.739 kN DH = P3 = 1.25P = (1.25)(6.957) = 8.697 kN
Determine shear force and moment for beam ABCD
- Cantilever Method
Page 9
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
At Point A VAB
FY+ = 0 P + VAB = 0 VAB = - P VAB = - 6.957 kN #
P = 6.957 kN
VAB = - 6.957 kN
MA
+=0
MA - VAB(2) = 0 MA
MA – (-6.957)(2) = 0 MA = - 13.914 kNm #
At Point B VBC
VAB = - 6.957 kN
FY+ = 0 - VAB + VBC + P1 = 0 VBC = - 6.957 - 3.479 VBC = - 10.436 kN #
P1 = 3.479 kN
VBC = - 10.436 kN
MB
+=0
MB - VBC (3) = 0 MB
MB – (-10.436)(3) = 0 MB = - 31.308 kNm #
- Cantilever Method
Page 10
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
At Point C VCD
VBC = - 10.436 kN
FY+ = 0 - VBC + VCD - P2 = 0 VCD = - 10.436 + 1.739 VCD = - 8.697 kN #
P2 = 1.739 kN
VCD = - 8.697 kN
MC
+=0
MC - VCD (4) = 0 MC
MC – (-8.697)(4) = 0 MC = - 34.788 kNm #
Bending moment diagram for beam ABCD 34.788 kNm
31.308 kNm 13.914 kNm
13.914 kNm 31.308 kNm
4.0 m
- Cantilever Method
6.0 m
34.788 kNm
8.0 m
Page 11
Example No. 1 The steel jointed building frame show in the figure carry the lateral load as shown. Analyse the frame and draw the bending moment diagram using the cantilever method. 10 kN
1.5 m
10 kN
2.0 m
2.0 m
2.5 m
3.0 m
Solution Step # 1 : Locate point of inflection X = 3.5 m
10 kN A
C
E
G 1.5 m
10 kN
2.0 m
2.0 m
2.5 m
3.0 m
Step # 2 : Locate the centre of gravity X=
- Cantilever Method
=
A(0) A(2) A( 4.5) A(7.5) 3.5m 4A
Page 1
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Step # 3 : Determine the axial force in each column 10 kN
P
P1
P2
P3
Since any column stress, , is proportional to its distance from the neutral axis, NA, we can relate the column stress by proportional triangles. P3 P2
P1 P P P1 , P1 = 0.429P = 0.429x0.833=0.357kN 3.5 1.5 P P2 , P2 = 0.286P = 0.238kN 3.5 1 P P3 , P3 = 1.143P = 0.952kN 3.5 4
Calculate the value of “P” using equation of equilibrium Mh
+=0
10 (0.75) + P1 (2) - P2 (4.5) - P3 (7.5) = 0 10 (0.75) + (0.429P)(2) – (0.286P)(4.5) – (1.143P)(7.5) = 0 P = 0.833 kN
- Cantilever Method
Page 2
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Step # 4 :
Determine Shear forces of each part of frame and determine member end moments
At Point “a” Vb 10 kN
Mb
Hb
+=0
Hh (0.75) = 0.833(1) Hh = 1.111 kN #
Hh
FY+ = 0
FX+ = 0
-P + VAB = 0
10 – Hb - Hh = 0
Vb = P
Hb = 10 - Hh
Vb = 0.833 kN #
Hb = 8.889 kN #
P = 0.833 kN
Vb
Ma
Ma
Hb
Ma
+=0
+=0
(down)
Ma = Vb(1)
Ma=0.75(1.111)
Ma = (0.833)(1)
Ma=0.833kNm(column)
Ma = 0.833 kNm (beam)
At Point “c” Vd
Vb = 0.833 kN
Md Hd
Hb = 8.889 kN
FY+ = 0
+=0
Hi(0.75)-0.833(2.25)
Vd-0.357-0.833=0 8.889-3.094-Hd
-0.357(1.25)=0
Vd = 1.19kN #
=0 Hd = 5.795kN #
Hi = 3.094kN #
Hi
FX+ = 0
P1 =0.357kN
Mc
Vd Mc
Mi
Mc=3.094(0.75)
Hd Md
+=0
Hi
Mc = 2.321kNm (column)
+=0
Mc=1.19(1.25)kNm
P1
Mc = 1.488kNm (beam) - Cantilever Method
Page 3
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
At Point “e” Vf
Vd = 1.19 kN
Mf Hf
Hd = 5.79 kN
FY+ = 0
+=0
FX+ = 0
Hj(0.75)-1.19(2.75) Vf-1.19-0.238=0 -Hf+5.79-4.84 Vf = 1.428kN #
-0.238(1.5)=0
Hf = 0.96 kN #
Hj = 4.84kN #
Hj
P2 =0.238kN
=0
Me
Vf
Mj
+=0
Me=4.84(0.75) Me
Hf Mf
Me = 3.63kNm (column)
Hj
+=0
P2
Me=1.428(1.5) Me = 2.142kNm (beam)
At Point “g” Vf Hf
Vf
FX+ = 0 Hk=Hf Hk
Hk = 0.96kN #
Hf Mf
P3 = 0.952 kN
Mg
+=0
Mg=1.428(1.5) Mg = 2.142kNm (beam) Mf
+=0
Mg=0.75(0.96) Mg=0.72kNm(column)
- Cantilever Method
Page 4
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Step # 5 : Draw bending moment diagram (Beam ACEG) 1.488kNm
2.14kNm
0.833kNm
0.833kNm
1.488kNm
2.14kNm
1.5 m
2.0 m
2.0 m
2.5 m
3.0 m
1.5 m
2.0 m
2.0 m
- Cantilever Method
2.5 m
3.0 m
Page 5
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Example No. 2 The building frame in figure below is subjected to horizontal forces of 100 kN, 80 kN and 50 kN at node D, H and L, respectively, and all columns has a dimension of 250 mm x 250 mm. Using the cantilever method; i. ii. iii.
establish the location of the centroidal axis of the frame calculate the axial forces for columns EI, FJ, GK and HL draw the bending moment diagram for beam ABCD
A
B
C
D
100 kN 3.2 m
E
F
G
H
80 kN 3.2 m
I
J
K
L
50 kN
4.0 m M
N
4.0 m
- Cantilever Method
O
6.0 m
P
8.0 m
Page 6
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
Solution i.
Establish the location of the centroidal axis of the frame
X=
=
A(0) A(4) A(10) A(18) 8m 4A
8m
4.0 m
ii.
4.0 m
2.0 m
8.0 m
calculate the axial forces for columns EI, FJ, GK and HL From the distribution diagram shown calculate the value of P1, P2 and P3 in term of P P P1
P2 P3
P P1 , P1 = 0.5 P 8 4 P P2 8m P , P2 = 0.25 8 2
100 kN
P P3 , P3 = 1.25P 8 10
3.2 m
80 kN 1.6 m 4.0 m
4.0 m
2.0 m
8.0 m
- Cantilever Method
P
P1
P2
P3
Page 7
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
By taking Moment at mid point of column EI, calculate the value of P MEI
+=0
P2 (10) + P3 (18) – P1 (4) – 80 (1.6) – 100 (4.8) = 0 (0.25P)(10) + (1.25P)(18) – (0.5P)(4) – (80)(1.6) – (100)(4.8) =0 2.5P + 22.5P – 2P – 128 – 480 = 0 P = 26.435 kN Therefore the value of axial forces for columns EI, FJ, GK and HL are as follows; EI = P = 26.435 kN FJ = P1 = 0.5P = (0.5)( 26.435) = 13.218 kN GK = P2 = 0.25P = (0.25)(26.435) = 6.609 kN HL = P3 = 1.25P = (1.25)(26.435) = 33.044 kN
iii.
draw the bending moment diagram for beam ABCD
- Cantilever Method
Page 8
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
8m 100 kN 1.6 m 4.0 m
P
4.0 m
P1
2.0 m
8.0 m
P2
P3
By taking Moment at mid point of column AE, calculate the value of P MAE
+=0
P2 (10) + P3 (18) – P1 (4) – 100 (1.6) = 0 (0.25P)(10) + 1.25P(18) – (0.5P)(4) – (100)(1.6) =0 2.5P + 22.5P – 2P – 160 = 0 P = 6.957 kN Therefore the value of axial forces for columns AE, BF, CG and DH are as follows; AE = P = 6.957 kN BF = P1 = 0.5P = (0.5)( 6.957) = 3.479 kN CG = P2 = 0.25P = (0.25)( 6.957) = 1.739 kN DH = P3 = 1.25P = (1.25)(6.957) = 8.697 kN
Determine shear force and moment for beam ABCD
- Cantilever Method
Page 9
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
At Point A VAB
FY+ = 0 P + VAB = 0 VAB = - P VAB = - 6.957 kN #
P = 6.957 kN
VAB = - 6.957 kN
MA
+=0
MA - VAB(2) = 0 MA
MA – (-6.957)(2) = 0 MA = - 13.914 kNm #
At Point B VBC
VAB = - 6.957 kN
FY+ = 0 - VAB + VBC + P1 = 0 VBC = - 6.957 - 3.479 VBC = - 10.436 kN #
P1 = 3.479 kN
VBC = - 10.436 kN
MB
+=0
MB - VBC (3) = 0 MB
MB – (-10.436)(3) = 0 MB = - 31.308 kNm #
- Cantilever Method
Page 10
FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275
At Point C VCD
VBC = - 10.436 kN
FY+ = 0 - VBC + VCD - P2 = 0 VCD = - 10.436 + 1.739 VCD = - 8.697 kN #
P2 = 1.739 kN
VCD = - 8.697 kN
MC
+=0
MC - VCD (4) = 0 MC
MC – (-8.697)(4) = 0 MC = - 34.788 kNm #
Bending moment diagram for beam ABCD 34.788 kNm
31.308 kNm 13.914 kNm
13.914 kNm 31.308 kNm
4.0 m
- Cantilever Method
6.0 m
34.788 kNm
8.0 m
Page 11
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