Post on 10-Apr-2015
“Structural Analysis”
ENG2036M
Dr. Alessandro Palmeri
<a.palmeri@bradford.ac.uk>
Lecture # 2
[2]
Outlines
Further on statically determined beams
Principle of superposition
Relations between transverse load (q), SF (V) and BM (M)
ExamplesBeams with internal hinges
Frames with sloping members
[3]
Principle of Superposition
This fundamental principle states:The total displacements {ux(xP), uy(xP),ϕ(xP)} or internal forces {N(xP), V(xP), M(xP)} at a particular point P of abscissa xP in a planar structure subjected to a set of external loads acting simultaneously can be determined by adding together (i.e. superimposing) the displacements or internal forces produced by each of the external loads acting individually
[4]
Principle of Superposition
Two requirements must be imposed for this principle to apply:
The material must behave in a linear elastic manner, i.e. the Hooke’s law is valid, and therefore displacements are proportional to external loads
The geometry of the structure must no undergo significant change when loads are applied, i.e. small displacement theory applies
[5]
Principle of Superposition
More concisely:
The response of a linear structure is the same if all loads are applied simultaneously, or if the effects of the individual loads are combined
[6]
Principle of Superposition
The principle of superposition may simplify the evaluation of support reaction and the construction of the internal force diagrams
FFL=M
2L 2L
F
q F L=
FL=C
2LL L
=
+ +
q F L=
[7]
Principle of Superposition
The principle of superposition may simplify the evaluation of support reaction and the construction of the internal force diagrams
FL2
FL
FL
5
2FL
+ + =
[8]
Relations between transverse load, SF and BM
Let us consider the beam segment depicted below, which is subjected to a distributed transverse load q=q(x), positive if upward, whose ordinates vary with the abscissa x, measured from the origin, O, at the left of the segment
( )q q x=
x
dx x+
O
[9]
Relations between transverse load, SF and BM
The beam element shown to the right can be thought as obtained by operating two ideal sections at abscissas x and x+dx
Since the length dx of this element is infinitesimally small, the slight variation of the distributed load can be neglected. Therefore:
We can assume that the distributed load q is constant over the length dxThe resultant of the distributed load, of magnitude q dx, is located at the midpoint of the element
q
dx
dV V+
dM M+M
V
dq x
d 2x
[10]
Relations between transverse load, SF and BM
In order to be in equilibrium, the conditions of statical equilibrium must be satisfied for the infinitesimal beam element
The condition of transversal equilibrium gives:
This equation states that the slope of the SF diagram, V=V(x), at a generic abscissa x, equals the ordinate of the distributed load at that abscissa, q(x)
dd d
d
VV q x V V q
x+ = + ⇒ =
q
dx
dV V+
dM M+M
V
dq x
d 2x
[11]
Relations between transverse load, SF and BM
Taking moments about the centroid of the right-hand face of the beam segment, the condition of rotational equilibrium furnishes:
This equation states that the the slope of the BM diagram, M=M(x), at a generic abscissa x, is given by the SF at that abscissa, V(x)
2dd d
2
xM M M V x q+ = + +
d
d
MV
x⇒ =
q
dx
dV V+
dM M+M
V
dq x
d 2x
[12]
Relations between transverse load, SF and BM
Differentiating both sides of the second equation with respect to the abscissa x, and then substituting the first equation into the result, one obtains:
According to our sign conventions, it follows that the BM diagram, M=M(x), at a generic abscissa x, is concave upward/downward if the distributed load at that abscissa, q(x), acts downward/upward
2
2
d
d
Mq
x=
q
dx
dV V+
dM M+M
V
dq x
d 2x
[13]
Relations between transverse load, SF and BM
Moreover, if the first two differential equations are integrated between two consecutive loading discontinuities, such as external concentrated forces or couples, or support reactions as well, one obtains:
( )dx
V q x x∆
∆ = ∫
Change in the SF
Area under the distributed load diagram
= Change in the BM
( )dx
M V x x∆
∆ = ∫
Area under the SF diagram
= −
[14]
Relations between transverse load, SF and BM
Similar relations hold in the cases of a concentrated transversal load F (positive if downward) or of a concentrated couple C(positive if anticlockwise):
V F∆ =
dx
V V+ ∆
MM
V
FM∆ = −C
dx
V
M M+ ∆M
V
C
[15]
FLVLM
RV
RM
LVLM
RV
RMC
LVLM
RV
RMq
LVLMRV RMLq
Rq
zero slope
zero slope
constant slope
increasing slope
LV
RV
LV RV
LV
RV
LMRM
LM RM
LMRM
LMRM
x
x
0x
xx
x
0x
0x
Free-body diagram(*) SF diagram BM diagram
Co
nce
ntr
ated
tr
ansv
. fo
rce
Co
nce
ntr
ated
co
up
leU
nif
orm
tr
ansv
. lo
adT
rap
ezo
idal
tran
sv. l
oad
F
Cconstant slope
piecewise linear
parabola (concave upward)q
VL
VR
VL VR
VL
VR
qL
LV
RV0x
qR > q
L
cubic parabola (concave upward)
(*) Internal forces V and M are drawn according to the positive sign convention
[16]
Structures with internal hinges
From a kinematical point of view, an internal hinge can be used in order to model a structural joint which eliminates any relative displacement between the two points which it connects (both horizontally and vertically), while it allows a relative rotation
From a statical point of view, an internal hinge transmits both horizontal and vertical components of NF and SF, while it does not transmit BM, i.e. the BM diagram must be zero at the position of an internal hinge
[17]
Structures with internal hinges
On the contrary, an internal fixed joint does not allows a relative rotation between the two points which it connects, and hence it is able to transmit BM (statical-kinematical duality)
[18]
Structures with internal hinges
Fundamentally, there are two alternative approaches to deal with structures with an internal hinge:
Kinematical approach: the structure is handled as composite of two rigid bodies; each one posses three degrees of freedom each and is subjected to the internal reaction forces arising from the internal hingeStatical approach: the structure is thought as made of a single rigid body, having in total three degrees of freedom, but in which a constructive condition imposes that the BM is not transmitted across the internal hinge
[19]
Structures with internal hinges
[20]
“KIN
EM
AT
ICA
L A
PP
RO
AC
H”
[21]
“ST
AT
ICA
L A
PP
RO
AC
H”
[22]
Structures with internal hinges
More generally, the statical approach is much more convenient, since it brings to a set of equations in fewer unknowns, which are simply the external reactions
In this case the degree of statical indeterminacy, d, is given by
where f is the number of unknown reactions, eis the number of available conditions of statical equilibrium (e=3) and c is the number of conditions of construction, i.e. the number of the internal hinges
d f e c= − −
[23]
Structures with internal hinges
Kinematically unstable
d= f−e−c= (2+1)−3−1= 3−4= −1
Statically determinate
d= f−e−c= (2+1+1)−3−1= 4−4= 0
Hyperstatic to the 2nd order
d= f−e−c= (3+1+2)−3−1= 6−4= 2
[24]
Frames with sloping members
Distributed loads acting on sloping members can be defined in different ways, depending on the situations, as demonstrated in the case of the roof in the three-pin arch depicted below
L
αw
cos( )w L α
L
Sq
Sq L
L
αWq′
W cos( )q L α′
α
L
Wq′′
W cos( )q L α′′
[25]
Frames with sloping members
The self weigh of the roof (dead load):
acts downward (i.e., in the direction of the field of gravity)is distributed along the length of the inclined beam
L
αw
cos( )w L α
[26]
Frames with sloping members
The snow load (live load):
acts downwardis distributed along the horizontal projection of the inclined beam
L
Sq
Sq L
[27]
Frames with sloping members
The wind load (dynamic load):
acts orthogonally to the inclined beamis distributed along its lengthproduces higher suction (negative pressure) on the windward side
L
αWq′
W cos( )q L α′
α
L
Wq′′
W cos( )q L α′′