INTRODUCTION
at the location of plastic hinge of a deformed str → Curvature = infinity
in a statically indeterminate structure, the formation of 1st plastic hinge will reduce the number of redundancies by one
Redundant structure → Completely closed ring, Fixed portal frame, Propped cantilever
Natural frequency of a mass m at the end of the cantilever beam → f=(1/2)3EI/ml3
Shail structure → Rigid curved surface, Built of sheets or plates
Degree of freedom or independent displacement component
Pin-joint of plane frame = 2 (translation)
Pin-joint Space frame = 3 (translation)
Rigid-joint plane frame = 3 ( 1 rotation + 2 translation)
Rigid-joint Space frame = 6 (3 rotation + 3 translation)
Static indeterminacy
External determinacy (Dse) → Related to support condition
internal determinacy (Dsi) → Geometry of structure
Ds = Dse + Dsi
Dsi = n - 3
Plane Truss → Ds = m + r - 2j = m - (2j - 3)
Space Truss → Ds = m + r - 3j
Rigid-jointed Plane Frames → Ds = 3m + r - 3j
Rigid-jointed Space Frames → Ds = 6m + r - 6j
m + r > 2j → this condition is Sufficient but not necessary Regarding stability of truss
Stability
Reaction should be parallel, concurrent &
External stability: neither concurrent, nor parallel & non-coplanar.
For stability: equilibrium equation should be satisfied concurrently at each & every joint of the structure.
For instability: at least one joint of the str should be able to displace without causing change in length of at least one member
2D = 3 reactⁿ & 3D = 6 reactⁿ
in a plane independent static equilibrium eqn = 03 ( Σ Fx, Σ Fy, Σ Mz)
Surface structure → Small thickness
Planer System
Partially constrained → Equilibrium eqn > No of force
for indeterminate structure & essential unstable → No. of reactⁿ > Equilibrium equation
Reaction < 3 → essentially unstable
Y = Force & X = parts/members
Kinematic Indeterminacy
Pin-jointed plane frame → Dk = 2j - r
Pin-jointed space frame → Dk = 3j - r
Rigid-jointed space frame → Dk = 6j - r
J → number of joint, r → unknown reaction
2D rigid frame/Beam
Dk = 3j - Re + Rr + m
Ds = 3m - 3j + Rr
Value of re (Unknown Reaction or Constraints)
Fix = 3
Free = 0
Hinged = 2 → free rotation support
Hinged with roller = 1
Value of Rr = 3 - Re
METHOD OF ANALYSIS
A joint where several members meet, any applied moment will be distributed among the various members → in proportion to their stiffness
For equilibrium of deformable body we need a minimum 3 equilibrium equation → Fx=Fy=M=0
indeterminate/Redundant Str → Equilibrium + Compatibility condition + Force-displacement relation require for analysis
Statically indeterminate/Redundant structure are always chosen over statically determinate structure
For a linear elastic structural analysis system minimization of potential energy → Yields compatibility condition
Principal of Superposition
Hooke's law valid ( σ ∝ ε) i.e. material should behave Linear-elastic, effect of temp changes is considered, structure is analysed for the effect of support settlement
Stress-strain relationship = linear
Small deformation
Resultant deflection due to all loads = Sum of individual deflection
Stiffness factor at joint = Sum of stiffness factors the member connected to the joint
FORCE METHOD
Betti's Law
P=Q
P12=P21
Maxwell's reciprocal theorem
it is a special case of Betti’s law
Validity → All structure with linear force displacement relation, any st obeying hooke's law
Use → Analysing indeterminate str, Can be applied to rotations caused by flexure, shear or torsion
pq=qp → Deflection at point P due to load at point Q = deflection at Q due to load at point P
Unit load method (Virtual work method)
Apply unit load in the direction in which deflection is needed
Based on consistent deformation
Virtual work → Work done by real forces due to hypothetical displacement or work done by hypothetical forces during real displacements
=∫M.mdx /EI
M → due to external load, m →due to unit load
Force assumed & can’t omitted → Concentrated load on the mid span of a simply supported beam
Derived from the castigliano’s theorem
Strain energy method (Real work method)
U=UOA+UOB
Castigliano's Theorem
Strain energy method of finding out slopes and deflection
1st theorem
Principle of superposition is valid
Linearly or non-linear Elastic structure
dU/d=P, dU/d=M
2nd theorem
Any type of structure but linearly elastic
dU/dP=, dU/dM=
The first partial derivative of the total internal energy in a structure with respect to the force applied at any point is equal to the deflection at the point of application of that force in the direction of its line of action
Clapeyron's Three moment thᵐ
For continuous beam
Any type of cross section, No discontinuity such as hinges within the span,
Eqn → Slope compatibility at that point
Column analogue method
Width of column = unity
Column dim = L x 1/EI → A = L/EI
Valid when Str has static indeterminacy up to 3
Flexibility matrix method
Unit force is applied at coordinate j & Displacement are calculated at all coordinate
δ12 → Displacement at 1 due to unit load at 2
jth Coordinate → A unit forces applied at co-ordinate J and the displacement are calculate at all coordinates
Square symmetric matrix
Order of matrix(n x n) = Ds = Redundant force
Elements → Dependent on the choice of coordinate, Always dimensionally homogeneous
Each flexibility matrix element = displacement
Elementij=Elementji
Diagonal elements = Always (+ve), ≠ 0, ≠ (-ve)
Valid for stable structure
Flexibility matrix = 1/ Stiffness matrix
Flexibility = 1/stiffness = 1/K = ∆/P = θ/M
Note → Stiffness matrix → Same property except order(n x n) = Dk
DISPLACEMENT METHOD
Stiffness Matrix Method
Analysis of indeterminate structure with lesser DS
Stiffness → Force require to produce unit displacement
K = P/∆ = M/θ
for wall → K = ∞
Kij → force at i due to unit displacement at j
Order of matrix (n x n) = Dk
Size of basic stiffness matrix for plane truss member = 2x2
Stiffness matrix doubled → Deflection will be half
Slope Deflection Method (SDM)
G.A. Maney method
Axial & Shear deformation are neglected
Best → For Continuous Beam
Use of Kinematic redundant
Number of simultaneous equation to be solved = Number of joints in the structure
MAB=MFAB+(2EI/L)(2A+B)-6EI/l2
MBA=MFBA+(2EI/L)(2B+A)-6EI/l2
Deformation are considered to be caused by Bending moment only
Joints are considered rigid → No change in value of angles between members
Fixed End moment (Fix beam)
i. P at centre = PL/8
ii. Udl all span = wL²/12, at centre = wL²/24 → Sagging, Point of contraflexure = L / 2√3 → from centre
iii. P eccentric = Pab²/L², Pa²b/L²
iv. ∆ settlement = 6EI∆/L²
v. M at centre = M/4
vi. UVL = wL²/20 (at w) & wL²/30 (at 0)
Moment Distribution method (MDM)
By prof. Hardy Cross
Stable & Statically indeterminate structure
Best → For rigid frame
It is a Trial-error method, an iterative method
Use of Carry over factor and Distribution factor
Absolute Stiffness (K)
K = Flexural rigidity/Length = EI/L
Far end fix → K = 4EI/L
far end hinged → K = 3EI/L
Far end is a guided roller → K = EI/L
Far end is roller → 0
far end free → K = 0
Distribution factor
DF=Ki / Ki
DF=1 → All member meeting at a joint
Displacement factor → Δ = - 1.5 x DF
For end span, on the fixed end side → DF = 0
Relative stiffness
Far end is fixed → RS = I/L
Far end hinge → RS = 3I/4L
Carry Over Factor
COF = Carry over moment/Applied moment
Carry over moment → Moment developed at one end due to the applied moment at another end
fix hinge or far end is fix = 1/2 → Propped cantilever
fix free or far end is Guided roller = -1 → = Cantilever
fix, hinge & hinged at mid span = a/b
Far end is simply supported or hinged = 0
Kani’s method
Successive approximation, Iterative approach for applying slope deflection method
Use of Rotational coefficient
If end of continuous beam is fixed → rotation contribution = 0
Used for → Fixed frame, Portal frame, Box frame
Any arithmetic error that creeps in will automatically get corrected
An overhang can be conveniently dealt with by regarding it as a member with infinite length
Note
Maxwell’s diagram → Forces in bar members
Hinsburg’s method → Concept of substitute members
Newmark’s method → Deflection by numerical procedure
Williot-Mohr method → Displacement of joints in a Truss
TRUSS
Zero load test → To check the geometrical unstable
Stanchions → Compression members in truss
Primary unit of truss → Triangle
Properties of truss
Compressive parts are thicker than tensile parts
Truss transmits load → in axial direction → as Tension & Compression
Truss is said to be completely analysed when all member forces & their corresponding stress are determined.
Simplest geometrical form of truss is Triangle
Can't use Concrete as material for truss
Assumption in truss
Members are joined by Smooth pins or friction less pins
Self wt. of truss members is neglected
Truss str is loaded only at the joints
Members of a truss do not carry any flexural load
Note → No assumption about all members have same cross-sectional area, Material Homogeneous isotropic elastic
Beam vs Truss
Beams can't transmit load in axial direction while trusses can
Beam is single member whereas truss is composed of many members
Beams, Frames → Flexural load → Shear & Bending
Truss → only Axial loads → Tension & Compression
Frame vs Truss
Trusses can't bend but frame can
BM is zero everywhere in the truss but frames have BM at the joints
Deflection/Displacement in truss
Angle wt method, Joint displacement method, Williot-Mohr method (Graphical method)
Deflection → Depends on Axial rigidity
Lack of fit → Causes both horizontal and vertical deflection
Total deflection → Force + Temp change + member too short or to long
Δ = nNL/AE + nLαΔT + ndL
N → Force due to external load , n → due to unit load
Force
F = P + kx
x=-(PkL/AE) / (k2L/AE)
Method of force analysis of truss/frame
3 methods → Graphical, Method of joints, Method of sections
i. Graphical
ii. Method of joints
Unknown force < 3 → Not greater than 2
The members of the truss are connected together at their ends by friction pins.
iii. Method of sections
For easy solution use it
Types of Truss
Forms of truss → Prott truss, Howe truss, warren truss and k-truss
i. Simple truss
A truss whose number of members is given by → m = 2j – 3
Consist entirely of a triangle, It can consists of any other shaped intermediate parts, as long as it is stable
ii. Compound truss
Formed by joining two or more simple trusses
Will be Rigid & determinant → if m = 3 + m1 + m2.
May be formed by connecting two simple rigid frames → by Three bars
King post truss
Stress carried by king post truss → Compressive stress
King post → Member used to prevent the sagging of beam and connected with the ridge
Hudson’s formula
Dead weight of the truss girder bridges → As a function of bottom chord area
W = 0.785 A
FRAME
Basic Perfect frame → Triangle
Appx analysis of building frames under vertical load → Point of inflection = 1/10 of span length from each end of the beam
Substitute method of Frame analysis → Frame is subjected to vertical dead load and live load
Malti-bay portal frame → SFintermediate column =SFend column
Point of inflection → anywhere in the frame
Moment resisting frame
A rigid frame detailed to provide good ductility and support for both lateral and gravity loads by flexure action
Classification → Ordinary, intermediate and Special MRF
Sway direction
No sway → Load & frame both are symmetric
Sway correction → Necessary when either the loading or the frame is unsymmetrical
Symmetric → Opp to load(P) from mid span or sway towards lesser load, Towards column of low stiffness, Towards hinged support
Unsymmetric → Side where ratio (l/L) of column is less
Stiffness → Far end fix = 4EI/L, far end hinged = 3EI/L, far end free = 0
ARCH
Arch subjected → SF, BM & Normal Thrust
Arch is a compression member
an arch subjected to pure compression due to a UDL → shall be a Parabolic Arch
In fixed arch structure → No hinges are necessary
An arch is used in long span bridges because → there is considerable reduction in Bending moment
Linear arch
Normal thrust only
LA Represents thrust line
Arch Components
Spanderal → Triangle space formed b/w the extrados and horizontal line draw through the crown of an arch
Skew back → An inclined surface or splayed surface on the abutment, from which the arch curve starts or ends
Haunch → Lower half of arch b/w the crown and skew back
Note for both 2H-3H
Parabolic 3H or 2H Arch → if fully UDL loaded on whole span → BM = Radial shear = SF = 0, But normal thrust ≠ 0
2H → ILD for Horizontal thrust = Parabolic (L/4h)
3H → ILD for Horizontal thrust = Triangular (25L/128h)
Line of thrust in parabolic arch → Parabolic
Load system entirely vertical → Ha = Hb = H → Horizontal thrust is same at each support
3-HINGED ARCH
Hinged at supports and anywhere in rib (generally at crown)
3H Arch → resist load by developing Thrust, radial shear & BM
Ds = 0 → Statically determinate
ΣFx = ΣFy = ΣMx = 0
To find Reaction(Va, Vb) → Use ΣM = 0 at other support
To find Horizontal thrust (H) → Use ΣM = 0 at crown
Radial Shear → Sx=RAcos-Hsin
Normal shear → Nx=RAsin+Hcos
Absolute Max hogging moment → When the point load is at crown
Max sagging (+ve) BM → 0.211 x L from support
Max hogging (-ve) BM → 0.25 x L from support
3H Parabolic Arch
y=4hx(l-x)/l2
Case-i → Point load at crown
V = W/2
H = WL/4h
Case-ii → UDL
Any Section → Subjected to normal thrust only
If Arch shape is Parabolic → Arch is Free from SF, BM
Va = Vb = wl/2
All over span → H=wl2/8h
Half span → H=wl2/16h
Case-iii → Supports not at same level (UDL)
H=wl2 / 2(h1 +h2)2
l1=(lh1 ) / (h1 +h2)
l2=(lh2 ) / (h1 +h2)
Case-iv → Supports not at same level (Point load at crown)
H=Wl / (h1 +h2)2
3H Semicircular Arch
Case-i → Point load at crown
V = W/2
H = W/2
Case-ii → UDL over entire Span
Va = Vb = wR
H = wR/2
BMx=R.x-wx2/2-H.y
BMx=(wR2/2)(sin2-sin)
BMMax=-wR2/8 → θ = 30°
BMc=0 =90°
Temp effect on 3H Arch
Due to temperature change → Stresses are not produced in the arch, but the horizontal thrust changes
Rise in temperature increases the length of the arc
Temp ↑es → H ↓es
New crown height → h=(h+l2/4h)T
Parabolic (UDL) → H ∝ 1/h → dH/H = - dh/h
2-HINGE ARCH
Ds = 1 → Statically indeterminate by one degree
Unknown force = 4 → Va, Vb, Ha, Hb
H=(Myds/EI) /(y2ds/EI)
2H Parabolic Arch
y=4hx(l-x)/l2
Case-i → UVL all over span
H = wl²/16h
Case-ii → UDL
All over span → H = wl²/8h → Same as 3H
Half span → H = wl²/16h → Same as 3H
Case-iii → Point load (W)
At crown → 25Wl / 128h
Other than crown → H=(5/8)(Wa(l-a)(al-a2+l2)/hl3)
2H Semicircular Arch
Case-i → UVL all over span
H = (2/3) x (wR/π)
Case-ii → UDL all over span
Va = Vb = wR
H = (4/3) x (wR/π)
Shear centre = 2R/π
Case-iii → UDL half of span either left or right side
H = (2/3) x (wR/π)
Case-iv → Point load (W)
At crown → H=W/ → Does not depends on span or radii of arch
Other than crown → H=(W/)sin2()
At crown → =90, At support → =0
Temp effect on 2H Arch
Temp ↑es → Horizontal thrust at support ↑es
Temp rise cause the max BM at the crown
Settle down → Horizontal thrust ↓es
H=(LT) /(y2ds/EI)
Parabolic → H=15EIT/8h2 → H ∝ 1/h2
Semicircular → H=4EIT/R2 → H ∝ 1/R2
Reaction locus of 2H arch
Semicircular arch → Straight line or horizontal line → at πR/2 distance
Parabolic arch → Curved (Parabolic) → at 1.6h (end), 1.28h (mid) distance
Eddy's Theorem
Valid for Vertical loads only
BMx ∝ Y
Y = Vertical intercept between the linear arch (or theoretical arch) and the centre line of the actual arch
In a 3H arch → The linear and the actual arch meet at least three points
JACK ARCH
Rise = 1/8 to 1/12 of the span
Joists are spaced 1 - 1.5 m
Composed of arches of either Bricks or lime, Concrete
CABLE & SUSPENSION BRIDGE
Economical span of bridge → Cost of supporting system of one span = Cost of one pier
Flutter → unstable vibratory motion due to combined bending and torsion, in flexible plate like structure ex.suspension bridge deck, sign board
Bridges are classified as → Deck type
Cables
Cable & Wires → Tension members
Catenary → Shape of cable due to self weight
Cable structure → Suspension bridge, cable stayed roof
Perfectly flexible cable → BM at every section = 0
Minimum Tension → Tmin=H → Where sag is max (at centre)
Max Tension → Tmax=V2max+H2 → At Support
Dip or Sag = (1/10 to 1/15) of Horizontal span
Cable length (S)
S=0l1+(dy/dx)2
Support at same level → S=l+8h2/3l
Support are not at same level → S=l1+l2+2h12/3l1+2h22/3l22
Symmetrical cable subjected to UDL on horizontal span
Profile/Shape → Parabolic
H=wl2/8h
y=(w/2H)x2
dy/dx=8hx/l2
T=H1+(dy/dx)2=H1+(8hx/l2)
Tmin=H=wl2/8h x=0 (at max sag)
Tmax=(wl/2)1+(l/4h)2 x=l/2 (at support)
Suspension bridge
Suspension bridge with two hinged stiffening girder → Statically indeterminate of 1 degree
INFLUENCE LINE DIAGRAM
Influence line for a forcing function(SF,BM,reaction, axial force) gives its variation → at given section/point → for any postⁿ of point load
ILD for SF or BM → Linear or triangle → For any type of beam/str
Point load → SF or BM at a point=Load x Ordinate of ILD of SF or BM
UDL → SF or BM at a point=Load intensity x Area of ILD of SF or BM
Ordinate of ILD → Always have dimension of Length
ILD for Stress function → Linear
The maximum ordinate of the ILD for BM at the fixed support in cantilever beam will occur at free end
Muller-Breslau Principles
Use → determine Shape of ILD, Ordinate of ILD, Part of structure to be loaded to obtain the max effect
Applicable for both Determinate & indeterminate
Material is within elastic limit and obeys hooke's law
It is a straight application of Maxwell’s reciprocal theorem
Proof → By principle of virtual work
Eff. of Rolling Load
Use above diagram for below condition → 1,2,3
1. Single Point Load
max SF : just left or right to that point
max BM : on the sectⁿ under the load
max -ve SF : right hand support
max +ve SF : left hand support
Absolute max BM : at centre or mid span
2. UDL < Span
Max BM → a/b = x / l -x → Ratio of span = ratio of load placed
max -ve SF → head of udl reaches at that point
max +ve SF → tail of udl reaches at that point
absolute max BM → CG of load at centre or mid span
absolute max -ve SF → Head at right hand support
absolute max +ve SF→ Tail at left hand support
3. UDL > Span
Use dimag from above
Train of concentrated loads
Absolute max BM → CG at centre
Max BM at a point → The point load under consideration and the CG of train of point loads equidistant from the mid span
Max -ve
ILD for deflection
Free end of cantilever → Cubic parabola with zero ordinate at fixed end and Max ordinate at free end
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