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Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
2
Lecture Outlines
Internal Loadings at a Specified Point Shear and Moment Functions
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
3
N N
V
V
M M
N
M
V
V
N
M
Positive Sign Convention
Internal Loadings at a Specified Point
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
4
Example 4-1
Determine the internal shear and moment acting in the cantilever beam shown in the figure at sections passing through points C and D.
5 kN 5 kN 5 kN 5 kN 5 kN
1 m 1 m 1 m 1 m 1 m20 kN•m
A BCD
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
5
SOLUTION
5 kN 5 kN 5 kN
1 m 1 m 1 m20 kN•m
BCVC
MC
NC0 =
Fy = 0:+ VC -5 - 5 - 5 = 0, VC = 15 kN
+ MC = 0: -MC -5(1) - 5(2) - 5(3) -20 = 0, MC = -50 kN
VD
MD
ND0 =
5 kN 5 kN 5 kN
1 m 1 m 1 m20 kN•m
BD
5 kN
Fy = 0:+ VD -5 - 5 - 5 -5 = 0, VD = 20 kN
+ MD = 0: -MD -5(1) - 5(2) - 5(3) -20 = 0, MD = -50 kN
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
6
Example 4-2
Determine the internal shear and moment acting at a section passing through point C in the beam shown in the figure.
9 m
20 kN/m
3 mC
A B
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
7
9 m
20 kN/m
C
AB
VC
MC
NC
3 m
)209
3(
30 kNC
1 m
10 kN
= 0
SOLUTION
(2/3)9 = 6 m(1/2)(9)(20) = 90 kN
ByAy
Ax
+ MA = 0: By(9) - 90(6), By = 60 kN
= 60 kN
Fy = 0:+ Ay - 90 + 60 = 0 Ay = 30 kN
30 kN =
0 =
Fy = 0:+-VC - 10 + 30 = 0, VC = 20 kN
= 20 kN
+ MC = 0:
MC + 10(1) - 30(3) = 0, MC = 80 kN•m
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
8
Example 4-3
The 40 kN force in the figure is supported by the floor panel DE, which in turn is simply supported at its ends by floor beams. These beams transmit their loads to the sumply supported girder AB. Determine the internal shear and moment acting at point C in the girder.
B
40 kN
2 m 2 m 2 m0.6 m 1.4 m
A
C2.5 m
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
9
C
SOLUTION
40 kN
0.6 m 1.4 m
12 kN28 kN
12 kN28 kN4 m
6 m
8 m 23 kN17 kN
B
40 kN
2 m 2 m 2 m0.6 m 1.4 m
A
C2.5 m
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
10
23 kNC
28 kN 12 kN2.5 m
17 kN
C
2.5 m
17 kN VC
MC
NC = 0
Fy = 0:+
-VC + 17 = 0, VC = 17 kN
+ MC = 0:
MC - 17(2.5) = 0, MC = 42.5 kN•m
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
11
x1
x2
x3
x1 x2 x3
w
P
AB C
D
w
P
AB C
D
OR
Shear and Moment Functions
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
12
Example 4-4
Determine the shear and moment in the beam shown in the figure as a function of x.
9 m
10 kN/m
x
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
13
x
109
x
45 kN
270 kN•m
SOLUTION (1/2)(9)(10) = 45 kN
(2/3)(9) = 6 m45 kN
270 kN•m
95)10
9)()(
2
1(
2xxx
3
xV
M
N = 0
Fy = 0:+
0459
5 2
x
V
459
5 2
x
V
+ Mx= 0:
027045)3
(9
5 2
xxx
M
27045)3
(9
5 2
xxx
M
9 m
10 kN/m
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
14
Example 4-5
Determine the shear and moment in the beam shown in the figure as a function of x1, x3, x3, and x4.
x2
x3x1
x4
60 kN/m265 kN
135 kN•m4 m
6 m
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
15
SOLUTION60 kN/m
265 kN
1350 kN•m4 m
6 m
Fy = 0:+ 505 - 60x1 - V = 0V = 505 - 60x1
Fy = 0:+ 505 -240 - V = 0V = 265 kN
+ Mx2= 0: 03420505)2(240 22 xxM
2940265 2 xM
505 kN
3420 kN•m
MV505 kN
3420 kN•m
x1
60x1
+ Mx1= 0: 03420505)2
(60 11
1 xx
xM
342050530 12
1 xxM
2 m x2 -2
240
x2
505 kN
3420 kN•m
MV
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
16
60 kN/m265 kN
1350 kN•m4 m
6 m505 kN
3420 kN•m
Fy = 0:+ 505 - 240 - V = 0
V = 265 kN
+ Mx3= 0: 03420)4(505)2(240 33 xxM
1880265 3 xM
M
V
240 kN/m
2 m505 kN
3420 kN•m2 m x3
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
17
60 kN/m265 kN
1350 kN•m4 m
6 m505 kN
3420 kN•m
265 kN
1350 kN•mx4
MV
Fy = 0:+ V - 265 = 0
V = 265 kN
+ Mx3= 0: 01350265 4 xM
1350265 4 xM
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
18
Example 4-6
Determine the shear and moment in the beam shown in the figure as a function of x.
30 kN/m
10 kN/m
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
19
9 m
4.5 m
Fy = 0:+
0])9
)(20(2
1[1075 Vx
xx
+ Mx= 0:
20 kN/m
90 kN
6 m
90 kN
10 kN/m
2
x
2
x3
x
10x
10 kN/m
xx
)9
)(20(2
1
mkNx
/9
20
M
V
105 kN75 kN
211.11075 xxV
0)3
]()9
)(20(2
1[)
2)(10(75 M
xx
xxxx
32 37.0575 xxxM 75 kN