# Structural Analysis 2

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Structural Analysis ENG2036MDr. Alessandro Palmeri

Lecture # 2[2]

OutlinesFurther on statically determined beamsPrinciple of superposition Relations between transverse load (q), SF (V) and BM (M)

ExamplesBeams with internal hinges Frames with sloping members

[3]

Principle of SuperpositionThis fundamental principle states:The total displacements {ux(xP), uy(xP), (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 SuperpositionTwo requirements must be imposed for this principle to apply:The material must behave in a linear elastic manner, i.e. the Hookes 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 SuperpositionMore 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 SuperpositionThe principle of superposition may simplify the evaluation of support reaction and the construction of the internal force diagramsq=F L

M = FL

F

=q=F L

L 2

L 2

F

+L

+

C = FL

L

L 2

[7]

Principle of SuperpositionThe principle of superposition may simplify the evaluation of support reaction and the construction of the internal force diagramsFL 2

FL

+

+

FL

=

5 FL 2

[8]

Relations between transverse load, SF and BMLet 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)

Ox

x+dx

[9]

Relations between transverse load, SF and BM dx 2The 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 M be neglected. Therefore:We can assume that the distributed load q is constant over the length dx The resultant of the distributed load, of magnitude q dx, is located at the midpoint of the element

q dx

q

M + dM

V

V + dV dx[10]

Relations between transverse load, SF and BM dx 2In order to be in equilibrium, the conditions of statical equilibrium must be satisfied for the infinitesimal beam element The condition of transversal equilibrium gives: dV V + q dx = V + dV =q dx 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)

q dx

qM

M + dM

V

V + dV dx

[11]

Relations between transverse load, SF and BM dx 2Taking moments about the centroid of the right-hand face of q dx the beam segment, the condition of rotational equilibrium furnishes: dx 2 M + d M = M + V dx + q 2 dM M =V dx This equation states that the the slope of the BM diagram, M=M(x), at a generic abscissa x, is given by V the SF at that abscissa, V(x)

q

M + dM

V + dV

dx[12]

Relations between transverse load, SF and BM dx 2Differentiating both sides of the second equation with respect to the abscissa x, and then substituting the first equation into the result, one obtains:

q dx

d2M =q 2 dx 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

qM

M + dM

V

V + dV dx

[13]

Relations between transverse load, SF and BMMoreover, 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:

V = q ( x) dxx

M = V ( x ) d xx

Change in the SF

=

Area under the distributed load diagram

Change in the BM

=

Area under the SF diagram

[14]

Relations between transverse load, SF and BMSimilar relations hold in the cases of a concentrated transversal load F (positive if downward) or of a concentrated couple C (positive if anticlockwise):F

V = FM

M = CM

CM + M

M

V

V + V

V

V dx

dx

[15] Free-body diagram(*)Concentrated transv. force

SF diagram

BM diagrampiecewise linear

M L VL

F x VR

MR

zero slope

VL F x VR M LVL

xMR

Concentrated couple

M L VL C x VR M L VL q

MR VL

zero slope

VR

VR ML

x C

MR

constant slope

M L VL qL

(*) Internal forces V and M are drawn according to the positive sign convention

parabola M R constant slope (concave upward) q VL x0 x0 VR M L MR V VR L VR qR increasing cubic parabola qL slope VR M R (concave upward) VL x0 x0 ML VR MR V L VR

Trapezoidal transv. load

Uniform transv. load

Structures with internal hingesFrom 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

> qR

qL

[16]

[17]

Structures with internal hingesOn 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 hingesFundamentally, 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 hinge Statical 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]

KINEMATICAL APPROACH

[21]

STATICAL APPROACH

[22]

Structures with internal hingesMore 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

d = f ecwhere f is the number of unknown reactions, e is 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

[23]

Structures with internal hingesKinematically unstabled= fec= (2+1)31= 34= 1

Statically determinated= fec= (2+1+1)31= 44= 0

Hyperstatic to the 2nd orderd= fec= (3+1+2)31= 64= 2[24]

Frames with sloping membersDistributed 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 beloww L cos( )

qS L qS

q L cos( ) W q W

q L cos( ) W q W

w

L

L

L

L

[25]

Frames with sloping membersw L cos( )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

w

L

[26]

Frames with sloping membersqS LThe snow load (live load): acts downward is distributed along the horizontal projection of the inclined beam

qS

L

[27]

Frames with sloping membersq L cos( ) WThe wind load q L cos( ) (dynamic load): W acts orthogonally to the inclined beam q W is distributed along its length produces higher suction (negative L pressure) on the windward side

q W

L