The Deflection of a Uniform Beam Subject to a Linearly Increasing Distributed Load Can Be Computed As

HK01- CIVIL ENGINEERING PROGRAMME FACULTY OF ENGINEERING UNIVERSITI MALAYSIA SABAH

COURSE KA20602 ANALYSIS BERANGKA

TITLE EXPLANATION ON MAXIMUM DEFLECTION ONLY ACTING ON THE FIXED END OF A CANTILEVER BEAM PREPARED FOR DR.HARIMI DJAMILLA

PREPARED BY ELDREN JAMEE FARAHAYNI YAHYA LLOYD EDMUND MOHAMED FARIHAN MAIL MOHD ASYRAF ISMAIL

a) Proposal and Background

BK13110098 BK13110111 BK13160574 BK13160569 BK13160580

Title : EXPLANATION ON MAXIMUM DEFLECTION ONLY ACTING ON THE FIXED END OF A CANTILEVER BEAM Background

What is a cantilever beam? Why does its maximum deflection only happens on the fixed end of the beam? A cantilever is a rigid structural element, where a beam is anchored at only one end to a support from which it is protruding. When subjected to a structural load, the cantilever carries the load to the support where it is forced against by a moment and shear stress. Cantilevers are widely found in a construction, notability in cantilever bridges and balconies. In cantilever bridges the cantilevers are usually built as pairs, with each cantilever used to support one end of a central section. The deflection of a uniform distributed load cantilever beam with its elastic curve equation computed as 2

y=

x w0 2 (x −4 Lx+ 6 L2 ) 24 EI

Thus, cantilever beams are important and it is important to measure the elastic curve of the beam since it will be involved in many construction such as bridges and more.

Methodology The advantages of a cantilever beam are; a) does not require a support on the opposite side. b) creates a negative bending moment, which can help to counter a positive bending moment create elsewhere. While the disadvantages of a cantilever beam are; a) large deflections, b) generally results in larger moments and c) have a backspan causing an uplift of the far support. For the superposition method, a cantilever beam with it being subjected to a uniformly distributed loading which its elastic curve is computed as x2 w 0 2 y= (x −4 Lx+ 6 L2 ) 24 EI While for the Golden Section method, a search algorithm for finding a minimum on an interval of XL and XU .Also uses the golden ratio ф = 1.618,to determine the interior points of X 1 and X2 by using the golden ratio, one of the interior points can be re-used in the next iteration. d=( ф-1 ) ( Xu - Xl ) X1 = Xl + d X2 = Xu - d Methods Various methods were used to calculate the result of the elastic curve of the cantilever beam. These various method were necessary to determine the point of the maximum deflection on the overhang beam. Three methods were used which were superposition method, Golden Section( calculating manually) and Golden Section ( using MATLAB). All these results were then used to prove the theory of whether the maximum deflection will occur on the fixed end or not.

b) Abstract The main objective is prove whether the elastic curve of a cantilever beam is at the end of the fixed end of the overhang beam or not. By using the method of Golden Section method through manually calculation and MATLAB program and the superposition method, which all methods in the end gave a slightly different answer. The results by three methods shows that the elastic curve of the cantilever beam were the same. Thus, as a conclusion of these methods proven that in the case of a cantilever beam, the maximum deflection will only happen at the fixed end of the overhang beam. Objectives To determine the elastic curve of a deflected cantilever beam. To determine which method was better, the superposition method or Golden Section method. Both method were used to evaluated for both correctness and creativity of the deflected cantilever beam. Introduction What is an elastic curve? It is the deflection of a beam or shaft must be often be limited in order to provide integrity and stability of a stability of a structure or machine, and prevent the cracking of any attached brittle materials such as concrete or glass. Furthermore, code restrictions often require these members not vibrate or deflect severely in order to safety support their intended loading. Most important is the deflection at specific points on a beam or shaft must be determined if one is to analyse those that are statically indeterminate.

The elastic curve represents the centerline deflection of a beam or a shaft. Its shape can be determined using the moment diagram. Positive moments cause the elastic curve to be concave upwards and negative moments cause it to be concave downwards. The radius of a curvature at any point is determined from 1 p

=

M EI

While for method of superposition is used to determined if the deflection or slope at a point on a member subjected to a combination of loadings. Cantilever beam is a beam supported on only one end. The beam transfers the load to support where it has manage the moment of force and shear stress. A moment of force is a force to twist or rotate an object while the shear stress is a stress which is applied parallel to the face of a material. nothing more, the cantilever beam bears a specific weight on its open end as a result of to support on its enclosed end, also in addition to not breakdown due to shear stress applied on the cantilever beam. It is often used in Civil Engineering buildings and its constructions allows for overhanging structures without external bracing or support pillars. Cantilevers can be also used with trusses or slabs. These cantilever design is famous in many kinds of architectural design and other kinds of engineering, where it is used in terms like end load, intermediate load and end moment to find out how mush a cantilever will hold. How does the executing a mathematical program for the numerical method is useful for Civil Engineering problems? The MATLAB is a very intuitive and easy to implement in any programming language which is the bisection method. The bisection method can be easily adapted for optimizing 1- dimensional functions with a slight but intuitive modification. Now, this

bisection method with the golden ratio that results in a faster computation for dealing with many Civil Engineering problems.

Theory By using the method of superposition where it enables us to determined the deflection or slope at a point on a member subjected to a combination of loadings where in this report, we used a cantilever beam with it being subjected to a uniformly distributed loading which is computed as x2 w 0 2 y= (x −4 Lx+ 6 L2 ) 24 EI

Only to determine the elastic curve point of the maximum deflection, but for this case, since it is a cantilever beam, therefore the elastic curve of the maximum deflection acting is on the fixed end of the beam where the x is equal to zero. As for the Golden Section method, a search algorithm for finding a minimum on an interval of XL and XU .Also uses the golden ratio ф = 1.618,to determine the interior points of X 1 and X2 by using the golden ratio, one of the interior points can be re-used in the next iteration. d=( ф-1 ) ( Xu - Xl ) X1 = Xl + d X2 = Xu - d

However, there are conditions to follow up when doing the Golden Section which is if the value of f(X 1) is smaller than f(X2), then X2 becomes the new lower limit and X1 becomes the new X2. While vice versa, if the value of f(X1) is larger than f(X2), then X1 becomes the new upper limit and X 2 becomes the new X1. In either case, only one new interior point is needed and the function is only evaluated one more time. Although are absolute errors in the Golden Section since the interior points ( X1 , X2 ) are symmetrical, thus using the normalisation computed as Error Criterion ԑa = (( 2- ф ) x ( | Xu - Xl |) / Xopt ) x 100% Results

This is an example of a cantilever beam we are going to examine for this research.

Golden Section method (Calculating without using any program)

The deflection of a uniform distributed load cantilever beam with its elastic curve equation computed as x2 w 0 2 y= (x −4 Lx+ 6 L2 ) 24 EI

Given that L=8 m, E=200 GPa, I=300 x 10-6 m4, and w0=10 kN/m, determine the point of maximum deflection using the golden-section search until the approximate error falls below Es=1% with initial guesses of xl=0 and xu=L.

i 1.000 0 2.000 0 3.000 0 4.000 0 5.000 0 6.000 0 7.000 0 8.000 0 9.000 0

Xl 0.000 0 3.055 7 4.944 3 6.111 4 6.832 8 7.278 7 7.554 3 7.724 4 7.830 0

f(Xl) 0.002 7 0.002 1 0.001 7 0.001 6 0.001 5 0.001 4 0.001 4 0.001 4 0.001 4

X2 3.055 7 4.944 3 6.111 4 6.832 8 7.278 7 7.554 3 7.724 4 7.830 0 7.894 4

f(X2) 0.002 1 0.001 7 0.001 6 0.001 5 0.001 4 0.001 4 0.001 4 0.001 4 0.001 4

X1 4.944 3 6.111 4 6.832 8 7.278 7 7.554 3 7.724 4 7.830 0 7.894 4 7.935 6

f(X1) 0.001 7 0.001 6 0.001 5 0.001 4 0.001 4 0.001 4 0.001 4 0.001 4 0.001 3

Xu 8.000 0 8.000 0 8.000 0 8.000 0 8.000 0 8.000 0 8.000 0 8.000 0 8.000 0

f(Xu) 0.001 3 0.001 3 0.001 3 0.001 3 0.001 3 0.001 3 0.001 3 0.001 3 0.001 3

D 4.944 0 3.055 7 1.888 5 1.167 3 0.721 5 0.445 7 0.275 6 0.170 0 0.105 6

Xopt 4.944 3 6.111 4 6.832 8 7.278 7 7.554 3 7.724 4 7.830 0 7.894 4 7.935 6

Note that the current maximum is highlighted for every iteration. After nine iteration, the point maximum deflection occurs at x=7.9356 m with a

Ea 61.80 85 30.90 49 17.08 34 9.911 7 5.902 2 3.567 1 2.174 4 1.333 6 0.818 3

function value of 0.001340824. Since this is a cantilever, the point maximum deflection should occur at x=0, which in this case, was only off by 0.0644m to achieving 0 m at the fixed end. Superposition method Displacement

w (x) =

− p x2 (6 L2 −4 xL+x 2) 24 EI

wmax = w (L) =

Slope

− p L4 8 EI

= - 0.0000853 m

2

θ (x) =

2

− px (3 L −3 xL+ x ) 6 EI

θmax = θ (L) =

−pL 6 EI

3

= - 0.0000142 rad

Moment and Maximum Bending Stress

M(x) =

1 p ( L−x) 2

2

2

L p 2

MMax = M (0) =

σMax = |MMax |

C I

=|

= 320 N.m

L2 p 2 Z |= 8.53 MPa

Shear

V (x) = -p (L - x) VMax = -pL = -80 N Thus, can be concluded that the elastic curve of the cantilever beam as shown as above. Also proven that the maximum deflection occurs at x=0m which is at the end of the fixed cantilever beam. Maximum deflection of the beam is 320 N.m and maximum shear stress is 80 N.

Golden Section( Using MATLAB ) The deflection of a uniform distributed load cantilever beam with its elastic curve equation computed as x2 w 0 2 2 y= (x −4 Lx+ 6 L ) 24 EI

Given that L=8 m, E=200 GPa, I=300 x 10-6 m4, and w0=10 kN/m, determine the point of maximum deflection using the golden-section search until the approximate error falls below Es=1% with initial guesses of xl=0 and xu=L. A file saved as fivedottwenty.m which the file contents is written as below; %FIVEDOTTWENTY max deflection clear; l= 8; e= 200000000000; i= 0.0003; w= 10000; syms x;

y= @ (x) (((x^2)*w)/(24*e*i))* ((x^2)-(4*l*x)+ (6*(l^2))); dydx= @ (x) (((x^3)*w)/(6*e*i))- (((x^2)*l*w)/(2*e*i)) + (((l^2)*x)/(2*e*i)); bisec (dydx,0,80,320,0.0000853); maxDeflection = y( bisec(dydx,0,80,320,0.0000853));

We then introduced the function of bisec to allow the program to run the mathematical problem. The two files needs to be in a different tab to allow the program to recognize the function of bisec( which is also known as the bisection method). The file which is saved as bisec.m and the contents of the file is written as; function ANS = bisec (f, x_l, x_u, iter, error) for i=1 : iter x_r(i) = (x_1 + x_u)/2; if((f(x_1)*f(x_r(i))) < 0) x_u = x_r(i); elseif ( (f(x_l)*f(x_r(i))) > 0) x_l = x_r(1); elseif ((f(x_l)*f(x_r(i))) ==0) break; end

if ((i>1) && (abs((x_r(i)-x_r(i-1))/x_r(i)) * 100)