HP 35s User Manual

							 
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HP 35s  Applications in Mechanical Engineering 
 
 
 
 
Applications in mechanical engineering 
 
Practice solving problems in mechanical engineering 
 
- Application 1: Stress on an element (Mohr circle) 
 
 
   
						

							 
hp calculators 
 
HP 35s  Applications in Mechanical Engineering 
 
hp calculators - 2 - HP 35s  Applications in Mechanical Engineering - Version 1.0 
 
Applications in mechanical engineering 
 
This training aid will illustrate the application of the HP 35s calculator to several problems arising in mechanical 
engineering. These examples are far from exhaustive, but do indicate the incredible flexibility of the HP 35s calculator.  
Practice solving problems in mechanical engineering  
Application 1: Stress on an element (Mohr circle)  
 The Mohr circle equations convert an arbitrary stress configuration to principal stresses, maximum shear 
stress, and rotation angle. It is then possible to calculate the state of stress for an arbitrary orientation !'. 
 Figure 1  
 The Mohr circle formulas are shown below, where s is the normal stress,  is the shear stress, sx is the 
stress in the x direction for Mohr circle input, sy is the stress in the y direction for Mohr circle input, xy is the 
shear stress on the element for the Mohr circle input, s1 and s2 are the principal stresses, ! is the rotation 
angle, and max is the maximum shear stress. Note that ! is the angle of rotation from the specified axis to 
the principal axis, and so should be thought of as a negative angle. This is opposite to the normal Mohr 
circle convention.  
Maximum shear stress  2xy
2
yxmax2
s-s  #$$
%
&

(
)* Figure 2 
 
Principal stress  maxyx12
ss s ##*  Figure 3 
 
Principal stress  maxyx22
ss s +#*  Figure 4 
 
Rotation angle  $$
%
&

(
)
+*
yx
xy1-
ss
2 tan2
1 !  Figure 5 
 
Example: If sx = 25,000 psi, sy = -5,000 psi, and !xy = 4,000 psi, compute the principal stresses, the angle of  
 rotation !, and the maximum shear stress.   
Solution: Solve for the maximum shear stress, max.   
						

							 
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HP 35s  Applications in Mechanical Engineering 
 
hp calculators - 3 - HP 35s  Applications in Mechanical Engineering - Version 1.0 
 
 
 In RPN mode: !###$###%&!()*###+
++++(),-(./+ 
 In algebraic mode:++-()4!###&###%0!0+
++++,()*###$(./  
 Figure 6  
 Solve for the principal stress, s1.  
 In RPN mode: !###$###%,!1/,  
 In algebraic mode:++4!###,###%0!,1/$+
+++++
 Figure 7  
 Solve for the principal stress, s2.  
 In RPN mode: !###$###%,!1/&  
 In algebraic mode:++4!###,###%0!&1/$  
 Figure 8  
 Solve for the rotation angle, !.   
 In RPN mode: !$*###2!###$###%+
++++&(3!  
 In algebraic mode:++(3!2*###4!###&###%+
++++00!$+ 
 Figure 9   
Answer: The principal stresses, s1 and s2, are 25,524 psi and -5,524 psi. The angle of rotation, !, is 7.4657 degrees. 
The maximum shear stress is 15,524 psi.  
   
						

							 
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HP 35s  Applications in Medicine 
 
 
 
 
Applications in Medicine 
 
Practice solving problems in medicine 
 
- Application 1: Beer’s Law 
 
- Application 2: Body Surface Area (BSA) 
 
- Application 3: Schilling Test for  
  Vitamin B12 Absorption 
 
 
 
   
						

							 
hp calculators 
 
HP 35s  Applications in Medicine 
 
hp calculators - 2 - HP 35s  Applications in Medicine - Version 1.0 
 
Applications in Medicine 
 
This training aid will illustrate the application of the HP 35s calculator to several problems arising in medicine. These 
examples are far from exhaustive, but do indicate the incredible flexibility of the HP 35s calculator. 
 
Practice solving problems in medicine 
 
Application 1: Beer’s Law 
 
 Beers law is a physical law which states that the quantity of light absorbed by a substance dissolved in a 
non-absorbing solvent is directly related to the concentration of the substance and the path length of the 
light through the solution. Beers Law describes how the intensity of light diminishes as it passes through 
an absorbing media. For many colored materials there is a linear relationship between a property of the 
materials called optical density and the concentration of the colored species. The linear relationship is 
called Beers Law. The formulas needed to solve Beer’s Law are shown below. 
 Figure 1 
 Figure 2 
 Figure 3  
 In these formulas, T is the percent transmittance, A is the absorbance (optical density) of the substance,  
 Au is the absorbance of the unknown, As is the absorbance of the standard, Cu is the concentration of the  
 unknown substance, and Cs is the concentration of the standard. 
 
Example: A standard solution with a solute concentration of 2 mg/ml is found to have an absorbance of 0.41 at 550  
 nm. An unknown from a patient is found to show 46% transmittance at the same wavelength. Convert this  
 percent transmission to absorbance. Also find the solute concentration in the unknown. 
 
Solution: Solve for the absorbance of the unknown. Note that T is entered as the percent multiplied by 100. 
 
 In RPN mode: !#$%& 
 
 In algebraic mode:((!%&#$ 
(
 Figure 4 
 
 Then solve for the solute concentration of the unknown. Note that the absorbance of the unknown is  
 available in the calculator’s display to continue the calculation.   
						

							 
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HP 35s  Applications in Medicine 
 
hp calculators - 3 - HP 35s  Applications in Medicine - Version 1.0 
 
 In RPN mode: )*#+,!- 
 
 In algebraic mode:((,)*#+-! 
 
 Figure 5 
 
Answer: The absorbance of the unknown is 34% and the solute concentration of the unknown is 1.65 mg/ml.  
 Figures 4 and 5 indicate the display in RPN mode. 
 
Application 2: Body Surface Area (BSA) 
 
 There are two primary methods used to estimate body surface area, the Dubois method and the Boyd 
Method. Each method uses inputs of a patient’s height and weight in metric units and estimates the 
patient’s BSA.  
 
 Dubois’ formula is shown below in Figure 6 and requires input of the height in centimeters and the weight in 
kilograms. Note that Dubois’ formula is undefined for children with a BSA less than 0.6 m2. Boyd’s formula 
should be used in these situations. 
  
BSA(m2) = Ht 0.725 x Wt 0.425 x 0.007184   Figure 6 
  
 Boyd’s formula is shown below in Figure 7 and requires input of the height in centimeters and the weight in 
grams. 
  
BSA(m2) = Wt (0.7285 – 0.0188 x LOG( Wt )) x Ht 0.3 x 0.0003207 Figure 7 
 
Example 1: A patient is 176 centimeters tall and has a weight of 63.5 kilograms. What is the patient’s BSA using the 
 Dubios formula? What is the patient’s BSA using Boyd’s formula? 
 
Solution: Solve for the BSA using Dubios’ formula. Figure 8 shows the answer in algebraic mode. 
 
 In RPN mode: +.$)*.!/0(
(((($1*/)*#!/0-(
(((()*)).+2#- 
 
 In algebraic mode:((+.$0)*.!/-(
(((($1*/0)*#!/-(
(((()*)).+2# 
 
 Figure 8   
						

							 
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HP 35s  Applications in Medicine 
 
hp calculators - 4 - HP 35s  Applications in Medicine - Version 1.0 
 
 Now solve for the BSA using Boyd’s formula. Note that Boyd’s formula requires the weight to be input  
 in grams. Figure 9 shows the answer in RPN mode. 
 
 In RPN mode: )*.!2/)*)+22(
(((($1/))%&345-0(
((((+.$)*10-)*)))1!).- 
 
 In algebraic mode:(($1/))04)*.!2/(
(((()*)+22-%&$1/))66-(
((((+.$0)*1-)*)))1!). 
 
 Figure 9 
 
Answer: Dubois’ method estimates a BSA of 1.78 m2, while Boyd’s method estimates a BSA of 1.76m2. 
 
Example 2: A patient is 176 centimeters tall and has a weight of 63.5 kilograms. What is the patient’s BSA using Boyd’s 
formula? Solve the problem by entering Boyds formula as an equation. 
 
Solution: To enter Boyd’s formula into the calculator as an equation, press the following keys on the HP 35s: 
 
(789%:8;04)*.!2/)*)+22-%&
8;66-89 
 
 The HP 35s SOLVER displays the first variable encountered in the equation as it begins its solution, in this 
case the variable W. The value of 0.0000 is displayed below if this is the first time the equation has been 
solved on the HP 35s calculator. If any previous equations have used this variable, it will display the value 
presently held in the variable. Enter the value of W.   
						

							 
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HP 35s  Applications in Medicine 
 
hp calculators - 5 - HP 35s  Applications in Medicine - Version 1.0 
 
  
 Figure 12 
 
(In either RPN or algebraic mode, press:  $1/))? 
(
 Figure 13 
 
(In either RPN or algebraic mode, press:  +.$? 
 
 Figure 14 
 
Answer: Boyd’s method estimates a BSA of 1.76m2. Additional problems can be solved using this equation, if 
desired. Figure 14 displays the result. 
 
Application 3: Schilling Test for Vitamin B12 Absorption 
 
 The Schilling Test determines the amount of a radioactive vitamin B12 intake that is absorbed. The equation 
for this calculation is shown in Figure 10. 
 
  Figure 15 
  
 where V is equal to 1 if the urine volume is less than or equal to 1 liter, or equal to the actual urine volume if 
greater than one liter. If the urine volume is less than 1 liter, it is assumed to have been brought up to 1 liter 
by the addition of water (and is indicated by DIL, the dilution of the standard). The background (BCPM), 
standard (SCPM) and urine (UCPM) counts per minute should be of equal volumes counted over equal 
time intervals (which need not be one minute). The patient being tested should not have received recent 
prior radioactivity. 
 
Example 1: A capsule of radioactive vitamin B12 is administered orally to a patient. Over the following 24 hours, a 
volume of 2.54 liters of urine is collected. A 20 milliliter sample of the urine is counted for 10 minutes to give 
1923 counts. A 1 milliliter of standard is diluted to 20 milliliters and counted for 10 minutes, giving 1757 
counts. 20 milliliters of tap water is used for a background count and over a 10 minute time interval 
produces 127 counts. Find the percent of the dose excreted.   
						

							 
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HP 35s  Applications in Medicine 
 
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Solution: V is 2.54 liters, DIL is 20, the Urine CPM is 1923, the standard CPM is 1757, and the background CPM  
 is 127. 
 
 In RPN mode: !*/#!),+@!1+!.(
((((+./.+!.,-+))-(
 
 In algebraic mode:((!*/#,!)-(
((((44+@!1+!.6,(
((((4+./.+!.66-+)) 
 
 Figure 16 
  
Answer: The amount excreted is 13.99%. The amount absorbed is 86.01%. Figure 16 shows the answer in  
 algebraic mode. 
 
Example 2: A capsule of radioactive vitamin B12 is administered orally to a patient. Over the following 24 hours, a 
volume of 2.54 liters of urine is collected. A 20 milliliter sample of the urine is counted for 10 minutes to give 
1923 counts. A 1 milliliter of standard is diluted to 20 milliliters and counted for 10 minutes, giving 1757 
counts. 20 milliliters of tap water is used for a background count and over a 10 minute time interval 
produces 127 counts. Find the percent of the dose excreted. Solve the problem by entering the formula as 
an equation. 
 
Solution: V is 2.54 liters, DIL is 20, the Urine CPM is 1923, the standard CPM is 1757, and the background CPM  
 is 127. 
 
 To enter the formula into the calculator as an equation, press the following keys on the HP 35s: 
 
(78A%:8B,8B-448C896,48D
8966-+)) 
 
 Figure 17 
 
 To verify proper entry of the equation, press 
 
(%= 
 
 and hold down the =key. This will display the equation’s checksum and length. The values displayed 
should be a checksum of BF1C and a length of 23, as indicated in Figure 18. 
 
 Figure 18 
  
 Release the =key and the display will return to the one shown in Figure 17. Now press:(