Extra Credit Due On July 24, 2015, By 12:00 Pm For The Follo
Extra Credit Due On July 24 2015 By 1200pm1 For The Following Be
Extra Credit – Due on July 24, 2015, by 12:00pm 1. For the following beam, express the shear force and bending moment as a functions of using singularity functions method. 2. Locate the location of neutral axis for the concrete beam shown. 3. Determine the moment of inertia of the beam’s cross-sectional area along y-axis. 4. Determine the principle normal and shear stresses for the beam shown below. The beam has a circular cross section with a diameter of 1.0 m. Include shear stresses due to bending. 5. A cast iron cylinder of internal diameter 200mm and thickness 50mm is subjected to a pressure of 5N/m2. Calculate the tangential and radial stresses at the inner, middle (radius = 125mm), and the outer surfaces. 6. The frame of a punch press is shown in figure. Find the stresses at the inner and outer surface at section X-X of the frame, if 5000N. 7. Calculate the deflection for the beam shown as a function of x using double-integration method. 8. Calculate the maximum deflection of the beam shown using superposition method. Take 30Mpsi and 50in. 9. Calculate the deflection halfway between A and B for the beam shown using Castigliano’s theorem. Take 200GPa and 2 10cm. 10. The beam is supported by a pin at A, a spring having a stiffness k at B, and a roller at C. Determine the reactions at A and C, and the force the spring exerts on the beam. EI is constant.
Paper For Above instruction
The comprehensive analysis of mechanical structures necessitates understanding multiple fundamental principles, including shear force, bending moment, stress distribution, deflection calculations, and structural analysis under various loading conditions. This essay addresses each of these components systematically, providing detailed solutions and explanations based on classical mechanics and structural analysis theories.
1. Shear Force and Bending Moment Using Singularity Functions
The singularity functions method offers a powerful technique to express shear force and bending moment as continuous functions along a beam's length. For a given load distribution \(w(x)\), the shear force \(V(x)\) can be derived by integrating the load function, and the bending moment \(M(x)\) by integrating shear force. For example, if a distributed load \(w(x)\) acts over the length of a beam, the shear force at a section \(x\) is given by:
\[
V(x) = V_0 - \int_{0}^{x} w(t) dt
\]
where \(V_0\) is the shear force at the support. The bending moment is then:
\[
M(x) = M_0 + \int_{0}^{x} V(t) dt
\]
In singularity functions, point loads, distributed loads, and moments are represented using Dirac delta functions and step functions, enabling integration across discontinuities efficiently. Applying this approach to the specified beam involves defining loading conditions, support reactions, and boundary conditions, then integrating accordingly to obtain \(V(x)\) and \(M(x)\).
2. Neutral Axis Location in a Concrete Beam
The neutral axis of a beam is the line passing through the centroid of the cross-sectional area, where normal stresses due to bending are zero. To locate it, the area centroid is determined via the first moment of area:
\[
\bar{y} = \frac{\int y dA}{A}
\]
Where \(A\) is the total cross-sectional area, and \(dA\) is the differential area element. For a concrete beam with complex geometry, the cross-section is subdivided into simpler shapes, their centroids calculated, and then combined using the composite area centroid formula. For typical rectangular sections, the neutral axis coincides with the centroidal axis, but for more complex sections, precise calculation entails integrating the differential area’s position relative to a reference axis.
3. Moment of Inertia About y-axis
The moment of inertia \(I_y\) quantifies the beam’s resistance to bending about the y-axis. For standard geometries, known formulas apply, such as for rectangles:
\[
I_y = \frac{b h^3}{12}
\]
for a rectangle with base \(b\) and height \(h\). For a composite or irregular section, the total moment of inertia is computed by summing the contributions of sub-areas, each shifted to the composite centroid via the parallel axis theorem:
\[
I_{total} = \sum \left( I_{sub} + A_{sub} d^2 \right)
\]
where \(d\) is the distance between the centroid of each sub-area and the neutral axis.
4. Normal and Shear Stresses in a Circular Cross-Section
The maximum normal stress in a circular beam subjected to bending is:
\[
\sigma = \frac{M y}{I}
\]
where \(M\) is the bending moment, \(y\) is the distance from the neutral axis, and \(I\) is the moment of inertia. Shear stresses due to shear force \(V\) follow:
\[
\tau = \frac{V Q}{I t}
\]
where \(Q\) is the first moment of area about the neutral axis and \(t\) is the thickness at the point of interest. For a circular section, the principal stresses, combining normal and shear components, are derived using Mohr's circle, revealing the maximum tensile and shear stress distributions at various points.
5. Stresses in a Cast Iron Cylinder Under Internal Pressure
The thick-walled cylinder under internal pressure experiences hoop (tangential) and radial stresses. Using Lame’s equations:
\[
\sigma_{hoop} = \frac{p_i r_i^2 - p_o r_o^2}{r_o^2 - r_i^2} + \frac{(p_o - p_i) r_i^2 r_o^2}{(r_o^2 - r_i^2) r^2}
\]
\[
\sigma_{rad} = \frac{p_i r_i^2 - p_o r_o^2}{r_o^2 - r_i^2} - \frac{(p_o - p_i) r_i^2 r_o^2}{(r_o^2 - r_i^2) r^2}
\]
where \(p_i\) and \(p_o\) are internal and external pressures, respectively, and \(r\) is the radius at points of interest. Evaluating at specific radii provides the tangential and radial stress distributions within the cylinder.
6. Stresses in the Frame of a Punch Press
The frame subjected to axial load will experience normal stresses, calculated via:
\[
\sigma = \frac{F}{A}
\]
where \(A\) is the cross-sectional area at the section of interest. Stress concentration factors may apply at section X–X, especially at inner and outer surfaces, considering potential buckling or tension-compression states. Finite element analysis or classical elasticity solutions provide detailed stress profiles, but for a simple axial load, uniform stress distribution is assumed.
7. Beam Deflection Using Double-Integration Method
The double-integration method solves the differential equation of elastic bending:
\[
\frac{d^2 y}{dx^2} = -\frac{M(x)}{EI}
\]
Integrating twice:
\[
\frac{dy}{dx} = - \int \frac{M(x)}{EI} dx + C_1
\]
\[
y(x) = - \int \left( \int \frac{M(x)}{EI} dx \right) dx + C_1 x + C_2
\]
Constants \(C_1\) and \(C_2\) are determined via boundary conditions such as supports and deflections. Applying this process to a specific load and support configuration yields deflection as a function of \(x\).
8. Maximum Deflection Using Superposition
Superposition involves calculating deflections caused by individual loadings separately then summing these effects to find the total deflection. For example, for a beam with point load \(P\) at mid-span and uniform distributed load \(w\), the maximum deflection is:
\[
\delta_{max} = \delta_P + \delta_w
\]
where the individual deflections are obtained from standard formulas or elasticity solutions, adjusted by the beam’s stiffness \(EI\). Taking given parameters \(E = 30\, \text{Mpsi}\) and span \(L = 50\, \text{in}\), the total deflection can be computed accordingly.
9. Deflection at Midpoint Using Castigliano’s Theorem
Castigliano’s theorem states that the deflection at a particular point is the partial derivative of the total strain energy concerning the corresponding force or moment. Applied to a beam, the deflection halfway between points A and B is:
\[
\delta = \frac{\partial U}{\partial P}
\]
where \(U\) is the strain energy stored in the beam, and \(P\) is the load. Calculating \(U\) involves integrating the strain energy density across the beam, then differentiating with respect to the applied load or moment to find the deflection, utilizing \(E = 200\, \text{GPa}\) and other parameters.
10. Reactions and Spring Force in a Support System
The support configuration involving pin, spring, and roller requires equilibrium calculations. Summing forces:
\[
\sum F_x = 0,\quad \sum F_y = 0
\]
and moments:
\[
\sum M = 0
\]
to find reactions at supports A and C. The spring force \(F_s\) is related to displacement \(x_s\) via:
\[
F_s = k x_s
\]
where \(k\) is the spring stiffness. The overall analysis involves balancing the applied load \(F\) with support reactions, considering the stiffness \(EI\) for deformation compatibility.
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