Transcript Title

SM1-03: Statics 2:Cross-sectional forces in bars
CROSS-SECTIONAL FORCES
IN BARS
M.Chrzanowski: Strength of Materials
1/23
SM1-03: Statics 2:Cross-sectional forces in bars
In statics we aim at evaluation of M, Q, N
in an arbitrary cross-section of a bar
M
.
M
i.e. in finding :
axial force
shear force
bending moment
q
Q
N = N(r0)
ro
Q = Q(r0)
M= M(r0)
as functions of cross section position.
M.Chrzanowski: Strength of Materials
z
N
M
x
y
2/23
SM1-03: Statics 2:Cross-sectional forces in bars
Instead of using general reference
coordinates {x,y,z} it is more
convenient to set them at bar axis.
Thus cross-sectional forces
become functions of one
cooridnate only:
q
Q
N = N(x)
P
Q = Q(x)
M= M(x)
The ultimate goal is to learn skills
allowing for drawing diagrams for
these functions – even without any
calculations!
M.Chrzanowski: Strength of Materials
N
M
3/23
SM1-03: Statics 2:Cross-sectional forces in bars
n
n
N
N
Q
Q
M
M
{wI} ≡ {ZII}
{wII} ≡ {ZI}
n
n
Q
N
N
Q
M
M.Chrzanowski: Strength of Materials
M
4/23
SM1-03: Statics 2:Cross-sectional forces in bars
n
n
N
o
Q
M
n
N
o
Q
M.Chrzanowski: Strength of Materials
M
5/23
SM1-03: Statics 2:Cross-sectional forces in bars
n
n
Q
N o
M
n
Q
N o
M
M.Chrzanowski: Strength of Materials
6/23
SM1-03: Statics 2:Cross-sectional forces in bars
According to the assumption made the beam axis represents the whole structure
n
N
Q
M
Q
N o
M
M.Chrzanowski: Strength of Materials
n
7/23
SM1-03: Statics 2:Cross-sectional forces in bars
Important observations
Equations for cross-sectional forces change at points of loading application
(„characteristic points”).
1 kN/m
2 kN
4 kNm
x
4 kN
2 kN
2m
2m
4m
To write equilibrium equations a global coordinate system has to be set
The set of loadings acting an a structure has to be in equilibrium (i.e. equal to zero
set of forces). Beforehand, kinematic boundary conditions have to be replaced by
static boundary conditions (c.f. Theoretical Mechanics)
M.Chrzanowski: Strength of Materials
8/23
SM1-03: Statics 2:Cross-sectional forces in bars
In global coordinates the sign of cross-sectional forces depends on the on the
assumption on which part of a structure is considered (that „from left” or „from right)
y
n
n
Q
N o
N
o
Q
M
x
M
To avoid this ambiguity system of coordinates has to be connected with the
normal outward vector n.
We adopt the following rules for N, Q axis:
Positive direction of N coincides with an outward direction of normal n and
positive direction of Q is such that couple {N, Q } is left-twisted.
M.Chrzanowski: Strength of Materials
9/23
SM1-03: Statics 2:Cross-sectional forces in bars
To define the sign of bending moment we introduce a special case of „axis”. It
will be given by pointing at upper or lower surface of a bar. It will be called
„bottoms” and marked by a dashed line drawn along the bar axis. We will
consider bending moment as being positive if its action results in extension
of „bottoms” .
n
n
Q
N o
N
o
Q
M
Q
N o
N
Q
n
M, Q, N > 0
n
o
All figures show:
M
M
M.Chrzanowski: Strength of Materials
M
10/23
SM1-03: Statics 2:Cross-sectional forces in bars
Example
1 kN/m
2 kN 4 kNm
x
4 kN
2 kN
2m
0<x<2
x
M(x)= +2x
n
Q(x)= +2
The results have
to be the same!
2m
4m
n
Q
N
N(x)= 0
M.Chrzanowski: Strength of Materials
N
Q
11/23
SM1-03: Statics 2:Cross-sectional forces in bars
Example
1 kN/m
2 kN 4 kNm
x
4 kN
2 kN
2m
2<x<4
4m
x
M(x)= +2x – 2(x 2)n= 4
Q
Q(x)= +2 – 2 = 0 N
N(x)= 0
M.Chrzanowski: Strength of Materials
2m
n
N
Q
12/23
SM1-03: Statics 2:Cross-sectional forces in bars
Example
1 kN/m
2 kN 4 kNm
x
4 kN
2 kN
2m
4<x<8
2m
4m
x
M(x)= +2·x2·(x 2)+4  1·(x
n  4)·(x 4)/2 = 8  (x 4)2/2
Q
Q(x)= +2 2  1·(x 4)= 4Nx
N(x)= 0
Q
M.Chrzanowski: Strength of Materials
n
N
13/23
SM1-03: Statics 2:Cross-sectional forces in bars
Example
1 kN/m
2 kN 4 kNm
x
4 kN
2 kN
x1
2m
2m
4m
0 < x1 < 4
M(x1)= +4·x1  1·x1·x1/2 = 4x
n 1  x12/2
Q
Q(x)= -4 + 1·x1=  4+ x1N
N(x)= 0
M.Chrzanowski: Strength of Materials
x1
n
N
Q
14/23
SM1-03: Statics 2:Cross-sectional forces in bars
Before we draw diagrams of cross-sectional forces for our example,
we make agreements according to positive directions of axes of the
functions M(x), Q(x), N(x)
Q(x), N(x)
M(x)
x
x
M(x)
1. Positive axis of M points towards „bottoms”
2. Positive axes of Q i N – can be chosen arbitrarily
M.Chrzanowski: Strength of Materials
15/23
SM1-03: Statics 2:Cross-sectional forces in bars
M>0
n
n
Q
N o
N
o
Q
M
M>0
M.Chrzanowski: Strength of Materials
M
M>0
16/23
SM1-03: Statics 2:Cross-sectional forces in bars
M>0
n
n
Q
N o
N
o
Q
M
M
M<0
Conclusion:
M<0
The diagram of M(x) appears always on the same side of a bar –
independent from assumption according to the choice of „bottoms”.
It is the side which fibres are subjected to tension.
M.Chrzanowski: Strength of Materials
17/23
SM1-03: Statics 2:Cross-sectional forces in bars
Notice
The above assumptions on signs and making diagrams of crosssectional forces are quite arbitrary.
These are often different (eg. in mechanical engineering or in some
countries: UK, USA):
M>0
M.Chrzanowski: Strength of Materials
18/23
SM1-03: Statics 2:Cross-sectional forces in bars
Example
1 kN/m
0<x<8
2 kN 4 kNm
0<x<2
x
4 kN
2m
M(x)= +2x
Q(x)= +2
M(0)= 0
M(2)= 4 kNm
2 kN
2m
N(x)  0
4m
2<x<4
0
2
4
6
8
M
Q
M(x)= 4
4<x<8
[kNm]
2
0
-2
-4
Q(x)= 0
M(x)= 8  (x 4)2/2
Q(x)= 4x
M(4)= 8 kNm
Q(4)= 0
M(8)= 0
Q(8)=  4 kN
[kN]
M.Chrzanowski: Strength of Materials
19/23
SM1-03: Statics 2:Cross-sectional forces in bars
Comments
1 kN/m
2 kN 4 kNm
x
4 kN
2 kN
2m
2m
1. In the light of the above assumptions
regarding sing convention – we do not need to
assign signs to the bending moments diagrams.
It is just enough to be sure that there are drawn
on the side of fibres subjected to tension. Thus
we need only to mark „bottoms” for writing down
the bending moment equations, but not for
making its final diagram!
M(x)
4m
0
2
4
6
8
4 [kNm] 4 [kNm] +
8 [kNm]
M
M(x)
Q
+
2 [kN]
M.Chrzanowski: Strength of Materials
-
2
0
-2
4[kN]
-4
M>0
M<0
2. In contrast – the signs of Q and N
diagrams have to be clearly indicated
3. Parts of diagrams marked in red have no
physical meaning: the values of corresponding
cross-sectional forces are not determined in
these points
20/23
SM1-03: Statics 2:Cross-sectional forces in bars
M-Q-q relationship
1 kN/m
2 kN 4 kNm
Analysis of the obtained diagrams implies
following observations:
x
4 kN
1. Linear part of M diagram corresponds to
constant Q and q=0
2 kN
2m
2m
2. Constant value of M corresponds to zero
4m
M
4
4
8
+
[kNM]
+
3. Linear part of Q diagram corresponds to
constant loading q but M diagram for this
part is second order parabola
Q
2
-
The above observations suggest that qdiagram is first derivative of bending moment
M and distributed loading q is first derivative of
tangential (shear) force Q.
4
[kN]
M.Chrzanowski: Strength of Materials
value of Q
Now we will proof these theorems.
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SM1-03: Statics 2:Cross-sectional forces in bars
M-Q-q relationship
q(x)
q(x+γΔx)
q(x)
Q(x)
M(x)
M(x+Δx)
x
A
x
Δx
0  ,  ,  1
Δx
x
Q(x+Δx)
x+Δx
M.Chrzanowski: Strength of Materials
}
dQ
Q( x)  q( x  x)  x  Q( x)  x
0
dx x  x
q( x )  
}
} }
 dQ

 dM

Q
(
x


x
)

Q
(
x
)


x
M ( x  x )  M ( x )  x 



dx
dx
x



x


x x 

dM
M ( x)  Q( x)  x  q( x  x)  x  x / 2 M ( x)  x
0
1
dx x x
dQ
dx
} x } x  0
Q( x) 
dM ( x )
dx
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SM1-03: Statics 2:Cross-sectional forces in bars
M-Q-q relationship
d

dx
dM ( x )
 Q( x)
dx
d 2 M ( x ) dQ

2
dx
dx
dQ
  q( x )
dx
M
dM/dx=Q>0
x
x
M
M.Chrzanowski: Strength of Materials
dM/dx=Q<0
d 2 M ( x)
  q( x )
2
dx
The problem of cross-sectional forces
determination is reduced to the solving
second order differential equation with
boundary conditions set for functions of
M and Q being sought.
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SM1-03: Statics 2:Cross-sectional forces in bars
stop
M.Chrzanowski: Strength of Materials
24/23