Ch6- Influence line - Lecture notes 6 PDF

Preview

Summary

Ch6- Influence line.pdf...

Description

Chapter-6

Dr. Kazi M.A. Sohel

Influence Lines for Statically Determinate Structures Objectives: Influence lines have important application for the design of structures that resist large live loads. The objectives are (i) Understand the moving load effect in simpler term (ii) Study various definitions of influence line (iii) Introduce to simple procedures for construction of influence lines for statically determinate structure.

Topics 

Influence Lines for Statically Determinate Beams



Computing Forces from Influence Lines



Influence Lines for Trusses



Solved examples and tutorials Textbook:StructuralAnalysis7th Edition byR.C.Hibbeler

Can my bridge survive?

What is Influence Line ?

1 2 6 (1 6 8 )

8 5 (1 2 7 )

1 (4 3 )

Force

Influence Lines Shear force and moment diagrams Fixed loads

V M

x V and M at different locations of the beam x Influence Lines

Load moves along the beam

V M

x

x V and M at a fixed location

P C

A x

B

Px RB  L

L=10

RA

P

RB P

B

A B

A

8 5

RA =0.5P

P

Dr. Kazi M.A. Sohel

P ( L  x) RA  L

RA=0.2P L=10

L=10

RB=0.8P

RB=0.5P

0.5P RA

x

Plot for RA

0.5P P

RB

Plot for RB

6-1 Influence Lines

x

RA

RB

0

1.0P

0

2

0.8P

0.2P

4

0.6P

0.4P

5

0.5P

0.5P

6

0.4P

0.6P

8

0.2P

0.8P

10

0

1.0P

Dr. Kazi M.A. Sohel

 If a structure is subjected to a moving load, the variation of shear & bending moment is best described using the influence line

 Influence lines play an important part in the design of bridges, industrial cranes, conveyors and other structures where loads move across their span. So what is an influence line? (see previous slide) An influence line represents the variation of either the reaction, shear, moment or deflection at a specific point in a member as a concentrated force moves over the member.  Once the influence line is drawn, the location of the live load which will cause the greatest influence on the structure can be found very quickly.  Influence lines represent the effect of a moving load only at a specified point on a member, whereas shear (SFD) and bending moment diagrams (BMD) represent the effect of fixed loads at all points along the member.

Dr. Kazi M.A. Sohel



Procedure to construct the influence line (three methods) 1) Tabulate Values 2) Influence-Line equations 3) Virtual work: Muller-Breslau principle (qualitative influence line)

Construction of influence line by using equn & tabulated values Example 1: Construct the influence line for the vertical reaction at A of the beam in Fig.1(a) x 10 m

Ay

Fig 1 (a)

Tabulate Values: A unit load is placed on the beam at each selected point x and the value of Ay is calculated by summing moments about B.

1 By

10 m

(b) Text book example 6-1

Dr. Kazi M.A. Sohel

x Ay

1

10 m

By

(c) Influence-Line Equation: When the unit load is placed a variable distance x from A, the reaction Ay as a function of x can be determined from M B  0;  Ay (10)  (10  x)(1)  0 Ay  1 

1 x 10

This line is plotted in Fig. (c)

Dr. Kazi M.A. Sohel

EXAMPLE 2: Construct the influence line for the vertical reaction at B of the beam in Fig. 2a. x

1 5m

Ay

Fig. 2a

5m By

Solution Tabulate Values: A unit load is placed on the beam at each selected point x and the value of By is calculated by summing moments about B. A plot of the values yields the influence line in Fig. c. Text book example 6-2

Dr. Kazi M.A. Sohel

x

1 5m

5m By

(c) Influence-Line Equation. Applying the moment equation about A,

 M A  0;

By (5)  1( x)  0 1 By  x 5

This line is plotted in Fig. (c)

Dr. Kazi M.A. Sohel

EXAMPLE 3: Construct the influence line for the shear at point C of the beam in Fig. 3a. Text book example 6-3 Tabulate Values: The unit load must be placed just to the left x=2.5- and just to the right x=2.5+ of point C since the shear is discontinuous at C, Figs. 6–3b and 6–3c. A plot of the values in Fig. 6– 3d yields the influence line for the shear at C, Fig. 6–3e.

Fig. 3a

Solution

0.25

(b)

(c)

Dr. Kazi M.A. Sohel

A

C

B

(e ) Influence-Line Equation. Here two equations have to be determined since there are two segments for the influence line due to the discontinuity of shear at C,

Vc  

1 x 10

Vc  1 

1 x 10

These lines are plotted in Fig. (e)

Dr. Kazi M.A. Sohel

EXAMPLE 4: Construct the influence line for the shear at point C of the beam in Fig. a. Text book example 6-4 1 x B

C

A

4m

4m

4m

Solution x

Tabulate Values: Using statics and the method of sections, get VC for different position of unit load.

Equation

Dr. Kazi M.A. Sohel

EXAMPLE 5: Construct the influence line for the moment at point C of the beam in Fig. a. x 1 B

C

A 5m

5m

(a)

Solution Tabulate Values:

A

B

C

B

Free-body of BC Text book example 6-5

Dr. Kazi M.A. Sohel

Tabulated values

(c)

Influence-Line Equations:

1

x

M C  0

C

A

1 M C  1(5  x)  (1  x)5  0 10 1 M C  x for 0  x  5m 2

5m

M C  0

B 5m

1.25

1.25

1 M C  (1  x)5  0 10 1 M C  5  x for 5m  x  10 m 2

Dr. Kazi M.A. Sohel

EXAMPLE 6: Construct the influence line for the momentr at point C of the beam in Fig. a. x 1 B

C

A 4m

4m

4m

Solution

x

Tabulate Values: Using statics and the method of sections, get MC for different position of unit load.

Equation

Text book example 6-6

Dr. Kazi M.A. Sohel

6-3 Qualitative Influence Lines Muller-Breslau Principle states that the influence line for a function (reaction, shear, moment etc..) is to the same scale as the deflected shape of the beam when the beam is acted upon by the function. -

A quicker way of constructing influence lines for beams is to apply the Müller Breslau Principle, which is derived from the principle of virtual work.

For example to obtain the influence line for the support reaction at “A” for the beam shown, remove the support corresponding to the reaction and apply a force in the +ve direction that will cause a unit displacement in the direction of Ay Notice that the deflected shape is linear (i.e. the beam rotates as a rigid body without any curvature). This is true only for statically-determinate beams.

1

1

Dr. Kazi M.A. Sohel



If the shape of the influence line for shear at C is to be determined, Fig 6.13(a), the connection at C may be symbolized by a roller guide as shown in Fig 6.13(b). 

Applying a +ve shear force Vc to the beam at C & allowing the beam to deflect to the dashed position, the influence line shape as shown in Fig 6.13(c)

Fig. 6-13

Dr. Kazi M.A. Sohel



If the shape of influence line for the moment at C Fig 6.14(a) is to be determined, an internal hinge or pin is placed at C

Applying +ve moment Mc to the beam, the beam deflects to the dashed line which is the influence line as shown in Fig 6.14(c)



Fig. 6-14

Dr. Kazi M.A. Sohel

The proof of the Muller-Breslau Principal can be established using the principle of virtual work Principle of Virtual Work States that for an equilibrium system, the work done by all forces upon a set of virtual displacement is zero.  

Work = a linear disp.  force in the direction of disp. Or work = rotational disp.  moment if the direction of the disp. x

For the FBD shown, the only forces having a corresponding virtual displacement are Ay and the unit load, thus:

1  Ay    y/  .1  0;  Ay   y/ FBDs of a beam and a virtually-displaced beam

1

A

C

Ay

B By

x

1

A C

y=1

y/ Ay

B By

This indicates that the influence line of Ay is numerically equal to the virtual displacement of the beam, when the virtual displacement is constructed with a unit displacement at Ay and no displacement at any forces except the unit load.

Dr. Kazi M.A. Sohel

Müller Breslau Principle The step-by-step process of applying the Müller Breslau Principle can be summarized as follows: 1. Expose the quantity of interest by a cut (or remove a support). 2. Impose a virtual displacement such that: (a) at the cut there is a unit displacement (or rotation). (b) the quantity of interest produces a positive work. (c) no other internal force produces any work. 3. The resulting displacement shape is the desired influence line. Notes 1. The Müller Breslau Principle provides a quick method for establishing the shape of the influence line. The generated shape is referred to as a qualitative influence line. Once this is known, the ordinates at the peaks can be determined by using the basic method introduced earlier. 2.

By simply knowing the general shape of the influence line, it is possible to locate the live load on the beam and then determine the maximum value of the function using statics

Dr. Kazi M.A. Sohel

Example:ShearforceandMomentatC 



If the beam is sectioned at C, the beam undergoes a virtual disp y at this point, Fig 6.15(c), then only the internal shear at C and the unit load do work The virtual work eqn is:

Vc  y  1 y '  0 If  y  1, then  Vc   y '

Fig. 6-15

Assume a hinge or pin is introduced into the beam at point C, Fig 6.15(d). If a virtual rotation  is introduced at the pin,  virtual work will be done only by the internal moment & unit load M c  1 y '  0 

If   1, then  M c   y ' Fig. 6-15 This indicates that the deflected beam has the same shape as the influence line for the internal moment at point C

Dr. Kazi M.A. Sohel

EXAMPLE 9: For each beam in Fig 6.16, sketch the influence line for the vertical reaction at A Text book example 6-9 SOLUTION The support is replaced by a roller guide at A since it will resist, but not Ay. The force Ay is then applied.

Again, a roller guide is placed at A and the force Ay is applied.

A double-roller guide must be used at A in this case

Dr. Kazi M.A. Sohel

EXAMPLE 10: For each beam in Figs. 6–17a through 6–17c, sketch the influence line for the shear at B. Text book example 6-10 (a)

The roller guide is introduced at B and the positive shear is applied. Notice that the right segment of the beam will not deflect since the roller at A actually constrains the beam from moving vertically, either up or down. (b)

(c)

Dr. Kazi M.A. Sohel

EXAMPLE 11: For each beam in Figs. 6–18a through 6–18c, sketch the influence line for the Moment at B. Text book example 6-11 Solution: A hinge is introduced at B and positive moments MB are applied to the beam. The deflected shape and corresponding influence line are shown.

(a)

(b)

Fig. 6-18 (c)

Dr. Kazi M.A. Sohel

6.2 use of influence line for Beams 

Once the influence line for a function has been constructed, it will be possible to position live loads on the beam which will produce the maximum value of the function. Two types of loadings will be considered 1) Concentrated force 2) Uniform load



Concentrated force  For any concentrated force, F acting on the beam, the value of the function can be found by multiplying the ordinate of the influence line at position x by magnitude of F.



Uniform load  value of a function caused by a uniform load = the area under the influence line  intensity of the uniform load

Dr. Kazi M.A. Sohel

Example 7: Determine the max +ve live shear that can be developed at point C in the beam shown in Fig 6.10(a) due to: A concentrated moving load of 4kN and a uniform moving load of 2kN/m. Solution: The influence line for shear at C has been established as shown in Fig 6.10(b)  Concentrated force  The max +ve positive shear at C will occur when the 4kN force is located at x = 2.5 m  The ordinate at this peak is +0.75, hence: V  0.75(4)  3 kN C Text book example 6-7

Dr. Kazi M.A. Sohel

Uniform load  The uniform moving load creates the max +ve influence for VC when the loads acts on the beam between x = 2.5 m and x = 10 m  The magnitude of VC due to this loading is:



1  VC   (10m  2.5m)(0.75) (2kN/m) 2   5.625kN m 

Total max shear at C due to concentrated load and uniform load :

(VC )max  3 kN  5.625 kN  8.625 kN

m

Dr. Kazi M.A. Sohel

EXAMPLE 8: The frame structure shown in Fig. 6–11a is used to support a hoist for transferring loads for storage at points underneath it. It is anticipated that the load on the dolly is 3 kN and the beam CB has a mass of 24 kg/m. Assume the dolly has negligible size and can travel the entire length of the beam. Also, assume A is a pin and B is a roller. Determine the maximum vertical support reactions at A and B and the maximum moment in the beam at D. Text book example 6-8

Solution:

Fig 6-11a

We first draw the influence lines for, reaction Ay, reaction By and moment at D. 

Max reaction at A: o The max value for Ay occurs when the dolly is at C o The dead load must be placed over the entire length:

1  ( Ay ) max  3000(1.33)  24(9.81)  (4)(1.33)  2   4.63 kN 

Max reaction at B:  Here the dolly must be at B:

1  ( By )max  3000(1)  24(9.81)  (3)(1) 2  1  24(9.81)  (1)( 0.333)   3.31kN 2  

Max moment at D: o Here the dolly must be at D:

 1 ( M D ) max  3000(0.75)  24(9.81) (1)(0.5)  2  1  24(9.81)  (3)(0.75)   2.46 kN  2

(b)

(c)

Dr. Kazi M.A. Sohel

EXAMPLE 12: Determine the max +ve moment that can be developed at point D in the beam shown in Fig 6.19(a) due to a concentrated moving load of 16kN, a uniform moving load of 3kN/m & a beam weight of 2kN/m Solution 2m

A hinge is placed at D & +ve moments MD are applied to the beam

The deflected shape & corresponding influence line are shown in Fig 6.19(b)

2m

2m

4m

(a)

x

Fig 6-19 (b)

Text book example 6-12

Dr. Kazi M.A. Sohel

 





The concentrated moving load of 16kN creates a max +ve moment at D when it is placed at D The uniform moving load of 3kN/m must extend from C to E in order to cover the region where the area of the influence line is +ve. Finally, the uniform weight of 2 kN/m acts over the entire length of the beam The loading is shown on the beam in Fig 6.19(c)

Fig. 16-9 (c)

Method 1: The maximum +ve moment at D can be determined using numerical values for the influence line as in Section 6.1. By inspection of Fig 6.19(b), only the peak value h at D must be computed. This requires placing a unit load on the beam at D in Fig 6.19(a) & then solving for internal moment in the beam at D. Show that the value obtained is h = 1.333

x Fig. 16-9 (b)

Dr. Kazi M.A. Sohel

By proportional triangles,

h '/ (4  2)  1.333 / (6  4) or h '  1.333 Hence, with the loading on the beam as in Fig. 6–19c, using the areas and peak values of the influence line, Fig. 6–19b, we have 1  1  M D  5  (10  4)(1.333)   16(1.333)  2  (4)(1.333) 2  2  M D  36 kNm

Method 2 In Fig 6.19(d), the reactions on BE have been computed  Sectioning the beam at D & using the segment DE, Fig 6.19(e), we have:



(d)

 M D  0;  M D  20(2) 19(4)  0  M D  36kNm

(e)

Dr. Kazi M.A. Sohel

6-4 Influence Lines for Floor Girders* Floor systems are constructed as shown in Fig (a) Floor loads from slabs  floor beams  side girders  supporting columns Stringer

(b) Idealizes model

*6‐4willnotbetestedinexams

Dr. Kazi M.A. Sohel

EXAMPLE 13: Draw the influence line for the shear in panel CD of the floor girder in Fig. 6–21a. Text book example 6-13 Solution: The unit load is placed at each floor beam location & the shear in panel CD is calculated.

Fig. 6-21 (c)

3m

3m 3m Fig. 6-21 (a)

3m

Fig. 6-21 (d)

Dr. Kazi M.A. Sohel

The unit load is placed at each floor beam location & the shear in panel CD is calculated  A table of the results is shown in Fig 6.21(b)  The details for the calculations when x = 0 and x = 6 m are given in Fig 6.21(c) and 6.21(d)  Finally a segment of the girder is considered & the internal panel shear VCD is calculated  The tabular values are plotted and the influence line is as shown in Fig 6.21(e) 

Fig. 6-21 (b)

Dr. Kazi M.A. Sohel

6.5 Influence Lines for Trusses In Fig 6.23, the loading on the bridge deck is transmitted to stringers which in turn transmit the loading to floor beams and then to joints along the bottom cord

Dr. Kazi M.A. Sohel

EXAMPLE 14: Draw the influence line for the force in member GB of the bridge truss shown in Fig. 6–24a. Text book example 6-15

Solution Tabulate Values: Each successive joint at the bottom cord is loaded with a unit load and the force in member GB is calculated using the method of sections, Fig 6.24(b). For example, placing the unit load at x = 6 m (joint B), the support reaction at E is calculated first. Then calculate force in GB.

(b)

  

Dr. Kazi M.A. Sohel

The values are tabulated as shown in Fig. 6.24 (b) Plotting the tabular data and connecting the points yields the influence line for member GB, Fig 6.24(d) Since the influence line extends over the entire span of truss, member GB is referred to as primary member