Cable

Cables are slender, flexible members made of a group of high-strength steel wires twisted together mechanically. Steel cables provide the simplest means for supporting loads.

Steel cables, which are economically manufactured from high-strength steel wire, have an ultimate tensile strength of approximately 1862 MPa. The wires of cables are formed by drawing the alloyed steel bars through dies. This process aligns the molecules of the metal so that wires with tensile strength reaching as high as 1862 MPa can be produced. A group of such high strength steel wires twisted together forming a strand are used to make the cables. The modulus of elasticity of steel cables is approximately equal to 179 GPa, which is lower than the modulus of elasticity of steel cable is approximately equal to 179 GPa, which is lower than the modulus of elasticity of structural steel bars (200 GPa). Steel cables are easily handled and placed in position, even for very long spans. They provide lowest cost-to-strength ratio of any common structural members. The self-weight of the cable is generally neglected.

Element of Suspension Bridge

Suspension bridges are used for highways, where the span of bridge is more than 200m. Basically, a suspension bridges consists of following elements:

• Cable
• Suspenders
• Decking including the stiffening girder
• Supporting tower
• Anchorage
• Windguy and Windties
• Anchorage

A typical suspension bridge and its components are shown below:

Equilibrium of Light Cable: General Cable Theorem

Let us consider general case of cable supported at two supports A and B, which are not at the same level. The cable is acted by the system of vertical forces i.e. P₁,P₂,P₃…P_n as shown in figure below.

Let L be the horizontal span of cable and α be the inclination of line AB, with the horizontal. Obviously the differences in elevation between two supports A and B is equal to Ltanα.

Let V_A and V_B be the vertical components of reactions at A and B. Since there is no horizontal loading on the cable, the horizontal reaction(H) at ends A and B will  be equal in magnitude but opposite in direction.

In order to find vertical reaction V_A, take moments about B;

-V_A * L – H tanα + ΣM_B = 0

∴ V_A = (ΣM_B/L)- H tanα ——————————————(1)

Where, ΣM_B = Sum of moments of all loads

             P₁,P₂,….P_n about B.

Consider any point M of horizontal distance x from A.

Assume that the cable is perfectly flexible so that the bending moment at any point on the cable is zero.

Now,taking moment of all forces to the left point of M is zero.

H*(xtanα – y_m) + V_Ay * x – ΣM_m = 0

Substituting value of V_A from equation(1); we get,

H*xtanα – H*y_m+ x/L * ΣM_B – H*xtanα– ΣM_m = 0

or, H*y_m = x*( ΣM_B/L) – ΣM_m

Also, from beam; we have,

ΣM_B=0

or, V_A*L- ΣM_B = 0

∴ V_A = (ΣM_B / L )

∴ Hy_m = V_A * x – ΣM_m

Where, Hy_m = Beam moment at M.

It states that at any point on the cable acted by vertical loads, the product of H and y_m ( vertical distance from that point to the cable curve) equal to the bending moment at same point M considering simply supported beam of same span and same loading at that of cable.

Cable Subjected To Uniformly Distributed Load and Its Equation: At Same Level

Let us consider a cable subjected to a uniformly distributed load of w per unit run on the horizontal span.

Let, L= Length of cable ( curve length )

       l = Span of cable ( horizontal length )

      V_A = Vertical reaction at support A

      V_B = Vertical reaction at support B

       h  = Central deflection

Now, vertical reaction V_A and V_B are equal.

V_A =  V_B = wl/2

Taking moment about right of C; we have,

 ΣM_C = 0

or, V_B * l/2 – H* h – w*l/2 *l*4 =0

or, H *h = wl/2 * l/2 – wl²/8

 ∴ H = wl²/8

Let, T be the tension at any point P of the cable. Cosider the section (i)-(i) at distance x from support A.

Let, θ be the inclination of the tangent at with the horizontal.

Now,

Tension(T) = (V²+H²)^1/2

But, V_max= wl/2 = V_A = V_B

∴ T_max= {(wl/2)² + ( wl²/8h)²}^1/2 = wl/2 * { 1 + l² /16h²}^1/2

i.e. tension is maximum at supports since vertical reaction (V) is maximum at support and H is constant throughout the cable.

Similarly, tension ( T) is minimum at point of maximum sag because “ V” is minimum at mid span i.e. V = 0

∴ T_min = (0 + H²)^1/2 =H

To get equation of cable

Applying equation of equilibrium on left portion of section (I)-(I); we get,

ΣF_x = 0 gives;

H= Tcosθ —————(1)

Also,

ΣF_y = 0 gives;

V_A = wx + Tsinθ

∴ Tsinθ = V_A – wx —————(2)

Dividing equation (2) by equation (1); we get,

tanθ = (w_A – w_B) / H

or, dy/dx = 1/H {wlx/2 – wx²/2} = wx(l-x)/2h

Substituting value of H; we get,

y= {(wx)/ (2* wl²/8h) } * (l-x)

Substituting value of H; we get,

y= {(wx)/ (2* wl²/8h) } * (l-x)

∴ y = 4hx*(l-x)/ l² ; which is equation of parabola.

Thus,shape of cable is parabola.

To find the length of cable

ds = {(dx)² + (dy)²}^1/2 = {1+ (dy/dx)²}^1/2 dx

∴ ds/dx = {1+ (dy/dx)²}^1/2

We have,

y= 4hx(l-x)/l²

∴ dy/dx = 4h*(l-2x)/l²

∴ ds/dx = {1+ ([dy/dx][l-2x])²}^1/2

( Neglecting the other term of expansion)

Now,

Curve length ( L) = ∫ds

Determination of the Horizontal and Vertical Reaction of Cable with Ends at Different Level Carrying Uniformly Distributed Load.

Consider cable ACB supported at different levels A and B and C is the lowest point carrying uniformly distributed load w/ unit length.

The point C is extended to A^! and B^! as shown by dotted lines such that AC = A^!C dip of h₂.

Let, the horizontal distance between A and C = l₁

And Horizontal distance between B and C = l₂

Now, we have;

l₁ + l₂ = l ————————-(1)

Considering AA^!

H= wl²/8h = wl₁²/ 8h₁ —————–(2)

Considering BB^!

H= wl²/8h = wl₂²/8h₂ ——————–(3)

Equating equations (2) and (3) ; we get

wl₁²/8h₁ = wl₂²/8h₂

l₁² = h₁/h₂ * l₂²

∴ l₁/l₂ = (h₁/h₂)^1/2  ————————(4)

Also,

Length of cable ACA^’ (S₁) = 2l₁ +8h₁² /(3*2l₁) =2l₁+4h₁²/3l₁

And,

Length of cable BCB^’ (S₂) = 2l₂+8h₂² /(3*2l₂) =2l₂+4h₂²/3l₂

∴ Actual length of cable (S) = S₁/2+ S₂/2

∴ S = l + 2h₁²/3l₁ + 2h₂²/3l₂

To determine H

We have,

l= l₁+l₂ = l₁+l₂*( h₂/h₁)^1/2  { From equation(4)}

∴ l₁ = [l*(h₁)^1/2]/[(h₁)^1/2+(h₂)^1/2]

Substituting value l₁ in equation (2); we get

H= (w/2h₁)* [l*(h₁)^1/2]/[(h₁)^1/2+(h₂)^1/2] = wl²/[(h₁)^1/2+(h₂)^1/2]

Which is required expression for horizontal reaction.

To find vertical reactions

Let,

V_A= Vertical reaction at support A

V_B= Vertical reaction at support B

Now, taking moment about C of the forces on the left sides of C;we get,

V_A*l₁ = H*h₁+ wl₁²/2

or, V_A= H*h₁/l₁ +( wl₁²/2) * 1/l₁

Putting value of H in above equation;we get,

V_A= wl₁

∴ V_A= wl₁

Similarly, taking moment about C of the forces on right sides of C; we get,

V_B*l₂ = H*h₂+ wl₂²/2

or, V_B= H*h₂/l₂ +( wl₂²/2) * 1/l₂

Putting value of H in above equation;we get,

V_B= wl₂

∴ V_B= wl₂

Suspension Bridge with Three-Hinged Stiffening Girder

When suspension bridge is subjected to a moving (or rolling) load system the funicular polygon will change its shape and hence the shape of cable is also change.

In suspension cable bridge it is necessary to maintain the parabolic shape of cable to satisfy this condition the load transmitted to the cable should be uniformly distributed load.

This is achieved by either two hinged or three hinged stiffening girder.

The main purpose of providing these stiffness girders is to reduce the sag under the rolling load.

The girders are suspended form cables through hanger cables.Hence, the uniformly distributed dead load of the roadway and stiffening girders is transmitted to the cables through hanger cables and is taken up entirely by the tension in the cables. The stiffening girder does not suffer any S.F. or B.M. under dead load as girder is supported by closely spaced hanger cables throughout.

Any live load on the bridge will be transmitted to the girders as point loads. The stiffening girders transmit the live load to the cable as uniformly distributed load. While doing so the stiffening girders will be subjected to S.F. and B.M. throughout their length.

Extra:

1. Explain  with neat sketches tower structure as well as wind cables and windguy.

Ans. Suspension bridges are often supported on towers. Towers provide means for the cable to change the direction. Normally three type of towers are used in suspension bridge system.

a. Tower with saddles

• Such types of tower are fixed at their bases and support the main cable through a carriage, which is free to roll horizontally on the tower top.
• In this arrangement, the two cables does not need to have same tension.

b. Tower with pulley at top

• Such type of tower acts like vertical cantilevers and offer some resistance to cable.
• Assuming pulley as frictionless, tension in anchor cable is taken as same as tension in the main cable.

c. Tower hinged at base

• Such types of towers are free to rock in the plane of the main cables, which are securely attached to the tower tops.
• Suspension bridges are necessary to be protected from the vibration caused by wind. The lateral stability of the bridge is thus achieved by windguy arrangement. A standard form of windguy arrangement is shown in figure.
• It essentially consists of the bridge with a cable in the lateral direction with parabolic shape.

Moving static loads

Structure which is used in bridges, gantry girders, crane beams etc. are subjected to loads which change their position often such loads are called moving static loads.

Few standard loads are:
 

1. Single concentrated load
2. U.D.L greater than the span
3. U.D.L smaller than the span
4. Two concentrated load with specified distance between them
5. Multiple concentrated loads(train of wheel loads)

Influence Line Diagram(I.L.D)

It is a curve the ordinate to which at point equal to the value of some particular function due to a unit load acting at that point. The function may be support reaction, shear force or bending moment.

Note:

I.L.D is always drawn for a concentrated unit load and it is independent of any type or system of loading.

Use of I.L.D in Civil Engineering:

1. To determine the structural quantity( Like reaction, shear force, bending moment) for a given system of load on the span of the structure.
2. To determine the position of like load system to have the maximum value of structural quantity.

Advantage of I.L.D:

1. It is a useful tool for dealing with moving loads.
2. It is a method of speedy determination of the value of structural quantity at any section under any complex system of loading.

Difference between I.L.D and B.M.D or S.F.D.

The ordinate of B.M.D or S.F.D gives the value of B.M or S.F. at the section where the ordinate has been drawn whereas the ordinate of I.L.D at any point gives the value of B.M. of S.F at the given section for which the I.L.D has been drawn and not at the point at which the ordinate has been drawn.

I.L.D for beams:

1. Simple supported beam

Consider a simply supported beam as below. Let unit load rolls from left to right.

Calculation of relations

I.L.D for R_A

I.L.D for R_B

I.LD for shear force (S.F) at C

Case I : When unit load is in portion AC i.e. (0 ≤ z ≤ x)

Considering right of C ( as the number of loads are less in right portion)

SF at C, V_c = -R_B = – z/L

Note: When rigid portion is considered upward force is taken negative.

Case II : When unit loads is in portion CB i.e. ( x ≤ z ≤ 1)

Considering left C.

V_C=R_A= 1-z/L

Note: Same result will be obtained by considering right of C.

• I.L.D for bending moment ( B.M.) at C

Case I : When unit load is in portion AC i.e.(0 ≤ z ≤ x)

Considering right of C.

Case II : When unit load is in portion CB i.e ( x ≤ z ≤ L)

Considering left of C.

2. Cantilever beam

Consider a cantilever beam as shown below. Let unit load rolls from left to right.

M_C = x- L

I.L.D is shown in the figure B4

3. Overhanging beam

Consider an overhanging beam as shown below. The unit load rolls from left to right.

Note: The I.L.D for overhang portion can also be determined by extending the I.L.D of simply supported beam up to the overhangs and their ordinates is determined by using similar triangle property.

I.L.D for Truss

The I.L.D for truss members can be easily drawn using method of sections. We know from method of sections that the external forces acting on one side of section AA( either left side of right of section of the truss shown below) and the internal forces in the members U₂U₃ , U₂L₃ and L₂L₃ cut by the sections are in equilibrium.

Applying principle of statics

For member L₂L₃

The moment of external forces about U₂ acting either left or right side of section must be equal and opposite to the moment of the force in the bottom chord member L₂L₃ about U₂.

(+↻)ΣMoment  about U₂ = 0

Let the moment of the external forces abour U₂ be M_U2

∴ Force in L₂L₃ = M_U2 /h

For member U₂U₃

Similar, force in member U₂U₃ can be obtained by taking moment about L₃ and applying:

(+↻)ΣMoment  about L₃ = 0

For member U₂L₃

Applying the sum of vertical component of all external forces on one side of section AA must be equal and opposite to the vertical component of the internal force in the member U₂U₃.

(+↑)Σ F_y = 0

Let the sum of vertical component of all external force is F_V.

Force is U₂L₃ = F_y/ sinθ

• The moving loads are never carried directly on the main girder but are transmitted across cross girders to the joints of bottom chord.
• Bridge Truss Floor System
• A typical bridge floor system is shown in Figure. As shown in Figure, the loading on bridge deck is transferred to stringers. These stringers transfer the load to floor beams and then to the joints along the bottom chord of the truss.
• It should be noted that for any load position; the truss is always loaded at the joint.

Types of Trusses

The main differences are that the Pratt truss has no force on the ends and the compression members are vertical. The Howe truss has no force on the center and the tension members are vertical instead.

Note :

• All top member compressive force and hence the I.L.D is negative.
• All the bottom experience tensile force and the I.L.D is positive.
• Vertical or inclined members experince tensile, compressive and both type of forces depending on the location with respect to the applied load on the truss.

I.L.D for Girders

Girder

Girder is a support beam which supports smaller beams. Girder is mainly used in bridges. In girder roof slab transmits the load to the cross beams and through cross beams to the girders. Hence, the girders are subjected to concentrated loads transmitted by cross beams.

The point at which the girder supports cross beam are referred as panel points.

1,2,3 and 4 are panel points.

The end reactions are not affected by the presence of cross beam. Hence, the reactions are same as that of simply supported beam.

The I.L.D for SF at any point within the panel is the same. Hence, I.L.D for SF is investigated with reference to panel but not any section.

Beam subjected to U.D.L longer than span

• Maximum negative shear force will be develop when head of udl reached considered section and absolute maximum negative shear force will developed when considered section is at B and the udl covers entire span.
• Maximum positive shear force will be develop when tail of udl reached considered section and absolute maximum positive shear force will developed when considered section is at A and the udl covers entire span.
• Maximum moment will be developed when the udl covers entire span and absolute maximum bending moment will developed when considered section is at mid span and load covers entire section.

Example:

1. Draw Ild for shear force and bending moment at 3m from right support when 6m udl of intensity 4kN/m crosses over 5m simply supported bridge from left to right. Calculate maximum negative, maximum positive shear force, maximum moment and also calculate absolute maximum positive, negative shear force and bending moment.

Maximum negative shear force will be develop when head of udl reached considered
section,

Maximum negative Shear Force = Intensity of load * Area under load in ILD
= 4(1/2)2*(2/5)
= 1.6 kN

Maximum positive shear force will be develop when tail of udl reached considered
section ,

Maximum positive Shear Force = Intensity of load * Area under load in ILD
= 4(1/2)3*(3/5)
= 3.6 kN

Maximum moment will be developed when the udl covers entire span

Maximum bending moment= Intensity of load * Area under load in ILD
= 4(1/2)5*(6/5)
= 12 kN m
Absolute maximum negative shear force will developed when considered section
is at B and the udl covers entire span.

Absolute Maximum negative shear force= Intensity of load * Area under load in ILD
= 4(1/2)5*1)
= 10 kN
Absolute maximum positive shear force will developed when considered section
is at A and the udl covers entire span.

Absolute Maximum positive shear force= Intensity of load * Area under load in ILD
= 4(1/2)5*1)
= 10 kN

Absolute maximum bending moment will developed when considered
section is at mid span and load covers entire section.

Absolute Maximum moment= Intensity of load * Area under load in ILD
= 4(1/2)5*6.25/5)
= 12.5kN m

Beam subjected to U.D.L shorter than span

• Maximum negative shear force will be develop when head of udl reached considered section and absolute maximum negative shear force will developed when considered section is at B and the head of udl reached at B.
• Maximum positive shear force will be develop when tail of udl reached considered section and absolute maximum positive shear force will developed when considered section is at A and tail of udl reached at A .
• Maximum moment will be developed when the udl is so placed that the section divides the load in the same ratio as it divides the span i.e x/d=a/l and absolute maximum bending moment will developed when considered section is at mid span such that x/d=a/l is same. Where x= distance between considered section to tail of load,d=length of udl, a=distance from left support to considered section, l= span length

Example:

1. A simply supported beam has span of 15 m, udl of 40 kN/m and 5 m long crosses the girder from left to right. Draw ILD for shear force and bending moment at 6m from left support and calculate maximum shear force and bending at that section.

Maximum negative shear force will be develop when head of udl reached considered
section ,

Maximum negative Shear Force = Intensity of load * Area under load in ILD
= 40(1/2)5*((1/15) +(6/15))
= 46.67 kN

Maximum positive shear force will be develop when tail of udl reached considered
section ,

Maximum positive Shear Force = Intensity of load * Area under load in ILD
= 40(1/2)5*((9/15)+(4/15)
= 86.67 kN

Maximum moment will be developed when the tail of udl lies at x distance from
considered section towards left support. So x= 5*6/15 =2m

Maximum bending moment= Intensity of load * Area under load in ILD
=( 40(1/2)2((36/15)+(54/15)))+ ( 40(1/2)3((36/15)+(54/15)))
= 600kN m

I.L.D due to series of concentrated loads

• Rolling loads are those loads which roll over the given structural element from one end to the another.
• For maximum negative shear force, most of the load lies in left portion of considered section so for that several hit and trial are made.
• Trial 1:Leading load w1 lies at considered section. Then,SF₁ =w₁*(a/L)+w₂ *Y₂ +w₃ *Y₃
• Trial 2 : W2 lies at considered section. Then SF₂ =- w₁*y₁+w₂*(a/l)+w₃*y₃+w₄*y₄

Check, If SF₁>SF₂,then maximum –ve shear force is SF₁ otherwise next trial is made by placing W3 over considered section. In this way we made hit and trial inorder to get maximum –ve SF. For maximum positive shear force, most of the load lies in left portion of considered section so for that several hit and trial are made.

Trial 1:Following load w4 lies at considered section. Then,SF₁ =w₄*(b/L)+w₃ *Y₃ +w₂ *Y₂ + w₁ *Y₁

Trial 2 : W₃ lies at considered section. Then SF₂ =- w₄*y₄+w₃*(b/l)+w₂*y₂+w₁*y₁

 Check, If SF₁>SF₂ ,then maximum +ve shear force is SF₁ otherwise next trial is made by placing W₃ over considered section. In this way we made hit and trial inorder to get maximum –ve SF.

For maximum positive shear force, most of the load lies in left portion of considered
section so for that several hit and trial are made.
Trial 1:Following load w4
lies at considered section. Then,SF₁ = w₄ *(b/L)+ w₃ * y₃ + w₂ * y₂ +
w₁ * y₁

Trial 2 : w₃ lies at considered section. Then SF₂
=- w₄ y₄ + w₃ (b/l)+w2y2+ w₁ * y₁

Check, If SF₁ > SF₂ ,then maximum +ve shear force is SF₁ otherwise next trial is made by placingW3over considered section. In this way we made hit and trial in order to get maximum –ve SF.

Maximum bending moment at a given section

• Maximum bending moment will be developed at a considered section when tilting load lies over that section.
• Tilting load: The load which changes the sign of avg. load on left side minus avg.load on right side when the point load passes over the section.

Tilting load can be determined as follows:

Here,w₃ changes the sign so tilting load is w₃ .Then tilting load is placed over considered section in order to get maximum moment.

Absolute maximum bending moment

• When a train of several loads crosses a simply supported beam, the absolute maximum bending moment under any given wheel occurs when this wheel load and the c.g of total resultant load are equidistance from the centre of beam.
• Steps to find absolute maximum bending moment
  • Calculate total resultant load by using,W= w₁+w₂+w₃+w₄
  • Calculate the position of resultant load by taking moment about the position of following load, i.e W*x͞ = w₄*0+w₃*x₃+w₂*(x₂+x₃)+w₁*(x₁+x₂+x₃) Here, a=b=L/2 – Assume w₃ is heavier and closer with resultant load W, so place w3 and resultant load W at equidistance from centre of beam.
  • Calculate Bending moment by using concept of ILD uses.
  • If w₃ is heavier but farther with resultant load W, so we made hit and trial by placing w₃ and resultant load W at equidistance from centre of beam for first case, and place w₂ and resultant load W at equidistance from centre of beam for second case then compare two cases, and adopt the values which is greater.

Strain energy and Complementary Strain Energy

• When an elastic member is deformed under the action of an external loading the member is said to have possessed or stored energy which is called strain energy.
• The strain energy stored by a member so deformed is equal to the amount of work done by the external forces to produce deformation.
• The area between the load-extension curve and the vertical axis is called complementary energy.
• For linearly elastic materials this area is equal to the area under the load extension curve i.e. strain energy.

Types of structure

The types of structure on the basic of materials used as follow:

• Steel structure
• Concrete structure
• Masonry structure
• Wooden structure
• Plastic structure

Structural Mechanics

Structure mechanics is the branch of mechanics that deals with forces and motion of structural system. It is the computation of deformations, deflections and of structural system. It is calculation of deformations, deflections and internal forces or stresses within structure for the estimation of existing structures.

Approaches of Structural Analysis:

Structural analysis can be classified are as follows:

1. Force method
2. Displacement method

1. Force method

In this method, the number of independent unknown forces ( both external and internal) is found and is compared with the number of independent equation of static equilibrium that can be written involving these unknown forces. If the number of unknown is equal to the number of independent equations, the unknown forces can be directly determined. But, if the number of unknown forces exceeds the number of equilibrium equations, a number of unknown forces in excess to the equilibrium equations are designated as redundant forces and are assumed to be equilibrium equations are designated as redundant forces and are assumed to be equilibrium equations to obtain a statically determinate residual structure called primary structure. For each redundant, one equation is written in which the displacement of point of application of redundant is expressed in terms of known forces and the unknown redundant forces. Finally, these equations are solved to get the unknown redundant forces.

2. Displacement method

In this method, the independent unknown displacement components involved in the structure are estimated and these are considered as the unknown to be found. The internal forces in the structure are then expressed in terms of these unknown displacements, using the stress displacement relations. For each unknown displacements, using the stress displacement relations. For each unknown displacement component, a corresponding equilibrium is written in terms of known external forces and unknown internal forces which are expresses in  terms of the displacement. These equations are solved to find the unknown displacement. Finally internal forces are calculated from the displacements.

Linear and Non-Linear Elastic Structure

If a system material has linear stress-strain relationships i.e. it obeys Hook’s law and undergo a small deflection of deformation is called linear elastic structure. For linear system, principle of superposition can be stationary applied.

Non-Linear system is the system is which is material does not have linear stress- strain relationships i.e. it does not obey Hook’s law and the deformation are so large that a change of geometry cannot be neglected in the analysis. Superposition principles does not hold good. If the non-linearity is due to stress-strain relationship, it is called material non-linearity and if the non linearity is due to considerable change in the geometry, it is called geometric non-linearity.

Computer Based Methods

• SAP( Structure Analysis Program)

This is a general purpose finite element program which is used for linear static and dynamic analysis of structures. Boundary condition can be assigned.

• STAAD Pro

It supports various steel, concrete and timber design codes. It can be used for static analysis geometric non-linear analysis or buckling analysis.

• NISA(Non-Linear Incremental Structural Analysis)

This software is used for non- linear analysis and buckling can be studied as well.

• EIABS

This software is mostly used for building (3-D analysis). Static earthquake loads can be considered in the analysis of structures.

Superposition Principles

It states that the displacement response resulting from number of force may be obtained by adding displacement resulting from individual force.

Where,

△_ij = Deflection at i due to load at j

△₁ = △₁₁ +△₁₂+△₁₃

△₂ = △₂₁ +△₂₂+△₂₃

△₃ = △₃₁ +△₃₂+△₃₃