In this post, we will cover the first example in our mechanics of materials course covering combined loading. In this example, we are given a cylindrical pressure vessel with a specified outside diameter and wall thickness. The maximum stress it can sustain whether its hoop or longitudinal is 25 ksi. Based on this limitation, we need to calculate the maximum internal pressure that this vessel can contain, and we also need to calculate the maximum internal pressure for a spherical pressure vessel of similar size.
Cylindrical or spherical pressure vessels are used as boilers and tanks in a variety of industries including oil and gas, petrochemical, and manufacturing. Material stress can easily be calculated if the vessel is classified as thin wall. If the inner radius to wall thickness ratio is greater than or equal to 10, then the pressure vessel can be classified as thin wall. When a vessel is classified as thin wall, we assume that the through thickness stress is constant because the stress distribution throughout its thickness does not change dramatically. One important detail to keep in mind is that the pressure we discuss refers to the gauge pressure, which is the pressure above the atmospheric pressure.
In this example, we consider the cylindrical and spherical pressure vessel to experience biaxial stress even though it also experiences radial stress. The radial stress on the inner wall of the vessel equals the internal gage pressure and decreases through the thickness to zero on the outside surface of the vessel. Because the radial stress is much smaller than the hoop and longitudinal stresses, we choose to neglect it in our calculations.
The first step is to check if the pressure vessel satisfies the thin wall assumption. We calculate the ratio of the inner radius divide by the wall thickness and check if its greater than or equal to ten. In our case, our ratio is greater than ten, so are our assumption is valid.
The second step is to calculate the maximum pressure based on the hoop or circumferential stress formula. The hoop stress is equal to the internal gage pressure multiplied by the internal radius divided by the wall thickness. We re-arrange the equation and solve for the pressure. The internal pressure is equal to the maximum hoop stress times the wall thickness divided by the internal radius.
The third step is to calculate the maximum pressure based on the longitudinal stress formula. The longitudinal stress is equal to the internal gage pressure multiplied by the internal radius divided by two times the wall thickness. We re-arrange the equation and solve for the pressure. The internal pressure is equal to the maximum longitudinal stress times two times the wall thickness divided by the internal radius.
The last step is to calculate the maximum pressure based on the stress formula for a spherical vessel. The stress is equal to the internal gage pressure multiplied by the internal radius divided by two times the wall thickness. We re-arrange the equation and solve for the pressure. The internal pressure is equal to the maximum stress times two times the wall thickness divided by the internal radius.
In this foundation design example, we will learn how to calculate the bearing pressure on a continuous shallow footing foundation subjected to a vertical wall loading. This is example number 2 in our foundation design course. The water table is below the bottom of the foundation, so the pore water pressure at the bottom of the foundation is zero.
Since this a continuous footing, the loads will be expressed as force per unit width.The first step is to calculate the unit weight of the foundation. It is equal to unit width of the footing times the depth of the footing times the unit weight of concrete. The foundation material is reinforced concrete, so the unit weight is assumed to be 23.6 kN/m.
Next we consider the pore water pressure acting on the bottom of the foundation. As we mentioned previously, this is equal to zero.
Finally, we compute the bearing pressure which is equal to the weight of the foundation per unit length plus the applied wall loading per unit length divided by the width of the foundation inus the pore water pressure.
The method of double integration is only applicable for elastic deformations in which the slope is beam is very small. Also, the double integration method only considers deflections due to bending. Since the deflection due to shear is much smaller than the deflection due to bending, it is typically omitted in engineering practice.
The first step is to define the variables. We call the load, “w”, and the length, “l”.
The next step is to calculate the reactions using statics.
The third step is to define a coordinate system. The coordinate system should be parallel to the undeflected beam with the origin at the left side of the beam and going positive to the right. A separate coordinate system will need to be established at locations of discontinuous loading.
The fourth step is to define the bending moment as a function of the previously defined coordinate system for each region of continuous loading.
The fifth step is to integrate the bending moment function. This is based on the relationship that the second derivative of the deflection of the beam with respect to the defined coordinate system is equal to the bending moment divided by the product of the Young’s Modulus and the moment of inertia of the beam. Each integration leads to a constant of integration, so in the end we have 2 constants of integration.
The sixth step is to solve for the constants of integration using the boundary conditions for the given beam. The boundary conditions pertain to the supports as well as the continuity conditions that ensure slope and displacement are continuous at points where two functions meet.
The last step is to solve for the required slope or deflection. This is done by plugging the numerical value for the location of interest into the slope and / or deflection equations. Positive values for the slope are counterclockwise and positive values for displacement are upward. It is always a good idea to compare the numerical values for the slope or displacement to a sketch of the assumed elastic curve.
This is example number 1 in our course for reinforced concrete design & analysis for flexural beam members. The problem statement is broken down into 3 parts. Part “A” asks us to calculate the steel reinforcement for the balanced case. The second part involves finding the maximum steel ratio for tension controlled and transition controlled section per ACI Code 318-11. For the last portion, we need to compute the location of the neutral axis and the depth of the equivalent compressive Whitney stress block for the tension controlled section in Part B. The compressive strength of the concrete is 4 KSI, and the yield strength of the steel reinforcement is 60 KSI.
In this video, we’ll cover the first part of the problem and then do the remaining parts in the next videos. Before calculating the balanced steel reinforcement quantity, it’s a good idea to understand what is meant by the balanced condition with regards to reinforced concrete beams. A balanced condition occurs when the steel yields at the same time as the concrete fails. The balanced strain is when the steel at first yield reaches a strain corresponding to its yield strength just as the maximum strain in the concrete at the extreme compression fiber reaches 0.003.
The first step is to calculate the balanced steel ratio. The balanced steel ratio formula is equal to 0.85 times beta 1 times the compressive strength of concrete times 87 times the distance from the extreme compression fiber to the extreme steel reinforcement all divided by the product of the yield strength, 87 plus the yield strength, and the distance from the extreme compression fiber to the centroid of the steel reinforcement. Since this is a singly reinforced concrete beam, the distance from the extreme compression fiber to the extreme tension steel reinforcement is equal to the distance from the extreme compression fiber to the centroid of the tension steel reinforcement. The balanced steel ratio for this reinforced concrete beam cross-section is equal to 0.0285. Once we have the balanced reinforcement ratio, we can multiply it by the effective cross-section to get the area of steel corresponding to the balanced condition. The effective cross section is equal to the product of the width and effective depth of the rectangular cross-section. The balanced steel area is equal to 11.63 inches squared.
Our example covers undamped free vibration of a single degree of free SDOF system in which the damping term and forcing function are both zero. Damping is the mechanism through which energy is lost and, and forcing is the cause of excitation for the system. This system will experience vibration in the absence of as externally applied force, and this phenomenon is known as free vibration.
The first step is to write the equation of motion. The equation of motion can be written as mass times acceleration plus damping coefficient times velocity plus stiffness times displacement equals some force or force function. The equation of motion is based on D’Alembert’s principle of dynamic equilibrium applied to a free body diagram.
The next step is to calculate the mass and stiffness. Mass is equal to the weight of the load divided by the gravitational acceleration. Stiffness is defined as the force required to cause a unit displacement of the specified mass. The stiffness depends on the loading and boundary conditions of our beam. It is equal to some constant multiplied by the product of the Young’s Modulus and moment of inertia of the beam divided by the length of the beam cubed.
The third step is to calculate the circular natural frequency. The circular natural frequency is equal to the square root of the stiffness of the SDOF system divided by the mass, and the units are in radians per second.
The fourth step is to calculate the natural period which is equal to two times pi divided by the circular natural frequency. The units of period are seconds.
The last step is to calculate the frequency which is simply the inverse of the natural period, and the units are in hertz.