Knovel Solutions

Every spring season, fans flock to baseball stadiums and cheer for their favorite teams and players. With an average attendance rate of 25,000 fans per game, stadium engineers need to create structures that ensure the safety and comfort of all attendees.

This problem comes from a design engineer working on replacing the sunshade support beams in a baseball stadium.

Per code, steel I-beams used in the supporting structure of a stadium sunshade should have the moment of inertia equal to 2.64 in4 or more.

The engineer would like to evaluate if the weight of the structure can be reduced by replacing steel with aluminum.

Aluminum alloys are a favored material in modern construction due to their lighter weight, fairly high strength-to-weight ratio, and resistance to corrosion.

Next Steps:

Search for the material properties of steel and aluminum
Calculate moment of inertia, section area, and nominal weight of both steel and aluminum
Compare the properties to determine if aluminum is an acceptable substitute for steel

Searching for Properties of Steel and Aluminum in Knovel

The engineer starts by searching for ‘material selection‘

and open Chapter 11, Selection of Material and Shape of the Materials Selection in Mechanical Design (3rd Edition).

In the bookmark, select Section 11.5 Exploring and Comparing Structural Sections and navigate to Figure 11.12 on page 306.

Using the Interactive Graph in Figure 11.12, find the typical properties of modulus of elasticity and density of steel and aluminum alloys.

Finding Nominal Weight of a Steel Beam

Taking into account that 2.64 in4 = 1.1×10-6 m4 and using the “Shape” portion of the Interactive graph, he finds that the typical section area for a steel section is 5.6×10-4 – 1.6×10-3 m2 (.00056 – .0016 m2).

Using the 7,905 kg/m3 density value for steel found earlier, and Knovel’s Unit Converter, this beam’s nominal weight is between 4.4 kg/m and12.6 kg/m (or 3.0 and 8.5 lb/ft).

Calculating Moment of Inertia for Aluminum

Since the beam stiffness has to be the same for both steel and aluminum, Is×Es = Ia×Ea, the moment of inertia of aluminum alloy beam is Ia = Is × Es/Ea.

Thus, Ia = 3.15×106 m4 (.0000011 x 203/71).

The same result could be obtained using the “Constraint” portion of the chart:

 Determining Aluminum Section Area

Returning to the “Shape” portion of the chart, determine the range of section areas for this Ia. It is between 1.2×10-3 and 3.7×10-3 m2.

Nominal Weight of Aluminum

Using the 2,640 kg/m3 density value for aluminum found earlier, the nominal weight of the required aluminum beam is then between 3.2 and 9.8 kg/m (or 2.2–6.6 lb/ft).

In one quick search, and use of Knovel’s interactive tools, the engineer was able to determine that aluminum could indeed replace the steel structure.

The expected reduction of the weight of the beam resulting from steel being replaced with aluminum is 1.2–2.8 kg/m. It will be 22–27% of the original weight.

The stress on concrete foundation will decrease due to the lower weight of the structure, resulting in a longer lifespan. Additionally, the structure made from aluminum will last longer and require less maintenance compared to steel, due to reduced corrosion.

In building commercial aircraft, it is imperative for engineers to be precise in their work. Even the smallest error can become a costly failure. Engineers in the Aerospace industry need reliable sources to find effective design solutions for their projects.

This problem comes from a structural engineer at a leading aerospace company. He used information found on Knovel to fix a defect in the airplane frame caused by improper repair.

A double hole was drilled in the web of the spoiler beam of an airplane instead of a single 3/16″ hole required to install a bolt. In an attempt to fix the problem, the mechanic cut a rectangular slot opening (see Fig. 1) to install filler and re-drill the hole.

Fig. 1 Rectangular slot in the spoiler web of an airplane

This solution causes high stress concentration in the sharp corners of the opening and could cause cracking of the beam in service. The structural engineer had to fix the defect properly.

Next Steps:

1. Propose a new shape for the opening to reduce stress concentration
2. Find an equation for the stress concentration factor for the new opening
3. Calculate the stress concentration factor to make sure its value is acceptable

Researching the Shape of the Opening to Reduce Stress Concentration:

The first question the engineer wants to answer is: What is the stress concentration factor for the beam with a single circular hole as specified? He starts by searching Knovel for ‘circular hole’ (Click to run search):

On the search results page, he opens Section 4.3.1 in Peterson’s Stress Concentration Factors (3rd Edition) and navigates to page 181 to read that in the “ideal” case of a circular hole in a panel of infinite width under uniaxial in-plane tension the stress concentration factor is 3:

However, in his case the beam at the point of the hole is in a state of pure shear. After flipping a couple of pages, the engineer reads on pages 186-187 that the stress concentration factor for pure shear in his case is 4 (see below). He will use this value as a base for further calculations.

Propose a new shape for the opening to reduce stress concentration

Now, the engineer has to find a shape for the opening that would lower the stress concentration as close as possible to the base level. The engineer knows that the stress concentration will decrease with increasing radii in the corners of the opening.

The maximum effect would be accomplished by having an opening with 2 semicircles on each end as shown in Fig. 2. The closest hole that approximates this shape is elliptical. The engineer searches Knovel for ‘elliptical hole‘ (Click to run search):

On the search results page, the engineer opens Section 4.4.3 of Peterson’s Stress Concentration Factors (3rd Edition) and navigates to page 220. It contains a case for an elliptical hole with pure shear stress. The element in his case can be assumed to be infinite because the dimensions of the opening are small in comparison to the dimensions of the beam.

At the point of the opening the beam is in a state of pure shear as shown in Fig. 4.25a above, and behaves as an elliptical hole under biaxial principal stresses as shown in Fig. 4.25b.

Find an equation for the stress concentration factor for the new opening

The stress concentration factor for this case is calculated as shown on page 223. Applying the values to the equation, the result is:

The opening has the following dimensions (see Fig. 2): a=0.3225 in b=0.154 in

Fig. 2 Dimensions of the proposed opening

Thus, the stress concentration factor is:

Calculate the stress concentration factor to make sure its value is acceptable

This value is greater than that for the circular hole by 14% (4.57 vs. 4) but was judged to be covered by margin of safety for durability.

An equation for the stress concentration factor was found and used to calculate this factor for the opening of given dimensions.

Based on the calculation, a solution was proposed and implemented as shown in Fig. 3 below. The rectangular opening was reshaped and filled with radial filler and a new hole was drilled at the lower end of the filler. The filler was secured with a radius block and another part on the opposite side of the beam.

Fig. 3 Filled opening and new hole

This problem comes from a marine engineer. As a part of the modernization plan at the shipyard where he works, the engineer needs to order new welding equipment. He decides to use Knovel to find images about advanced welding technologies suitable for marine applications.

The shipyard uses large volumes of plates and T shapes to make decks, bulkheads, shells and other structural components. As the need for weight and cost reduction grows, T-beams are being manufactured with stronger materials and smaller cross-sections. These thin materials are subject to significant distortion as they are welded. Some estimates place the cost of distortion at 30% of the structural cost of the ship.

The engineer must research processes and equipment for the mechanized welding of T- beams and flat panels with supporting girders in a way that minimizes distortion.

After compiling his data, the engineer will report back with his findings. The report should contain examples of successful implementations of the selected technology in marine applications.

Next Steps:

Research mechanized welding technologies used in marine applications and select the technology that meets the requirements

Report his findings

Researching welding technologies

The engineer starts by searching Knovel for ‘automatic welding’ (Click to run search):

On the search results page, he opens Chapter 9.2.1.2 Automatic Welding with Cored Wires in Ship Construction (6th Edition). The selection (below) states that SAW (Submerged Arc Welding) is the most commonly used welding technology for mechanized welding in shipbuilding.

Browsing further, the engineer reads that LBW (Laser Beam Welding) and HLAW (Hybrid Laser Arc Welding) are new welding techniques applicable for shipbuilding applications.

HLAW is of special interest to the engineer because this technique is recommended for the welding of T- beams and girder-supported flat panels for decks and superstructures to minimize the distortion of metal of moderate thickness. It is also very important that this technology is suitable for a wide range of steels and aluminum alloys.

Learning more about HLAW and application examples:

Now, the task of the engineer is to find out more about the technology, i.e., process parameters (speed of welding, metal grades, range of thicknesses, forms, positions, metal preparation, gaps, etc.), equipment, power consumption, advantages and disadvantages. The engineer searches Knovel for ‘hybrid laser arc welding’ (Click to run search):

On the search results page, he opens Chapter 9.1 Introduction in Failure Mechanisms of Advanced Welding Processes and navigates to pages 219-223 to read more about laser welding and HLAW.

Click here to read the engineer’s findings.

Going back to the search results page, the engineer opens another selection, Hybrid Laser-Arc Welding, to learn more about the implementation of HLAW technology to the welding of high-strength steel, Zn-coated steel, stainless steel, aluminum alloys and even dissimilar metals. He continues to browse this selection and finds a wide range of examples of successful application of HLAW in shipbuilding at shipyards in different European countries.

Finally, at the end of this chapter, he reads:

Searching for welding processes on Knovel, the engineer finds that:

1. LBW (Laser Beam Welding) is not suitable for specified applications. The most promising new welding technology that meets his shipyard’s requirements is HLAW (Hybrid Laser Arc Welding).
2. It is desirable to implement a more efficient Nd:YAG or fiber laser than CO2 laser, though the latter was also implemented successfully.
3. HLAW technology has the following advantages: good penetration, very high speeds, ceramic backing is not required for butt-welding, and most importantly, minimum distortion due to high concentration of heat and narrow melting zone.
4. HLAW is suitable for welding of different steel and aluminum grades. It could be used in butt and T welding. It can also be used to make metallic sandwich panels of different types.
5. Successful implementation of HLAW technology at leading European shipyards was reported.

On the basis of information gathered from Knovel, the engineer recommends the implementation of HLAW technology at his shipyard. His next steps are to start contacting suppliers and write the report.

We strive to develop use cases that reflect the everyday problems faced by engineers. Tell us how Knovel has helped you solve a problem and if your story is selected, you will be awarded a new iPad.

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:

1. The part failure was due to stress concentration at the shoulder
2. There were no problems with materials or dimensions

Next Steps:

1. Research best design of bicycle axle
2. Design and test the design to ensure it will increase stress resistance

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.

(Click to enlarge)

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.

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:

1. The length of the OD 2-7/8, ID 2.151 drill pipe is 3240 ft
2. The depth of the 4 ¾” hole is 250 ft

Next steps:

1. Calculating the total weight of the pipe
2. Calculating* the weight of the cement required for a 250-ft plug in a 4 ¾” hole

*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.

Calculating the weight of the pipe

We start with a search for ‘drill pipe properties’ (Click to run search)

and open the Standard Handbook of Petroleum and Natural Gas Engineering (2nd Edition), 4.6.2 Drill Pipe, Table 4.6.8 Dimensional Properties of API Drill Pipe Tubes:

(Click to enlarge)

From this table, the nominal weight of the pipe is 10.40 lb/ft resulting in the total weight of 3240 ft × 10.40 lb/ft = 33,696 lb.

Calculating the weight of the cement

Now let’s find the equations for calculating the number of cement bags needed to make the plug.

A quick search for ‘cement plug’ (Click to run search)

retrieves the chapter Setting a Balanced Plug in Formulas and Calculations for Drilling, Production, and Workover (2nd Edition).

There is an equation for calculating the hole or casing capacity on page 61:

Plugging in our own value for the hole size,

Hole Capacity = (4 ¾”)² /183.35 = 0.1231 ft³/ft.

Now, we will use the formula below to out how many bags (sacks) of cement we will need.

Before we continue, let us quickly look at an example calculation:

We’ll begin by calculating the hole or casing capacity.

Then, we simply input all our values into the formula for determining the number of cement bags (sacks).

Unlike the example above, the slurry yield is not given to us. We will need to calculate this value ourselves.

By searching for ‘slurry yield‘ in Knovel, (Click to run search)

we find the desired formula in chapter 4.17.5 Properties of Cement Slurry and Set Cement, Standard Handbook of Petroleum and Natural Gas Engineering (2nd Edition).

Using Table 4.17.1 we find that the weight per bag (sack) of cement is 94 lbs and the Absolute Volume is 3.59 gal/bag.

By looking at Table 4.17.2, we find that there is 5.20 gal of water/bag (sack).

We can now use Formula 4.17.3 (above) to find out the slurry yield.

Our Calculations:

Slurry yield = (3.59+5.20)/7.48 = 1.175 ft³/bag

Gathering our Plug Length, Hole Capacity, Excess Allowance (25%), and Slurry Yield we can calculate the number of cement bags needed by plugging values into:

Bags (sacks) of Cement = 250 ft × 0.1231 ft ³/ft × 1.25 / 1.175 ft³/bag = 33 bags.

Weight of cement needed is 33 bags × 94 lbs (see Table 4.17.1) = 3102 lb.

Using Knovel, we were able to calculate the weight of materials needed for the well drilling operation:

1. The weight of 3240 ft of the OD 2-7/8, ID 2.151 drill pipe is 33,696 lb.
2. 33 bags (or 3102 lb) of cement needed for a 250-ft plug in the 4 ¾” hole.
   

We now know the transportation capacity required in order to transfer materials to the drilling site. We can schedule the delivery accordingly and reduce costs by bringing in only what we need

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:

1. What are the recommended dimensions of the shaft key?

2. Will the 1/4×1/4×1 in steel shaft key with destructive shear stress of 30,000 psi shear off and fully protect the shaft and the brush hub?

3. Is an aluminum or softer key needed to protect the driven hub in a locked-rotor condition with full torque impact load?

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:

1. The recommended dimensions of the key for the shaft of 11/16 in diameter are 3/16 × 3/16 in.
2. The 1/4 × 1/4 ×1 in steel key will not protect the hub and the shaft. It is stronger than these components.
3. A softer key material, e.g., 1060 H16 aluminum alloy, would protect the hub and the shaft.

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.

The engineer decides that he needs to determine the temperature of the steam at 2 atm and starts by searching Knovel for “steam tables”.

In Knovel Steam Tables, he opens Table 1a. Saturation Properties vs. Temperature – SI Units. Steam pressure in this table is given in MPa. Using the Unit Converter, the engineer determines that 2 atm are equal 0.2 MPa:

According to Interactive Table 1a. Saturation Properties vs. Temperature – SI Units, the temperature of steam at 0.2 MPa is 120 ºC:

His next step involves a search on Knovel for ‘cantilever beam flexure temperature AND mathcad’

he opens Roark’s Formulas for Stress and Strain (7th Edition)[Mathcad-Enabled], and reviews Interactive Table 8.1 Shear, Moment, Slope, and Deflection Formulas for Elastic Straight Beams.

In the text, case 6a reflects the given boundary conditions. The engineer opens the Mathcad file with Metric units to begin his calculations (click to view a larger image):

Now, he needs to change the predefined input values available in the Knovel Math file to match his requirements. To find the thermal expansion coefficient of copper, he uses Data Search, a search tool that acts as a shortcut and only retrieves numeric and other tabular data contained in Knovel’s interactive graphs, equations and tables (click to view a larger image):

Woldman’s Engineering Alloys (9th Edition) comes up in the results as having the highest relevancy so he opens Interactive table Table 2. Physical Constants of Principal Alloy-Forming Elements and finds that thermal expansion coefficient of copper is 16.8×10-6 1/ºC:

The same table shows the Young’s modulus of copper as being equal to 16×106 psi. The engineer used the Unit Converter again to convert this value to the required SI units.

As his final step, the engineer needs to finish gathering the calculations needed to design the temperature triggger and calculate the moment of inertia. He browses Roark’s and expands Appendix A: Properties of a Plane Area, and opens Interactive table Table A.1 Properties of Sections.

He quickly finds the desired Mathcad worksheet:

Substituting the variables, he can calculate the moment of inertia for his case:

With the completion of this calculation, the engineer can now input all of the variables in the original case and then and solve it using the following equations:

The results of his work are presented in the tabular form as well as in the graph form shown below:

Using Knovel’s Interactive Tables, Data Search and Knovel Math, the engineer was able to gather all the information and calculations needed to fully design the temperature trigger for a given application.

What We Know:

The shop has a 30-ton press brake. Two V-dies are available with 2” and 1” openings, respectively. The sheet metal is 1/8″-thick AISI 4130 low-alloy steel. The engineer wants to calculate the bending capacity of the break for both dies in order to figure out the optimal cutting pattern for the sheet metal during fabrication.

After not finding a suitable calculation procedure in the manuals he has at-hand, the engineer turns to Knovel.

He begins by entering the search term ‘bending capacity’

The engineer finds an equation for calculating the press capacity along with a calculation example in the title ASM Handbook, Volume 14B – Metalworking: Sheet Forming in section 30.4.1 Sample Calculation.

To use the equation, the engineer has to know the tensile strength of AISI 4130 steel so he looks up ”AISI 4130 tensile strength”

And finds the desired value in an Interactive Table – Design Mechanical Properties of Military Handbook – MIL-HDBK-5H: Metallic Materials and Elements for Aerospace Vehicle Structures (Knovel Interactive Edition):

He now knows that the tensile strength of the material is 75 ksi  and converts 75 ksi into tsi as required by the equation he found earlier. Using Knovel’s Unit Converter, he uses the result of the conversion: 5400 tonf/ft² and divides it by 144. The final result is 37.5 tsi and he can now use the equation:

The engineer rewrites equation 1 to calculate the maximum length of the bend:

The die opening of 2″ is equal to 16t, where t is metal thickness. For a 16t opening, the die-opening factor k is 1.2. Thus, the bend length, l, is

If the die opening is 1″ or 8t, the factor k is 1.33. Then, the bend length is

Using the ASM Handbook and other Knovel resources, the engineer was able to calculate the maximum bend length for a steel sheet for 1” and 2” die openings. He then used this information to determine the optimal dimensions of the sheet to be bent and specify the appropriate die for fabrication.

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.

Calculating the Flow Rate

The Engineer starts by searching ‘pr valve and gas and sizing’.

(Click the search box image to execute the search and follow along)

An equation for the flow rate is found in the title Pressure Safety Design Practices for Refinery and Chemical Operations, on p. 182 of Section 8.5.2 Sizing for Vapor – Critical Flow:

* For the purposes of this example, natural gas is assumed to be pure methane.

He knows the inlet parameters for the valve are P1, T1 and that the orifice is diameter (D1). They are 3000 psia, 150ºF and 1 in (25.40 mm), respectively. A can be calculated as π x D12/4. However, he still needs to determine M and.

To make this determination, he needs to find the molecular weight of methane. He uses Data Search starting with the Search query (material or substance name  =  methane)  and (molecular weight EXISTS)

This search retrieves a number of highly relevant references, among them Knovel Critical Tables . He opens an interactive table called Physical Constants and Thermodynamics of Phase Transitions (highlighted):

The table contains 137 hits for methane and its derivatives. To hone in on methane, he uses the Filter, matching on the Entire Cell:

and finds the molecular weight of methane to be equal 16.04:

(click to view details)

His next step is to find and calculate the compressibility factor of Methane.

The compressibility factor, = z, is found by performing a search on ‘compressibility factor’

and opening the title Section 2.5 of Natural Gas Engineering Handbook on p. 22.

The compressibility factor can be calculated using Brill-Beggs equations below:

However, he chooses to work with the data right away by using the interactive Excel spreadsheet on p. 23 instead. He opens this interactive spreadsheet by clicking within the blue frame of the PDF page:

(click to work with the interactive spreadsheet)

After inserting the given pressure and temperature values, the compressibility factor is calculated to be 0.8329.

His last step is to calculate the Mass Flow Rate for natural gas. The mass flow rate of natural gas is calculated using equation 4 as follows:

The flow rate was calculated using Mathcad 14 but can be calculated using Excel or a scientific calculator. Note that the symbol for specific heat ratio was changed from K to g (K stands for degrees Kelvin in Mathcad) and that for compressibility factor -
from to z. The Ф-parameter was introduced into equation 4 to allowing him to readily change the inlet flow parameters (pressure, temperature and orifice area).

The Engineer was able to find a practical solution for pressure relief valve sizing based on the calculated flow rate. He is now free to choose a manufacturer from a potential list of suppliers of bellows-type pressure relief valves at his disposal.