MANUAL EAA203 MAKMAL STRUKTUR RINGAN.doc

EAA 203 – LIGHT STRUCTURE

S1 – CURVED BAR TEST

SCHOOL OF CIVIL ENGINEERINGUNIVERSITI SAINS MALAYSIA

ENGINEERING CAMPUS

Figure 1

OBJECTIVES OF THE EXPERIMENTS

The objectives of the experiments is to measure the deflections of a set of curved bars, of known geometry. Then to compare the empirical data with values calculated from theory.

1

EAA 203 – LIGHT STRUCTURETHE APPARATUS

The apparatus is illustrated in Figure 1. It consist of a set of four specimens

arranged upon a bench mounted baseboard. The specimens consists of a quadrant, a davit

and a semi-circle. These may be loaded in a turn by dead weights and the horizontal and

vertical deflections measured by dial indicators. The apparatus is protected by a clear

guard. Access to the specimens is via the ends of the apparatus around the guard fascia.

THEORY

Stress and deflection problems in circular ring and curved bars of narrow rectangular cross section are best described by polar co-ordinates. The position of a point in the middle of a plane is defined by the distance from the origin O and by angle between r and a certain axis Ox fixed in the plane.

2


Bending of Curved Bars by a Force at the End

A bar with a narrow cross section and a circular axis is constrained at the lower end and bent by a force F applied at the upper end in a radial direction. The bending moment at any cross section is proportional to sin θ, and the normal stress σθ, according to elementary theory on the bending of curved bars, is proportional to the bending moment. Assuming that this also holds for the exact solution we find that the stress function Ø should be proportional to sin θ, taking:-

(a)

Ø = ƒ(r) sin θ (b)

we find that ƒ(r) must satisfy the following ordinary differential equation

(c)

After transforming into a linear equation with constant co-efficient the general solution is :-

f(r) = A r3 + B + C r + D r log r (d)

Where A, B, C, D are constants of integration and must be determined from the boundary conditions. This solution may be substituted into (b) for the stress function and using the general formula (a) we find the following expressions for the stress components:-

(*)

The inner and outer boundaries are free from external forces so for r = a and r

= b or from equations (*) :-

3


(e)

The last boundry condition is that the shearing forces on the upper and of the bar should equal the applied force F. Take the width as unity or F as the load per unit width we obtain :-

=

or (f)

from equations (e) and (f) we get :-

(g)

in which

Substituting the values (g) of the constants into equation (*) we can get expressions for the stress components.

(**)

For the upper end of the bar where , we get :-

4


(h)

for the lower end

(k)

Displacements of the Bar

Now consider the displacements due to the applied force F. The stress components (*) can be written as:-

(l)

By integrating the first equation we obtain :-

(m)

is a function of only.

Substitute (m) into the second equation (l) with the expression for and integrate.

(n)

In which F(r) is a function of r only. Substituting now into the third equation (l) we obtain :-

5


which is satisfied by putting :-

F(r) = Hr (p)

in which H,K and L are arbitrary constants, determined from the conditions of constraint. The components of displacement from (m) and (n) are then :-

(q)

The radial deflection of the upper end of the bar is obtained by putting in the expression for u.

(r)

The constant L is obtained from the condition for the built in end. For

we have , hence from the second of equations

(q)

(s)

Using (g) the deflection of the upper end is therefore

(15)

When b approaches a and the depth of the bar h=b-a is small in comparison to a we can use :-

6


Substituting this in the equation of deflection of the upper (**) which is

(**)

and neglecting terms of higher order we obtain:-

which co-incides with the elementary formula for this case.

7

EAA 203 – LIGHT STRUCTUREStandard Deflection Equations for the Specimens

General :- E for steel = 207 GN/m2

where b = beam width and d= beam depth

Semi-Circle

R = 100mm d = 3.175mm b = 12.7mm MAX. LOAD 10N

Davit

R = 100mm L = 100mm d = 3.175mm b = 12.7mm MAX. LOAD 10N

( + L) + (r +L)

Quadrant

R = 100mm d = 3.175mm b = 12.7mm MAX. LOAD 15N

=

r r r

a

(a) (b) (c)Semi circle Davit Quadrant

PROCEDURE

8


The experimental procedure is essentially the same for all the specimens. Place the stirrup onto the bar end; the ring stirrup is captive. Adjust the dial indicators so that they bear onto the special round end of the specimen to be tested. The dial gauge is mounted on an x-y-z system capable of wide range of adjustment. The dial gauges for the ring are fixed in place. Ensure that the ring rests against the stop on the left of the bracket. The vertical ring dial gauge should have the load stirrup between its anvil and the ring specimen.

Now place the load hanger onto the stirrup. Tap it lightly with a fore finger and then note the dial indicator readings. Then add a load weight to the hanger. Again tap lightly and note the readings. Keep adding weights up to the maximum recommended load for each specimen as given in ‘Standard Deflection Equations for the Specimens’.

The horizontal measurement on a semicircle will require extra care. This is the weakest specimen and thus has the greatest deflections for the lowest forces. The return spring within the dial gauge is strong enough to limit the specimen deflection and thus give false readings. Gently use a finger or small screwdriver to ‘lift’ the dial gauge spindle away from the specimen until it only just touches the specimen end. Then note the reading. It may require two people, one to move the spindle and one more to take the reading. If this technique is not used, errors of 20% may be incurred.

It is good experimental practice to calculate the expected values from theory before performing the experiments. This method not only gives you an idea of what to expect but also enables a cross check to be undertaken during the experiment. Thus any potential doubtful points can be checked during experiments.

9

EAA 203 – LIGHT STRUCTURERESULTS

Tabulate your results as shown below and add the theoretical prediction by using

the ‘Standard Deflection Equations for the Specimens’.

Specimen Theoretical ValuesLoad Dial Gauge Reading

H V

Then plot a graph of deflection against load clearly distinguishing between theory and experiment.

CONCLUSIONS

Comment on the agreement between theory and experiment. Account for any discrepancies and suggest possible remedies.

10

EAA 203 – LIGHT STRUCTUREAppendix 1

Examples of Experimental Results

Figure 1: Semicircle

Figure 2: Davit

11


Figure 3: Quadrant

12


S1 – CURVED BAR TEST FORM

S1 – DATA SHEETS

Name: ____________________________________________________Matrix No. : _____________________________________________________Experiment Date : _____________________________________________________

Data Sheet 1

(a) Semi-Circle Curved Bar

Reading 1 Reading 2 Reading 3 AverageWidth (mm)Thick (mm)Radius (mm)Moment of inertia (mm4)

Load(N)

Vertical Deflection, (mm)

Horizontal Deflection, (mm)

0246810

Load(N)



0246810

13


Data Sheet 2

(b) Davit


Load(N)



0246810

Load(N)



0246810

14

EAA 203 – LIGHT STRUCTUREData Sheet 3

(c) Quadrant


Load(N)



0246810

Load(N)



0246810

15


S2 – CONTINUOUS BEAM TEST


ENGINEERING CAMPUS

Figure 1


The objectives of this experiment are to find the support reactions and central moment for a simply supported continuous beam and to compare them with values predicted from theory.

16


THE APPARATUS

The apparatus is illustrated in Figure 1. It consists of a frame to which three support brackets are attached. These may be set at any point along the frame. The bracket position is shown by a pointer which reads directly onto the scale on the top of the frame. Each bracket has a spring balance with a fine screw adjuster and a stirrup to hold the test beam. Additional stirrups are provided to attach hangers for a point load. Alternatively the beam may be loaded with the saddle weights to provide a uniformly distributed load. The number and spacing of the saddle weights will determine the loading per meter. An additional bracket is provided which is used to accurately determine the position of the applied loads.

THEORY

A beam which rest on more than two supports is said to be continuous and an example is illustrated below.

17

Figure 2

Figure 3

Figure 4


Changes of curvature occur in each span due to negative bending moments at each support. In the example above, the supports are taken to be at different levels from a horizontal datum AB. The loading on the beam is such that the free and fixing moments diagrams are shown in Figure 2 and with the resultant shown in Figure 4. The are of the resultant diagram on the span and A1 and its centroid is 1 from vertical line through the left –hand support. The area on the diagram for span and A2 with the centroid 2

from a vertical line through the right hand support.

If we know a tangent to the beam at the central support. Let α be the angle between the tangent and the horizontal, then:-

z0 =

= v1 – v0 + 1

-z2 = = - 2 + (v1 – v2)

and - = - + (1)

so, + = + EI (2)

when all the supports are level. v0 = v1 = v2 and so :-

+ = 0 (3)

If the areas of the free bending moment diagrams are S1 and S2 and the centroid of S1 is at the distance x1 from vertical through the left hand support, and similarly for S2 where the distance is x2.Then from equation (3) above :-

S1x1 –M0 1 - 1

(4)

18


+ - - ( 1 + 2) - = 0 (5)

Thus:-

(6)

Clapeyron’s Theorem

The above is Clapeyron’s theorem of three moments to which some simplifications are usually made to in corporate physical phenomena of the beam to aid solution.

If the beam is simply supported then, M0 = M2 = 0 and the equation (6) reduces to:-

M1 = + (7)

Consider the following example in Figure 5 :-

w1 w2

A B C

Figure 5

We have a beam on three supports where MA and MC =0. The bending moment diagram is shown in Figure 6

19


A B MB C

Figure 6

If we take moments about B for the left hand portion of the beam, in order to determine the fixing moment at B we find:-

M(B) + = - Ra (8)

The maximum bending moment for a point load is . So from the bending moment

diagram we get:-

S1 = x S2 = x

= = (9)

The above data enables us to evaluate equation (7) to give us a value for MB from theory whilst equation (8) yields a value from our experimental data.

Now consider equation (8) which may be written as:-

Ra 1 = - MB (10)

Ra = - (11)

If we take moments about B for the right hand part of the beam in a similar manner, we find that:-

RC = -

(12)

Now resolve vertically to determine the reaction at B:

20

EAA 203 – LIGHT STRUCTURERa + Rb + Rc = w1 + w2

Rb = w1 + w2 - Ra - Rc (13)

From equation (11) and (12), we obtain:-

Rb = w1 + w2 - w1/2 + MB/ - w2/2 + MB/ (14)Thus:

Rb = 0.5 ( w1 + w2) + MB (15)By using equations 1,12 & 15, we can compare calculated reactions with those found from experiments.

PROCEDURE

Set the two outer brackets to a convenient position, say 0 and 1200 mm. Next set the middle bracket to 500mm. Ensure that both top and bottom clamp bars are engaged with the frame. Tighten the clamp levers sufficient to hold the bracket in position- do not over tighten.

Place some stirrups on the beam and then place a stirrup over each spring balance hook. Move the stirrup along the beam as necessary so that the spring balance hangs vertically.

A wide variety of loads can be applied. First of all, try w1 = 20N at 250mm and load w2 = 40N at 850mm. Use the position bracket to set the load in the correct position. Now remove the loads and their hangers, but leave the stirrup in position.

Set the beam level by sighting along the top surface of the beam next to each stirrup and adjusting the spring adjuster until the same reading is visible on the scale on the bracket behind the spring balances.

21

EAA 203 – LIGHT STRUCTURETake initial reading of each spring balance and note this value. Now apply the

loads to the beam. If you think the stirrup may have moved, check its position with the position bracket.

After loading again, adjust the screw adjusters to set the beam level. It does NOT have to be at the same scale reading as in the unloaded condition. All that is necessary is that all three suspension points are at the same height.

Now read the spring balances and note your observations.

RESULTS

Tabulate your results as shown below.

Spring balance Initial Reading Final Reading Actual Reading123

Note the actual reading is the final – initial readings.

Draw a diagram of the beam with the positon of the supports and the values of the loads applied.

From the diagram (Figure 7) we see that, in this example

w1 =20N w2 =40N

Thus, by evaluating equations (9) we get

22

Figure 7


S1= 0.625 S2=2.45

Which can now be substituted into equation (7) to obtain a theoritical value of the moment at the central support.

= 3.84Nm

Our experimental observations yield:-

Ra= 2N Rb= 43N Rc= 14.5N

If we use equation (8) for the right hand half of the beam we obtain :-

= 3.85Nm

This compares very well with that predicted from theory. Now to compare the reactions. Equation (11) gives :-

Ra=

= 2.32N

Equation (12) gives:-

Rc=

= 14.5N

To obtain Rb we may use either equations (13) or (15). (13) yields:-

Rb = 20 + 40 - 2.32 - 14.5= 43.18N

23


Conclusions

Comment on the agreement between theory and experiment. Account for any discrepencies and suggest possible remidies.

S2 – CONTINUOUS BEAM TEST FORM


ENGINEERING CAMPUS

S2 – DATA SHEETSName : _____________________________________________________Matrix No. : _____________________________________________________Experiment Date : _____________________________________________________

RESULT

Data Sheet 1 (w1= 20N, w2= 40N)

Dial Gauge Intial Reading Final Reading Actual Reading

24


123

Actual Reading = Final Reading – Initial Reading

Reaction Theoritical ValueExperimental

Value% Difference

Ra

RB

Rc

MB



123



Value% Difference

Ra

RB

Rc

MB



25


123



Value% Difference

Ra

RB

Rc

MB

S3 – SAG AND TENSION TEST FOR CABLE


ENGINEERING CAMPUS

26


Figure 1


The objective of the experiments is to investigate a cable suspended between two points subjected to different conditions of load, tension, sag and support height.

THE APPARATUS

The apparatus is illustrated in Figure 1. It consist of two brackets, one being a

mirror image of the other (almost). The cable is attached at either end to spring balance

which fitted to an adjuster. The cable passes over pulleys to aid in setting up different

configurations. In addition, the right hand pillar can be moved along the base of the

apparatus. The cable is loaded by weights which clip onto the cable. The position of the

cable may be measured by using the horizontal scale on the apparatus and the try-square

provided.

27

EAA 203 – LIGHT STRUCTURETheory

Flexible cables are commonly used for a wide range of applications such as electricity transmission lines, cableways and suspension bridges.Consider a flexible cablE of constant section with a loading of w per unit length. The loading is uniformly distributed over the length of the cable on a horizontal base as shown in Figure 2.

Figure 2

The end of the cables, A and B are at different heights above the lowest point,O which is taken as the origin coordinate. If we section the cable at the origin and draw a free body diagram for the right hand portion, we find that equilibrium is satisfied by the triangle of forces TB, T0 dan WX1. (refer to Figure 3).

28


Figure 3

The position of the lowest point O, and also x1 is not known. To determine x1 we take moments about B:

T0y2 – wx1 = 0 [1]

Which may be written as:-

y = [2]

where x and y are the coordinate of any point on the cable relative to 0. Equation [2] is the equation of a parabola which is thus the shape taken up by the cable.

If we consider the case where A and B are at the same height then y1 = y2 and due

to symmetry the lowest point O is in the middle of the span. Thus, x1= . Equation [2]

can be written as:

y = [3]

where y is also known as the sag. Equation (3) may be simply rearranged to determine the horizontal tension, T0.

[4]

29


The length of the wire can be determined by considering an element of the curve of length δs, then:-

ds = dx

from equation [3] and [4] we obtain:

so, ds = dx 1 +

so, ds dx 1 +

If we intergrate from x=0 to x= and multiply by 2, we obtain:-

s 1 1 +

[5]

If we again consider the free body diagram and resolve horizontally, we obtain:-

-T0 + TB cos [6]

From which the tensile force in the cable at any point is given by:

T = [7]

As we are often not given To we may approach the same question, that of the maximum

and minimum tensions, by resolving vertically.

TB sin - wx1 = 0 [8]

however

Hence,

30


[9]

The value of TA may be obtained by substituting the appropriate x and y values into equation [9]. The minimum value of tension at 0 may be obtained by subtituting y2 and x1 into equation [2].

If we require a certain stress level the tension is divided by the cross sectional area, a of the cable as we assume that the stress is uniformly distributed across the section. Thus:-

[10]

We may know consider an alternative case to the one above. A cable is

freely suspended between two points A and B. It is subjected to a uniform loading of w

per unit lenght ‘along the wire’. This may be due to self weight or to some extra external

load such ice or snow. The result of such loading is to cause tension in wire itself. The

cable cannot sustain any bending moment thus at any given point the axis of the wire

must co-incide with the direction of the tension force within the wire. Consider the free

body diagram as shown in Figure 4.

Figure 4

For horizontal equilibrium, we obtain:-

T cos Ø = T0 [11]

31


Vertical Equilibrium:-

T sin Ø = ws [12]

If we eliminate the slope of the wire at C , we obtain:-

[13]

or T2 = To2 + w2s2

which is the equilibrium equation for the wire.

The shape of the wire may be obtained by eliminating T from equations [11] and [12]:

S = T0 tan Ø [14]w

The above equation, (14), describes the catenary

The maximum tension again occurs at the supports A and B. If the wire is m long, then

s = and T = Tmax at s = .

Tmax = [15]

PROCEDURE

The end bracket spacing, cable sag and suspension point heights will be given in the experiments. The cable is loaded by the special weights which have a small hook in one end. Hook the weight over the cable from the front. Twist the weight anti –clockwise and again pass the open end of the hook over the cable as shown in Figure 5. This procedure will effectively get a cable loop around the hook. On releasing the weight the hook will be at about 45o to the cable when viewed from the top. To move the weight on the table twist it so that the hook is that 90o to the cable. Hold the cable in one hand on the side to which the weight is to be moved. Gentle pressure will then slide the weight along the cable. Again release to lock in position. A little practice will show how rapidly a weight can be placed onto the cable and moved.

32


Figure 5

Experiment 1: Variation of Tension with Sag

This experiment is to investigate how the tension in the cable varies as the sag is adjusted.

Set the two top pulleys to the same height by using the combination square. Set the span of the cable to 1m. Slacken the toggle clamps move the scale bar so that the zero mark is where the cable leaves the pulley. Use the square to aid in positioning. Clamp the

left hand toggle clamp. As required move the right hand bracket by slackening the two knurled nuts and move the bracket along the base. Again use the square to position the bracket then tighten the knurled nuts and the toggle clamp. Now place some weights on the cable at a horizontal pitch of say 100mm. Use the square to assist and start at the left hand end. Some adjustment may be necessary when adjusting the sag and when adding the final weights.

Tabulate your results as below:-

Sag Tension from spring balance

x1 y2 TB

25 9.7 0.5 0.25 10.050 5.4 0.5 0.05 5.1100 2.19 0.5 0.1 2.7150 2.15 0.5 0.15 2.0

33

EAA 203 – LIGHT STRUCTURE200mm 1.7 0.5 0.2 1.6

With the weights at 100mm spacing , we have a loading of 2N/m, so w = 2. The span is fixed at 1m so x1 = 0.5TB is then calculated from equation (9).

Figure 6

The sag is easily adjusted by moving the bottom left-hand pulley. Further adjustment can be obtained by using the spring balance adjusters.

Experiment 2 - Shape of the Cable

Use the same cable loading as in previous experiment and this time measure the x, y coordinates of the cable using the scale and combination square.

Scale(Distance) Square

x1 y

y(from Eqn.2)

Square y

y(from Eqn.2)

0 200 -500 100 100 200 150 150100 162 -400 62 64 144 94 96200 135 -300 35 36 102 52 54300 115 -200 15 16 74 24 24400 103 -100 3 4 55 5 6500 100 0 0 0 50 0 0

34

EAA 203 – LIGHT STRUCTURE600 103 100 3 4 55 5 6700 115 200 15 16 74 24 24800 136 300 36 36 103 53 54900 161 400 61 64 145 95 961000 200 500 100 100 200 150 150

Note: To is first calculated for each experiment using the span, sag and loading, w. Use equation [4]. Equation [2] is then used to determine the theoretical values of the y coordinate for every x coordinate.

Theoretical and experimental points are co-incident on the scale of the graph.

Figure 7

Conclusions

1. How well do your experimental result agree with theoretical predictions?

2. Are the differences between the parabolic form and the catenary form significant?

35


S3 – SAG AND TENSION TEST FOR CABLE FORM


ENGINEERING CAMPUS

36

EAA 203 – LIGHT STRUCTURES3 - DATA SHEETS

Name : _____________________________________________________Matrix No.: _____________________________________________________Experiment Date : _____________________________________________________

Data Sheet 1 – Variation of Tension with sag

Cable Length, ℓ = 1mLeft Support Height, h1 =Right Support Height, h2 =Uniformly Distributed Load, w =

Sag(mm)

Tension(Dial gauge) x1

(mm)y2

(mm)TB

(Eqn. 9)Left Right Average

406080100120

Data Sheet 2 – Cable Shape

Set 1:-


x1 y

y(from Eqn.2)

0100200300400

37

EAA 203 – LIGHT STRUCTURE5006007008009001000

Set 2:-


x1 y

y(from Eqn.2)

01002003004005006007008009001000

S4 – EXPERIMENTS ON STRUTS WITH SIMPLE GEOMETRY


ENGINEERING CAMPUS

38



(i) Determination of flexural rigidity ( EI ) of the strut /column.

(ii) Demonstration of crippled shape.

(iii) Determination of load - deflection curves and crippling load for a strut with various end conditions.

APPARATUS

39


Figure 1Checklist

Code Item1 Guide Rods2 RH cross support3 RH cross member4 Centre cross member5 LH cross member6 Load cell and cross support7 Legs 4 off8 Stirrup12 Dial gauge15 50 gm hanger

THEORY

Introduction

A compression member whose length is considerably greater than the least radius of gyration of the cross-section is called a column or strut. The two words are now taken

40

EAA 203 – LIGHT STRUCTUREto have the same meaning but originally a column was taken to be vertical. The terms PILLAR and STANCHION are also used to describe vertical columns.

Critical Loads for Struts

The theory which follows is known as Euler’s theory and is based on the following assumption:-

i) the strut is initially perfectly straightii) the strut is applied axiallyiii) the length is large compared to the radius of gyrationiv) the assumptions made in the theory of bending apply.

Case 1: Critical Load for a strut with both ends pin jointed (hinged)

This case is also referred to as that for a strut ‘position fixed at each end’. One end , at least must be assumed free to move in the direction of application of the load. Neither end can move laterally but there is no rotational constraint at either end.

Assume that the strut takes up the deflected form shown in Figure 2. Let the deflection at a point C on the center line of the strut be y.

Figure 2

The bending moment at C is:

[1]

From the theory of bending :

41


[2]

By substituting equation[2] into equation [1] gives:

[3]

Whence,

[4]

where

This is a second order homogeneous differential equation for which the solution is:

y= C cos ax + D sin ax[5]

Differentiating [5], the slope is given by:

[6]

The slope is zero when x is zero therefore from equation [6], D must be zero and the equation reduces to:-

[6a]

And equation [5] reduces to:

[5a]

The maximum deflection occurs, y’ occurs at mid-span and therefore from equation [5a], and hence,

[7]

At the two ends of the strut, y is zero, therefore:-

42


[8]

For y to have a real value, must be zero. The only solution of

practical importance is that for which:-

Thus, [9]

Hence,

[10]

If the strut bends and just remains bent under the action of load P then P is the EULER CRIPPLING LOAD i.e:

[11]

The Euler formula can be applied only to ideal struts and one would expect the actual crippling load to be lower than P because a practical strut is never absolutely straight. Also, in practice there will be some eccentricity of the load.

The Euler formula takes no account of direct stress, a state of affairs which is acceptable in the case of long slender struts but definitely not acceptable in the case of short stocky struts (short and stocky is the exact opposite of long and slender) in which case it is the compressive stress that is of prime importance.

Case 2 : Critical Load for a strut with one end clamped and other end free

Referring to Figure 3 and assuming that P is the critical load i.e the load at which instability occurs, the moment induced at the fixed end will be:-

43


M= Pe (1)

At a distance x from the fixed end let the deflection of the strut be y then the bending moment at this point will be P(e-y) and from the theory of bending:

(2)

Whence

(3)

Figure 3

The solution of this second order differential equation is

y= e + C sin ax + D cos ax (4)

where a2 = and C & D are constants of integration

Now y = 0 when x = 0, therefore D = -e

Also when x=0

Whence,

y= e(1- cos ax) (5)

At the free and of the strut y=e and x = L which, substituted into equation (5) gives:-

-e cos aL =0 (6)

If e is non-zero then cos aL must be zero in which case:-

aL=

44

EAA 203 – LIGHT STRUCTUREwhich gives the least value of P to be:-

(7)

Where

Note that the crippling load given by equation (7) is exactly one quarter that given by

equation . It can be deduced therefore that equivalent length of a strut with one

end hinged and the other clamped is 2L (ie a strut with both ends hinged and length 2L has the same crippling load).

The mid-span deflection will be given by setting in equation (5):-

(8)

Case 3 : Critical Load for a strut with both ends clamped

Referring to Figure 4, the theory follows the same procedure as in Case 1 except that in the present case a bending moment is induced at each of end of the strut due to clamping.

45


At the point C, the bending moment is :-

M= –Py (1)

Hence, (2)

Let, and then equation (2) can be written as:-

(3)

Figure 4

Equation (3) is of exactly the same form as equation (3) (Case 1) and its solution is therefore :-

Y= C cos ax + D sin ax (4)

Hence,

y= C cos ax + D sin ax + (5)

The constants of integration C and D are found using the end conditions of zero slope and zero deflection and the condition of zero slope at mid span.

46

EAA 203 – LIGHT STRUCTUREDifferentiating (5) gives,

(6)

At x= 0 dy/dx =0

Thus, D=0 and equation (5) becomes :-

y= C cos ax + (7)

At x= 0 and y= y’

Hence, C= (8)

Substituting equation (8) and (7) gives the equation of the deflected centerline of strut :-

y= cos ax + (9)

Now, since the slope at the ends must be always be zero differentiating (9) and equating to zero gives:-

0= - sin (aL/2)

(10)

=0 implies that the maximum deflection is constant for all loads. ( This include

zero load and therefore the strut remains straight).

The alternative solution of equation (10) is that sin (aL/2) is zero in which case aL/2= π and the crippling load is:-

(11)

The crippling load given by equation (11) is the same as that for a pin-ended strut of length L/2.

4.0 PROCEDURE

A. Determination of Flexural Rigidity

The part numbers referred to are indicated in Figure 1.

47

EAA 203 – LIGHT STRUCTURE(i) Having selected the required strut, set the knife edge to the

correct position by moving the rear specimen beam to a position so that the strut can rest on the knife edges near to its ends.

(ii) Attach the dial gauge to the vertical pillar. Remove the ball end from the dial gauge stem and exchange it with the flat end which will be found screwed into the top of the stem.

(iii) Release the central cross member and slide it to a position so that the dial gauge stem is mid-way between the knife edges.

(iv) Rest the strut on the knife edges with equal overhang at each end and the dial gauge positioned on its centre-line.

(v) Open the latch of the stirrup, and thread it onto the strut. Close the latch and position the stirrup so that the foot of the dial gauge rests on the flat cut- out in the top of the latch.

(vi) Attach the weight hanger to the stirrup

(vii) Measure the span carefully and check that the dial gauge is at mid-span.

(viii) Adjust the dial gauge to read zero then attach loads to the hanger reading the dial gauge after each increment. Tap the top of the dial gauge stem gently before taking each reading.

(ix) Plot a graph of load against dial gauge reading. (Remember that each division on the dial gauge scale presents 0.1mm deflection).

(x) From the slope of a graph of loads against deflection, which should be linear passing through the origin, determine the value of EI and compare this with the value obtained assuming E= 2 x 1011 N/m2.

Typical Result

48

EAA 203 – LIGHT STRUCTUREStrut No. 6 width 19mm, thickness 3.2mm, length 550mm, span 450mm

RESULTS

LOAD (g)

100 200 300 400 500 600 700 800

GAUGE READING (div.)

2.5 4.8 7.0 9.1 11.3 13.4 15.5 17.6

The graph of load against dial gauge reading is shown in Figure 5.

Figure 5: Load against deflection for strut No. 6

49


Examples of Calculations

For a simply supported beam with appoint load at mid-span the deflection under the load is given by:

Therefore the flexural rigidity is given by:-

From the graph:-

B. Demonstration of Crippled Shape

(i) Ensure that the clamps of the specimen holders are tightened.

(ii) Select the required strut and adjust the rear specimen beam to the correct position and insert the dowels pins.

(iii) Turn the dial gauge so that the stem does not impede the insertion of the strut.

(iv) Examine carefully the strut carefully straighten if it necessary. Inserts the strut with its ends in the ‘vee’ grooves of the specimen holders. The edge of the strut will rest against the stops at the bottom of the holders. It may be necessary to unscrew the loading knob to reduce the load to zero after the strut is inserted.

(v) Lay the meter rule across the specimen holders in order to find the mid- span point on the strut. Turn the dial gauge so that its

50

EAA 203 – LIGHT STRUCTUREstem is perpendicular to the strut and the foot (ball end) is on the axis of the strut. Release the two locks on the central cross member and set it so that the dial gauge will measure the mid-span deflection of the strut.

(vi) Adjust the meter rule so that the mid-span reading is a convenient whole number (e.g 30cm). It might be convenient to ‘fix’ the rule with adhesive tape so that the original position is not accidentally ‘lost’.

(vii) Adjust the bezel of the dial gauge to indicate zero.

(viii) Slacken the locks on the central cross member and slide it carefully towards one end by 2cm increments. At each increment read the dial gauge.(Note: It is not possible to take measurements close to the ends of the strut).

(ix) Repeat (viii) sliding the cross member towards the other end. This establishes the original shape of the strut and allows for any miss- alignment of the specimen holders.

(x) Apply a load to a strut biasing the deflection away from the dial gauge. A suitable load is one which gives a central deflection of about 60 divisions (6mm).

(xi) Repeat steps (viii) & (ix) checking frequently that the load remains constant.

(xii) Repeat the experiment for the strut with the right hand end ‘pinned’ and the left hand end clamped. It is necessary to adjust the rear cross member in order to use the clamp.

(xiii) Repeat the experiment for the strut with both ends clamped. It is again necessary to adjust the rear cross member.

(xiv) Plot graphs and analyze the results

51


S4 – EXPERIMENTS ON STRUTS WITH SIMPLE GEOMETRY FORM


ENGINEERING CAMPUS

S4- DATA SHEETS

Data Sheet 1

Type of Specimen

Length(mm)

Thick(mm)

Width(mm)

1 2 3 Average

1 2 3 Average

1 2 3 Average

ALUMINIUM

STEEL

BRASS

Data Sheet 2

Dial gauge Reading (mm)Type of

SpecimenLoad, P(N)

100 200 300 400 500 600 700 800

ALUMINIUMSTEEL

52

EAA 203 – LIGHT STRUCTUREBRASS

Appendix

Figure 6: Shape of loaded strut. Test No. 1d= dial gauge reading

x= distance from centre

53




Appendix



54


S5 – THREE HINGED ARCH TEST


ENGINEERING CAMPUS

Figure 1: Three Hinged Arch Apparatus

55


INTRODUCTION

The three hinged arch is commonly used for medium span bridges where the abutments are on ground liable to settle under the end reactions. Being a statically determinate structure the analysis is simple, and the temperature changes are accommodated without additional stresses being set up.

The model symmetrical arch used in this experiment shows the typical construction with a circular arch rib supporting spandrels on which the flat road deck is carried. The elevation of the arch rib is a compromise between affording headroom under the bridge and the line of thrust of all the loads as near to the arch shape as possible.

For a uniformly distributed load the best theoretical shape for the arch is a parabola. A heavy point load at quarter span causes the greatest divergence of the line of thrust from the arch centre line.

It would be unusual to find an unsymmetrical arch in real life, but their analysis, brings a greater understanding of the forces acting in arches and portal frames. Hence, the

experiment provides an opportunity to study the interaction of the horizontal and vertical reaction for an unsymmetrical three pinned structures.

ACCESSORIES

The complete set of accessories (Figure 1) consists of:-

1- HST.401 Short bridge section1- HST.402 L.H. bridge section1- HST.403 R.H. bridge section1- HST.404 bracket assembly1- HST.405 track plate assembly1- HST.406 Reaction load hanger1- HST.409 Rolling Load (50N & 25N)

Optional extras4 sets – HST.408 Uniformly Distributed Load of 25N per set.

56


APPARATUS

The model bridges are made up from cast aluminium sections hinged at the crown. The right hand section is 500mm span, 200mm rise and 725mm radius with a pair of ball bearings at the springing. The left hand section of the symmetrical arch has the same span and rise: for the unsymmetrical model it has a span of 250mm, 125mm and 312.5mm. Both left hand sections can be pinned at the springing to bracket fixed to the vertical side of the HST.1 frame. The right hand bearing runs on a horizontal track plate fitted with a true span maker and a system for applying a horizontal force toward the center of the arch.

Two types of loading are available. The one consist of a set of steel bars to stimulate a uniformly distributed load. The other is a pair of a set of steel bars to stimulate a vertical load (supplied standard).

OBJECTIVES

A set of experiments can be carried out to investigate:-

(i) The value of the horizontal thrust at the arch springing(ii) The line of thrust of the arch(iii) The influence line for the horizontal thrust(iv) The reaction locus

THEORY

For a symmetrical three pinned arch the application of the overall conditions of equilibrium yields two equations for a vertical load as shown (Figure 2):-

Moment about AP.x – VB.L = 0

Vertical EquilibriumP- VB- VA = 0

To find the horizontal thrust an extra equation can be written because the moment at C is 0.

Moments about C for right hand sectionHBh – VB.½L = 0

It can be seen that VB is a linear function of the load position x, hence H is also a linear function of x.

57


Figure 2 : Symmetrical Three Pinned Arch

When the three pinned structure is unsymmetrical (Figure 3) the approach is the same but the solution of the equations is more complicated

Moments about AP.x – VB.L + HB(hB – hA) = 0

Vertical Equilibrium P –VB – VA = 0

Moments about C for right hand section HBhB – VB.ℓB = 0

As before, horizontal equilibriumHA – HB = 0

For a uniformly distributed load, w over all the span of a symmetrical arch

Moments about Aw.L. ½ L – VB. L = 0

VB = ½wL = VA

Moments about C for right hand sectionw.½L.¼L + HBh – VB.½L = 0

HB = wL2 / 8h

58


Figure 3: Unsymmetrical Three Pinned Arch

PROCEDURE

Part 1

Set up the symmetrical arch as shown in the diagram above, checking that the span is 1m. The span is measured from the left hand pin to the marker on the track plate for the right hand bearing. Add weights to the reaction load hanger to balance the self weight of the bridge, so that the axis is in line with the span marker. Treat this as the datum for the experiment.

Take the 50N load and place it over the left hand springing, where it should have no effect on the horizontal reaction. Move the load hanger to restore the bridge to its true span. Repeat this across the bridge until 50N load is over the right hand springing.

Join the two rolling loads 100mm apart with the coupling link provided and for two or three positions of this composite load find the horizontal reaction. Using the special distributed load bars or loose weights lay three or four different values of uniformly distributed load across the whole span and find the horizontal reaction.

Part 2

Set up the unsymmetrical bridge: the left hand springing should be 177mm above bottom members of the frame, and the span is 750mm. Add weights to the reaction load hanger to balance the self weight of the bridge. Treat this as a datum for the experiment.

Place the 50N load 50mm from the left hand springing and add weights to the reaction load hanger to find the value of horizontal thrust that restores the bridge to its true span. Move the load 50mm to the right and find the new reaction, repeating this until the load is at the crown pin. Then change the load movement to 100mm rightward, carry until the bridge has been transverse by the load.

59


Join the two rolling loads with the coupling link and position the load where it is expected to cause the maximum horizontal reaction and the corresponding position of the load. By adjusting the load position and the horizontal reaction, find the maximum horizontal reaction and the corresponding position of the load.

Take a uniformly distributed load 500mm long and of 50N total weight, a repeat the foregoing procedure to find the maximum horizontal reaction and the corresponding position of the uniform loading which is shorter than the span of the bridge.

RESULTS

Plot an influence line for the experimental value of horizontal reaction on each of the bridges. Add the theoretical influence line for a 50N load crossing the bridge.

Use the influence line to predict the horizontal reaction for the various loading placed on the bridges and compare them with the experimental values.

For one of the composite loadings used in Part 1 , make scale drawing of the arch centre line and construct the forces diagram in a similar way to the following Figure 4 below :-

Figure 4

60


S5 – THREE HINGED ARCH TEST FORM


ENGINEERING CAMPUS

S5 – DATA SHEETS

CASE 1 : Symmetrical Arch Length of Arch = Height of left support = Height of right support = Height of Arch = Vertical reaction (self weight) = Point Load =Rolling Load System = Uniformly Distributed Load =

Data Sheet 1 : Point Load 50 N

SN Location of point load from left support

Horizontal Thrust

Maximum Horizontal

Thrust

Influence Line Ordinate

1 02 2003 4004 5005 6006 8007 1000

61


Data Sheet 2: Rolling Load 75 N


Horizontal Thrust

Maximum Horizontal

Thrust


1 3002 4003 500

Data Sheet 3: Uniformly Distributed Load 75 N/m


Horizontal Thrust

Maximum Horizontal

Thrust


1 1002 2503 400

CASE 2 : Unsymmetrical Arch

Length of Arch = Height of left support = Height of right support = Height of Arch = Vertical reaction (self weight) = Point Load =Rolling Load System = Uniformly Distributed Load =

Data Sheet 1: Point Load 50 N


Horizontal Thrust

Maximum Horizontal

Thrust


1 02 1253 2504 3755 5006 6257 750

62


Data Sheet 2: Rolling Load 75 N


Horizontal Thrust

Maximum Horizontal

Thrust


1 2002 2503 300

Data Sheet 3 : Uniformly Distributed Load 75 N/m


Horizontal Thrust

Maximum Horizontal

Thrust


1 502 1003 150

63


S6 – INFLUENCE LINE TEST FOR BEAM DEFLECTION


ENGINEERING CAMPUS


The objectives of the experiments are to find the influence line for beam and to compare them with value predicted from theory.

INTRODUCTION

Influence line for the deflection of a point on a beam is a line showing the relationship between the deflection of that point and the position of application of a unit load acting on the beam. The deflection of the point due to the application of any load to the beam is obtained by applying the principle of superposition after multiplying the ordinate of the influence line by the magnitude of the load. (See the example which of application of the influence line for w=5).

64


APPARATUS

Two load cells, dial gauge, load hanger, weights and one beam

PROCEDURE

EXPERIMENT 1

Before starting the experiment, measure the length of the beam and mark it at mid –span and at the ¼ -span points for easy reference.

1) Choose a suitable reading on the upper scale of the apparatus for the mid-span of the beam. (One of the 10cm markers is most convenient).

2) Set one of the load cells so that it is ¼ spans to the left of the marker chosen in step 1. (Do not forget to take account of any offset in the position cursor).

3) Set up the second load cell ¼ spans to the right of the mid span reading. Lock the knife edge.

4) Place the beam in position with ¼ span overhangs at either end.5) Position two hangers equidistant from the mid- point of the

beam. (The cursors may press lightly against the scale).6) Place a dial gauge in position on the upper cross member so that

the ball end rests on the centre-line of the beam immediately above the left hand support. Check that the stem is vertical and the bottom O-ring has been moved down the stem. Adjust the dial gauge to read zero and then lock the bezel in position. Move the dial gauge to a position above the right hand support, check that the beam is parallel to the cross member. Then adjust the height of the knife edge so that the dial gauge reads zero.

7) Remove the dial gauge and unlock knife edges. Adjust the load cell indicators to read zero.

8) Apply loads to the hangers in a systematic manner, tap the beam very gently and take readings of the load cells.

9) Process the results and plot graphs to illustrate the theory.10) Vary the value of W for chosen value a and b (Figure 1) and

record the dial gauge readings.

65


W h

b a

L/4 L/4 L/4 L/4

Distance,a=200mmDistance,b=(refer to Figure 1.1 – Figure 1.5)

Figure 1: Schematic of Experimental Set-up

11) For each set of readings, determine the mean deflection per increment (see the result).

12) Plot the influence line for ‘unit’ load. The unit may be defined as the incremental load.

13) Apply load system and compare the measured deflection with that given by calculation based on superposition and the influence line.

14) The value of k can be calculated as follows:

k = deflection / 5N load = { (h at w = 5) – ( h at w = 0 ) ÷ 5 ] +

{ (h at w = 10) – (h at w = 5 )} ÷ 5 ….etc.

where :- h = deflection(mm) a = 200mm

These results can be presented in graphical form as Figure 2

66


Figure 2: Influence line for W= 5N and W = 10N

CONCLUSIONS:

Comment on the agreement between theory and experiment. Account for any discrepancies and suggest possible remedies.

APPENDIX 1EXPERIMENT 2

Example of Application of the Influence Line For W=5

10 N applied at b = 450 mm together with 5N at b= 250 mm.Deflection due to 10 N = 2 x k at 450 mmDeflection due to 5 N = k at 250 mmTotal Deflection = 2 x k at 450 mm + k at 250 mm

Measured Deflection is then compared with the total deflection obtain above. It should be found that the figures are in close agreement.

5 N applied at b = 400 mm together with 10N at b= 200 mm.Deflection due to 5N = k at 400 mmDeflection due to 10N = 2 x k at 200 mmTotal Deflection = k at 400 mm + k at 200 mm

The measured deflection is should be very close agreement with the total deflection.

10N 5N Dial gauge

67


200mm Load cell

450mm

L/4 L/4 L/4 L/4

Set 1

Figure 1.1

5N 10N Dial gauge

200mm Load cell 350mm

450mm

L/4 L/4 L/4 L/4

Set 2

Figure 1.2

5N 5N 5N Dial gauge


450mm

68

EAA 203 – LIGHT STRUCTURE L/4 L/4 L/4 L/4

Set 3

Figure 1.3

5N 10N 5N Dial gauge


450mm

L/4 L/4 L/4 L/4

Set 4

Figure 1.4

10N 10N 10N Dial gauge


450mm

69


L/4 L/4 L/4 L/4

Set 5

Figure 1.5

S6 – INFLUENCE LINE TEST FOR BEAM DEFLECTION FORM


ENGINEERING CAMPUS

S6- DATA SHEETS

Name :_____________________________________________________

Matrix No. : _____________________________________________________

Experiment Date :_____________________________________________________

Group :_____________________________________________________

Data Sheet 1 Type of Specimen

Length, L (mm)1 2 3 Average

Beam

70


Data Sheet 2

b (mm) 450 400 350 300 250 200W (N) h h h h h h0510152025K

where :-

h= deflection (mm)

Data Sheet 3

Experiment 2Theoritical Deflection

Experimental Deflection

% Difference

Set 1Set 2Set 3Set 4Set 5

71

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