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Knovel Solutions are the stories of engineers who have used Knovel to solve real life problems.
FEA (Finite Element Analysis) methods often underestimate the impact of stress concentration on the strength of mechanical components. The example below walks through how stress concentration methodology found on Knovel can be used in the design phase to increase the strength of a mechanical part without FEA.
This common engineering problem comes in the form of a story from a bicycle component manufacturer. The manufacturer recently received a number of claims from customers related to the failure of pedal axles. In order to correct their design, an engineer was brought in to assess the problem and offer a solution.
An engineer was assigned the task of redesigning the axle to reduce stress by 10-15%, which was deemed to be sufficient in preventing failure of the axle.
(Configuration of axle at the point of failure, dimensions in millimeters)
What We Know:
Next Steps:
Researching Designs That Reduce Stress
The engineer starts with a search for ‘stress concentration’ (Click to run search)
Open the Peterson’s Stress Concentration Factors (3rd Edition) and review the table of contents. Chapter 3 of this book is dedicated to concentration factors at shoulder fillets:
(Click to enlarge)
And Section 3.4 Bending contains equations for calculations of stress concentration factors and recommended methods for reduction of stress concentrations.
Method 1: Streamline Fillet
One of these methods involves using a streamline fillet.
Looking through Chapter 3 for more information on streamline fillets, the engineer finds a table with proportions for these fillets (Table 3.1 below).
Unfortunately, the fillet designed according to this table would have increased the diameter of the axle shank to at least (1.475 x 12 mm) + 2 mm = 19.7 mm, where 1.475 is the fillet to axle diameter ratio (see highlighted value in the Table 3.1), 12 mm is the axle diameter, and 2 mm are added to account for a shoulder required to mount the bearing.
Such an increase in the axle diameter is not acceptable because of extra weight and design considerations.
Method 2: Variable Radius Fillet
Another method, suitable for any type of loading (including torsion), is recommended on page 140. It involves a compound radius fillet (also called double radius fillet):
(Click to enlarge)
Design & Testing of Variable Radius Fillet
At this point, the engineer decides to determine if he can reduce stress by using a double radius fillet (see Figure 3.11). He turns to Section 3.5.3 Compound Fillet.
Specifically, he wants to find out if using the radius r2=6 mm and r1=1.4 mm would reduce maximum stresses in the axle.
Two distinct maximum stress concentrations have to be calculated: one for the circumferential line I and another for the circumferential line II.
The drawing of the resulting axle with the diameter at circumferential line I 12 mm and diameter at circumferential line II 13.6 mm is shown in Fig. 2 below:
The chart (click to enlarge) for stress concentration factor for this case is provided on the page 165 of the book:
Since this chart is interactive on Knovel, the engineer uses it to quickly determine stress concentration factors for the axle with modified design:
Using Knovel Graph Digitizer for Chart 3.10 (see picture below), let’s determine stress concentration factors for original condition and for circumferential line I and calculate stress reduction:
(Click to enlarge)
Now let’s determine stress concentration factor for circumferential line II:
The stress concentration factor at this section is obviously higher. However, since the diameter is significantly larger, the nominal stresses are smaller and, as a result, the concentrated stresses are smaller than those for the circumferential line I:
These preliminary calculations have shown that the nominal stresses at the circumferential line II are 68.7% the nominal stresses at the circumferential line I. Thus, the maximum stresses are 88.4% of the maximum stresses at the circumferential line I.
The engineer would like now to verify these calculations because one point was extrapolated outside the range of the chart and therefore could be inaccurate. To calculate, he uses the equations provided at the top of the chart. For the original design with single radius r=2 mm (see Fig. 1):
And then for the double fillet design with radius r2 increased to 6 mm as shown in Fig. 2:
The calculations (click here to see all calculations) demonstrate that replacing a fillet with single radius r = 2 mm with a double radius fillet having r2 = 6 mm decreases the stresses by about 16% which is sufficient to solve the problem.
Alternate Methods
A few more methods of reducing stress concentration are proposed in Section 3.6 Methods of Reducing Stress Concentration at a Shoulder. The most interesting method for the pedal axle is a relief groove.
The advantage of this method is that the configuration of the axle remains the same.
The calculations demonstrate that replacing a fillet with single radius (r = 2 mm) with a double radius fillet (with r2 = 6 mm) decreases the stresses by about 16%, which is sufficient to reduce the stress of the axle to acceptable levels.
Knovel Solutions are the stories of engineers who have used Knovel to solve real life problems.
Because test drilling for oil and gas often occurs in hard to reach areas of the world, the logistics become an important component of planning, and all aspects of the primary materials (drill pipe and cement) chosen must be considered.
In this issue, we are concerned with estimating the weight of the materials to be delivered to the drilling site.
The engineer responsible for logistics at a drilling site needs to calculate the weight of materials to be ordered for a well drilling operation
What We Know:
Next steps:
*To perform the cement calculations, the engineer needs to know slurry yield for an A class cement with recommended water-to-cement ratio and allow for 25% excess of material without taking into account the additives.
Knovel Solutions are the stories of engineers who have used Knovel to solve real life problems.
In this issue, we help to answer a question submitted by an engineer working at one of the world’s leading manufacturers of sustainable industrial floor maintenance equipment. Their products clean high traffic industrial surfaces such as airports, warehouses and even the floors of the White House and Yankee Stadium.
For two weeks, the engineer who submitted this problem researched the best ways to find a solution.
He needed to protect the steel shaft and die-cast aluminum brush hub of a hard floor scrubber machine from failure in a locked-rotor condition. He knew that one method of protection is to design a shaft key that fails before there is any damage to more expensive parts.
His question:
Can a steel key protect the shaft and the brush hub of a floor scrubber from damage?
What We Know:
The shaft of a hard floor scrubber machine is made of steel and has a diameter of 11/16 in. The output torque is 250 in-lbf at 200 rpm. The driven brush hub is made of die-cast aluminum. The destructive shear stress for the shaft material is 32,000 psi, and destructive compressive stress for die-cast aluminum is 24,000 psi.
The questions are:
Finding the recommended dimensions
We start with a search for ‘key size for shaft diameter’. (Click to run search)
And open Table 1, Key Size Versus Shaft Diameter in Machinery’s Handbook (27th Edition) & Guide to Machinery’s Handbook:
From this table, the recommended dimensions of the key are 3/16 × 3/16 in.
Now let’s find the equations for calculation of the torque moment for the shaft and shaft key.
A quick search for ‘shaft key’ retrieves the chapter Design of Transmission Shafting in Machinery’s Handbook (28th Edition) & Guide to Machinery’s Handbook.
There is an equation for calculation of the shaft diameter on page 301:
Using this equation we can calculate the torsion moment corresponding to 80% of the dangerous stress in the shaft.
From Table 1 on page 302, we find that the Kt factor is 1.5 and factor B=1 because the shaft is solid. Opening chapter 17., Design of Shafts and Keys for Power Transmission on page 155, we find equations for the torque moment versus dimensions of the key:
In the equation, “2″ represents the compressive stress of the hub material. It is used because the value is smaller than the destructive stress of the steel.
When we appy the values to the equations, the result is:
As we can see, the torque moment at which the shearing stress in the key is dangerous is greater than the torque at 80% of dangerous compressive stress in the hub — and even greater than the torque at 80% of dangerous shear stress in the shaft. So the steel key of these dimensions will not protect the hub and even the shaft.
Let’s calculate the destructive shearing stress for the material of the key required to protect the hub and the shaft. In order not to damage the shaft and the hub, the torque moment of key shear should not exceed the moment Tc:
Therefore, to protect the shaft and the hub with the key of these dimensions, we have to use a material with a destructive shear stress that is not more than 9,600 psi. The allowable shear stress should not to be less than….
to handle the maximum output torque of 200 lbf-in. We then use Data Search to restrict our results to interactive properties data. Entering the field name shear strength and our value of 9,600 psi,
(Click to enlarge)
we find the interactive table Estimated Minimum Mechanical Properties of Wrought Aluminum Alloys – (engineering units) in the Aluminum Alloy Database.
In this table, we learn that 1060 H16 tempered alloy satisfies our requirements:
(Click to enlarge)
This alloy has an ultimate shear stress of 8,000 psi and a yield shear stress of 6,000 psi, which is higher than 2327 psi, allowable shear stress in the key calculated earlier for 200 lbf-in torque. Hence, this alloy is able to handle output torque of 200 lbf-in safely. Reviewing the properties of this material, we notice that the key thickness has to be increased to prevent failure due to crushing.
Our conclusions:
An engineer needs to design a temperature trigger to be used in a steam control valve. The valve is installed on a live steam line in a tank with water that is being pre-heated from 20 to 80ºC. The pressure of the steam is 2 atm.
Calculating boundary conditions and the thermal expansion coefficient of copper.
What We Know: The main component of the trigger is a 15-mm-long cantilever beam made from copper. It has a rectangular cross section of 1 x 3 mm (depth and width, respectively). It bends due to a temperature difference between the top and the bottom surface of the beam. The engineer has to calculate flexural deformation 2 mm from the free end of the beam. (more…)
Relief valves protect pressure vessels and other equipment from being subjected to pressures that exceed their design limits.
In this scenario, a bellows pressure relief (PR) valve needs to be sized in line with the mass flow rate for a gas pipeline with natural gas under critical flow conditions, i.e., when total superimposed pressure plus built-up back pressure is equal to or less than the critical flow pressure. An Engineer, a PE and mechanical engineering consultant in Florida, turned to Knovel. (more…)
Knovel Math provides fully documented Mathcad worksheets for engineering calculations from trusted reference works, reducing the time it takes to find and solve equations, and document calculations.
In this solution story, we’ll walk through a step-by-step use case showing you how Knovel Math can help design a stage platform.
For more information about Knovel Math, or to view a short video, click here.
Designing a metal platform for a stage
A theatre requires a light metal platform in order to support stage lights and some scenery equipment.
The platform must accommodate up to 2 stage hands. The weight of the lights and scenery equipment may reach 2600 lb and could be assumed to be distributed evenly along the platform. To achieve this, the structure should be 20 ft long and 2.6 ft wide and suspended by 3 cables on each side attached 10 ft apart. A stage engineer must now determine the dimensions of the structural components and the load on the cables.
Modeling The Platform
The engineer envisions a platform made of 2 parallel long beams tied by short cross-beams with a light deck. The deck should be attached in a way that prevents it from being subjected to any bending load born by the beams. The platform will be made from the aluminum alloy 6063-T6 which has a density of 167.6 lb/ft3 and a design tensile yield strength of 25 ksi.
The weight of the lights, scenery equipment, stage hands and the structure is assumed to be distributed evenly between 2 parallel beams with a coefficient of 1.15 introduced to account for the possible unevenness in load distribution. As a result, each of the 2 parallel beams could be calculated as one continuous beam in bending. Moreover, by neglecting elongation of the supporting cables, the beam could be calculated as simply supported.
The maximum bending moment could be near the middle of the span or at the middle of the beam at the point where the cable is attached, if the weight of 2 people is applied to one of the beams near to the middle of the span.
The preliminary calculations have shown that this structure would have an approximate weight of 400 lb, and the cross section of a long beam would be 2.0 in2.
The solution can be reduced to that of a 3-span continuous beam with an evenly distributed load of:
and a concentrated load of W2 in one of the spans. Assuming the weight of a stage hand to be 260 lb, the concentrated load can be calculated as:Consequently, the load diagram of the beam could be drawn as:
Where w1 – distributed load
______W2 – concentrated load
______l1=l2=10 ft – spans
______a1 – position of the concentrated load
This beam is a multi-span continuous beam that is statically undetermined. The most commonly used solution for these beams utilizes the Three Moment Equation method. The load diagram to be used for the solution is:
Running a basic search on Knovel, the engineer searches using the keywords “three moment equation” (click on the image to run the search)
and retrieves Roark’s Formulas for Stress and Strain (7th Edition), a comprehensive handbook containing a Mathcad-enabled worksheet for Three-Moment Equation in Example 8.3.1 (Nondimesional).
Although Example 8.3.1 differs from the specific problem the engineer wants to solve, he can use it as a basis for his solution. To account for a different loading, the engineer has to modify the equations 1 and 2 in the Example 8.3.1 and use cases 1e, 2e and 3e from Table 8.1 (Also Mathcad-enabled on Knovel.)
The engineer can use these cases to calculate the slopes (to be used in the equations 1 and 2) at the right and left ends of the beam for all span loads. The series of calculations, schematically outlined below, are done in Mathcad by copying the equations from Knovel Math files for cases 1e and 2e (for span a).
and for the case 2e (for span b):
Click on the images to zoom in to the details.
The engineer’s goal is to calculate the bending moment at the middle support (M2) as a function of the position of the concentrated load W2:
Once the bending moment is calculated, the engineer can calculate the reactions at the supports by superposing the cases 1e, 2e, and 3e:
Then the engineer can derive the equations for the shear force and the bending moment of the whole beam as a function of their distribution along the beam (x) and the position of the concentrated load (a):
From the calculated maximum bending moment the engineer selects a suitable shape, using Knovel Math file for the case 3 from the Table A.1 of Roark’s to calculate section properties:
The bending moment of parallel beams has a maximum value (1.667 kip ft) when both stage hands are standing on the same side of the platform about 4.05 ft from its end.
To remain stable at this moment, the platform was built with long beams made from an aluminum alloy 6063-T6 channel with standard dimensions RT 1¾×4½×1/8. The maximum load on the middle cable is 1.579 kip when both stage hands are standing on the same side in the middle of the platform.
To see a 7-page report with the complete Mathcad solution click here.
