Case 1: Uniform beam under distributed load.
In the shown Figure, a uniform beam subject to a linearly increasing distributed load. The deflection \(y(\mathrm{~m})\) can be expressed by \(y=\frac{w_{o}}{120 E I L}\left(-x^{5}+2 L^{2} x^{3}-L^{4} x\right)\)
Where \(E\) is the modulus of elasticity and \(I\) is the moment of inertia \(\left(\mathrm{m}^{4}\right), L\) length of beam.
Use the following parameters \(L=600 \mathrm{~cm}\), \(E=50,000 \mathrm{kN} / \mathrm{cm}^{2}, I=30.000 \mathrm{~cm}^{4}, w_{\mathrm{o}}=2.5\)
\(\mathrm{kN} / \mathrm{cm}\), to find the requirements
1) Develop MATLAB code to determine the point of maximum deflection by using numerical method (bisection, false position method,β¦.). Hint(The value of x where dy/dx=0).
2) Plot the point of maximum deflection versus iteration number.
3) Plot the values of the relative approximate error of the point of maximum deflection (ππ,π₯) versus iteration number.
4) Plot the following quantities versus distance along the beam a) Displacement (y). b) Slope π(π₯) = ππ¦/ππ₯. c) Moment π(π₯) = πΈπΌπ2π¦/ππ₯2. d) Shear π(π₯) = πΈπΌπ3π¦/ππ₯3. e) Loading π€(π₯) = βπΈπΌπ4π¦/ππ₯4.
Figure P5.13a shows a uniform beam subject to a linearly increasing distributed load. The equation for the resulting elastic curve is (see Fig. P5.135) Use bisection to determine the point of maximum deflection (that is, the value of x where dy/dx = 0). Then substitute this value into Eq. (P5.13) to determine the value of the maximum deflection. Use the following parameter values in your com- putation: L = 600 cm, E = 50,000 kN/cmΒ², I = 30,000 cm, and w0...
The deflection of a uniform beam subject to a linearly increasing distributed load can be computed by using the following equation: y = ( 120EIL Given that L 600 cm, I 30,000 cm, wo-2500 N/cm, and E 50,000 KN/cm2 2. Develop a Matlab code that would implement the Golden-Section search method to find the maximum deflection until the error falls below 1% with initial guesses of Xi = 0 and Xu-L. Display all of the following: xl, xu, d, x1...
Use bisection method to determine the point of maximum deflection of the beam subject to a linearly increasing distributed load shown in the figure below (the value of x where dy/dx= 0). Then substitute this value into the equation to determine the value of the maximum deflection. Use the following parameter values in your computation: L = 600 cm, E=50,000 kN/cm2, I=30,000 cm4, and w0 =1.75 kN/cm.
(a). A rectangular cross section at a location along a beam in bending is acted upon by a bending moment and a shear force. The cross section is \(120 \mathrm{~mm}\) wide, \(300 \mathrm{~mm}\) deep and is orientated such that it is in bending about its major axis of bending. The magnitudes of the bending moment and shear force are \(315 \mathrm{kNm}\) and \(240 \mathrm{kN}\) respectively. Determine the maximum bending and shear stresses on the cross section. Plot the bending and...
QUESTION 4 (25 marks) A simply supported beam is loaded by an uniform distributed load, wkN/m, over the span of the beam, L, as shown in Figure Q4. (a) Determine the end reactions at point A and B in terms of w and L. (4 marks) (b) At an arbitrary point, x, express the internal mom (c) Show that the deflection curve of the beam under the loading situation is ent, M(x), in x, w, and L. (5 marks) 24EI...
Need help!! 1. (25 Points) In the figure below, figure (a) shows a uniform beam subject to a linearly increasing distributed load which starts a 0 at the left end and increases to Wo on the right end. As depicted in (b), the beam deflection can be computed with 4 120EIL where E is the modulus of elasticity [kN/cm2] and I is the moment of inertia [cm]. Calculate each of thee following quantities (take the derivatives by hand) and plot...
The simply supported beam is subjected to a uniform distributed load, w of 30 kN/m in the negative y-direction and a point load, P of 15 kN in the negative z-direction. The total length, L of the beam is 6 m. Answer the questions that follow: 'n Eenvoudige opgelegde balk word belas met 'n uniform verspreide belasting van w 30 kN/m in die negatiewe y-rigting en 'n puntlas P = 15 kN in die negatiewe z-rigting. Die totale lengte, L...
2. The governing differential equation that relates the deflection y of a beam to the load w ia where both y and w are are functions of r. In the above equation, E is the modulus of elasticity and I is the moment of inertia of the beam. For the beam and loading shown in the figure, first de m, E = 200 GPa, 1 = 100 Γ 106 mm4 and uo 100 kN/m and determine the maximum deflection. Note...
The deflection y, in a simple supported beam with a uniform load q and a tensile load T is given by dx2 El 2EI Where x location along the beam, in meter T-Applied Tension E-Young's Modulus of elasticity of the beam 1= Second moment of inertia of the beam Applied uniform loading (N/m), L- length of the beam in meter Given that T-32 kN, q = 945.7 kN/m, L = 2.0 meter, E = 206 GPa and 1 4.99 x...
4. (25 pt.) The beam subjected to a uniform distributed load as shown in Figure 4(a) has a triangular cross-section as shown in Figure 4(b). 1) (6 pt.) Determine mathematical descriptions of the shear force function V(x) and the moment function M(x). 2) (6 pt.) Draw the shear and moment diagrams for the beam. 3) (5 pt.) What is the maximum internal moment Mmar in the beam? Where on the beam does it occur? 4) (8 pt.) Determine the absolute...