Structures. Shainal Sutaria

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1 Structures ST Shainal Sutaria Student Number: Wednesday, 14 th Jan, 011 Abstract An experiment to find the characteristics of flow under a sluice gate with a hydraulic jump, also known as a standing wave is to be concluded. To acquire the properties of the hydraulic jump, a graph of the energy line in a rectangular channel through regions of gradually varied flow and rapidly varied flow is to be plotted. The value of the Froude number is also to be determined upstream and downstream of the hydraulic jump from the results. Thus the energy head loss can be calculated due to the jump. Then the theoretically estimated and measured values of downstream depth and energy head can be compared.

2 Contents Page Number 1. Introduction 3. Theory 4 3. Experimental Procedure and Results Experimental Procedure Objectives Experimental Procedure Apparatus and Method 5 3. Results 3..1 Test Data Tables Example Calculations Graph of Energy Head against Depth Calculation of Loss of Specific Energy Head Hydraulic Jump Classification 8 4. Discussion 9 5. Conclusions References 10

4 Fig. 1. x Assume that v0 a sin L where a is the initial central displacement of the strut. This equation complies with the boundary conditions that v 0 =0 when x=0, and x=l, and also d v 0 /dx=0 at x=l/. Thus the assumed deflection is therefore practical. Due to the initial curvature in the strut, a axial load P instantly create bending of the beam and consequently additional displacements, v occur which is measured from the original displaced position. The bending moment, M at any point in the section is M P v v ). (.01) ( 0 When the strut is initially unstressed the bending moment at any section is proportional to the change in curvature at that section from its initial arrangement and is not its absolute value. d v d v P Thus M EI, and hence ( v v ) 0. (Note that P is not the buckling dx dx EI load for the strut.) Substituting for v 0, we get. dv dx (.0) P EI v The solution of this equation is, a x v Acos x B sin x sin, L L (.03) P in which,. EI P EI x a sin L

5 From the initial boundary conditions in equation (.01), where v=0 at x=0, we get A=0 and v=0 x=l we get 0 Bsin L The strut in this case is in stable equilibrium as only bending is involved, sinμl cannot be zero. Therefore B=0, which gives the equation, a x v sin, L 1 L (.04) Since A=0 and B=0 P Substitute from above equations that and EI vo (.04) we get the equation v EI 1 PL x v0 a sin into equation L EI Now L Hence Pcr, is the buckling load for a perfectly straight pin-ended strut. v v P cr P (.05) 1 The effect of the load P will increase the initial deflection by a factor of P cr 1 P As P approaches P cr, v tends to infinity. This is not possible in practice as the strut would breakdown before reaching Pcr. Due to this we consider the displacement at the mid-point of the strut Pcr vc a P 1 (.06) This can be rearranged to give 0 1

6 v c P (.07) cr vc P This is in the form of a straight line equation y mx c, which represents a linear relationship between v c and v c /P. A graph of v c against v c /P will produce a straight line as the critical condition is approached. The gradient of the straight line represents P cr and the intercept on the v c axis is equal to a (initial central displacement). This graph is acknowledged as a Southwell plot. When a strut is loaded centrally and without an end load, the strut becomes a simply supported beam. Thus the mean deflection can be calculated as, 3 WL deflection, where W is the weight of the central load, L is the length of the 48EI beam, E the Young s Modulus and I the second moment of area. Due to the 3 WL graph of load against deflection being a straight line, tan. Therefore 48EI tanθ=deflection. (where θ=angle between deflection angle and the line of the graph). a Fig.. Part :Unsymmetrical Bending. Unsymmetrical Bending If the plane containing the applied bending moment is not parallel to the principal axis of the section, the simple bending formula cannot be applied to find the bending stress. For unsymmetrical sections the direction of the principal axes has first to be determined. In this experiment a beam made up of angle section is used and it is required to find the principal axes.

7 Let, OX and OY Perpendicular axis through the centroid. OU and OV Principal axis. δa is a elemental area at distance u and v from the OU and OV axis respectively. The rectangular second moment of area IUV is given by (13) Where, u = x cos θ + y sin θ and v = y cos θ x sin θ Thus giving (14) The condition for a principle axis state that IUV = 0, Thus (15) Tanθ =

8 The second moment of area for the X and Y axis has to be determined. This can be done by the use of the following equation: (16) Ix= Iy= Ixy=bdxy Where x and y in this case is the distance from the surface of the beam to the neutral axis. b The breadth of the strut d The Depth of the strut Ix and Iy Second moment of area..3 Shear centre When a beam or section does not have a vertical axis of symmetry, then the horizontal shear stressed on the beam are not in complete equilibrium, they will create a couple causing the section to twist. The twist leads to the development of torsional stresses in the member. The shear centre is the point where the torsional stresses are zero, and only bending occurs. In a channel section, there is no vertical axis of symmetry, and so, the shear centre needs to be determined For a channel section under a shearing force, F, at a distance d from the centre of the web, the shearing stress at any point is given by: = (3) This can help derive the equation for d, the distance from the centre of the web, which will give the position of the shear centre [at d, there will be no torque, just pure bending]: d

10 3. Results 3..1 Test Data Tables Length of strut, L (mm) 1 3 Mean Width of strut, b (mm) 1 3 Mean Thickness of strut, d (mm) 1 3 Mean Table Second moment of Area of Strut, I (mm 4 ) = bd E = 70x10 9 N/m EI (70 10 ) ( ) P cr N L 0.99 Table End Load (P) Central Deflection (mm) Deflection, V c v c /P (mm/n) (N) Load Unload Mean (mm)

12 Load, W (N) Deflection, Vc (mm) Load, P (N) Graph of P against v c Deflection, v c (mm) Graph of Vc against Vc/P Vc/P (mm/n) Graph of Weight against Deflection P = 0 kg P = 0 kg Deflection, v c (mm)

13 Fig Example Calculations From the theory in the previous section it can be shown that: Length of strut, L= ( ) / 3 = 990 mm Width of strut, b = ( ) / 3 = 5 mm Thickness of strut, d = ( ) / 3 = 6.38 mm Second moment of Area of Strut, I 3 bd mm 4 P cr EI L 9 (70 10 ) ( For End Load, P = 300 N: Central Deflection = ( ) / = 35.8 mm Deflection, V c = = 9.66 mm v c /P = 4.68 / 300 = 0.03 mm/n For End Load, P = 0 kg, Load, W = 5 N: Deflection, V c = = 3.68 mm 1 ) N 4. Unsymmetrical Bending b = 36.3 mm d = 30.5 mm t1 = 3.5 mm t= 3. mm The centroid of the beam is determined: y = ((36.3 x 3. x 1.6) + (( ) x 3.5 x 15.5) / ((36.3 x 3.) + (7.3 x 3.5)) = mm x = ((30.5 x 3.5 x 1.65) + (33.05 x 3. x 18.15) / ((33.05 x 3.) + (30.5 x 3.5)) = mm

14 The second moment of area (Ix, Iy and Ixy) is calculated where the values of x and y is the one calculated above: Ix =1.483* mm4 Iy =.356* mm4 Ixy =8.440* mm4 The angle is calculated by: Tanθ=1.933 θ=6.65 θ = By theoretical result θ was obtained to be X-Direction Deflections Y-Direction Load(N) Load Unload Mean Load Unload Mean Deflections Deflections Tanθ=dy/dx=3.571 Θ=37.17

15 4.3 Shear centre Two dial gauges had been kept in contact with the surface, each gauge is rested near the edge of the upper flange. The initial readings of the dial gauge were then taken. Then A weight was placed along the horizontal bar that was attached to the channel.changes in the dial gauge deflections were then recorded as the weights were moved through set distances on the horizontal bar. t=3.5 mm h= mm k= mm L= 86 mm The angel of twist is given by the equation: Angel of twist = (d-d1)/l The above equation is applicable as at the point of zero torsion, the dial gauge reading for both the dial gauge are the same, i.e. d1=d. 5. Discussion For the axial compression and transverse bending experiment the value of tan θ was found to be.107 and the angle of deflection was determined to be o, and for P =0 kg with the application if central loading, the value of tan θ was found out to be 1.14 and the angle of deflection was found out to be o. The theoretical value of θ was also determined to be 63.54o. There was an error of 1.071o and 15.09o when value from graph was found our of p = 0 and p = 0. The error may be due to defect in the dial gauge or the reading may have been miss read. For the experiment on unsymmetrical bending the The slope was found out to be 37.17o when the deflection in y axis was plotted against deflection in x axis.the theoretical value of θ was found to be 31.33oand an error of 5.84o was found out.

16 There maybe many reasons for the errors such as failure to apply load in both the ends at the same time,error in dial gauge, etc. For the experiment on shear centre he theoretical value for d was found out to be mm. From the graph plotted the experimental value of d was determined and was 15 mm. There was an error of 1.71 mm.there may be several reasons like the incorrect measurement of the point where the load has to be applied,defect in dial gauge, etc. 6. Conclusion The experiment was carried out successfully and the expected results were achieved. The behavior of action of load on a strut was studied. The values of bulking load and deflection angle were calculated and were found to be correct. The experimental values explained the effect of load on the deflection angel so if the load increases so does the deflection angel. The value of buckling load was calculated using two different ways and was compared to its theoretical values which were found to have a slight error. These errors could be caused due to various reasons such as defect in dial gauge, defect in weights, incorrect method of applying weights. 7. References Lab report notes

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