Program XBEAM for Android Smartphones. April 19, 2014. By Prof. Frank Rieg, Chair for Engineering Design and CAD, Fakulty for Engineering Science, University of Bayreuth, Germany.

This is a small and fast – yet quite powerfull – program for lectures and training courses in elastostatics. The program is based on an old TI59 program from 1980, which was designed by Prof. Otto Ewald and Dipl.-Ing. F. Merlino. XBEAM is the English version of already released German program XBALKEN (eng. Beam = ger. Balken).

 

Do not compute safety-relevant structures and do not work with XBEAM if you are not well-trained in elastostatics!

 

XBEAM works with the transfer matrix method for plane beams, see Decker:Maschinenelemente. 18th edition, Carl Hanser Verlag: Munich 2011 (in German language). The beam can be arbitrarily statically over-determined, and may have spring supports and so-called Gerber joints. For extensive frameworks in space and for continuum elements I recommend our freeware FEA program Z88Aurora from our University institute (www.z88.de).

 

Procedure: Always do two passes. During the first pass enter the left support, all sections of beams or shafts, single loads as forces, bending moments, angles or deflections, intermediate conditions as joint supports or Gerber joints, and the right support. Just proceed from left to right while entering every feature. Second pass: Start again with the left support and enter all the features you did in the first pass. But now you may compute the reactions, i.e. deflection w, bending angle psi, bending moment M and shear force Q. The sections may now be shorter for computing intermediate results.

 

Hints:

1)       If you want to investigate another beam: Exit the program and re-start it in order to reset all internal values.

2)       After entering numerical values close the numeric keypad with the OK key of the numeric keypad before finishing the entering of numeric data by Okay of XBEAM. This gives you a better view.

3)       Enter a minus sign by double-clicking the .- key.

4)       Enter exponential numbers like 2.06x105 as normal floating-point values, e.g. 206000 .

5)       Warning: There is no Undo, hence, you cannot correct wrong entries – only by Exit and a new start. Thus, you must work accurately.

6)       Cancle this help by another push onto Manual.

 

It could be a good idea to work through the following examples:

 

Example 1: statically over-determined shaft. d1 to d4 = 80 mm, F= 33218 N, E= 210000 N/mm2

 

1.Pass: Data Entry

1) Left Support: Free , Okay

2) Single Load: Force, Value of Load -33218, Okay

3) Shaft Section: Length 150, Young’s M. 210000, Diameter 80, Distrib. Load no, Okay

4) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support, Okay

5) Shaft Section: Length 240, Young’s M. 210000,  Diameter 80, Distrib. Load no, Okay

6) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support, Okay

7) Shaft Section: Length 200, Young’s M. 210000,  Diameter 80, Distrib. Load no, Okay

8) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support, Okay

9) Shaft Section: Length 65, Young’s M. 210000,  Diameter 80, Distrib. Load no, Okay

10) Inter.Cond./Right Sup.: Right Support, Okay, Free, M  0, Q  33218, Okay

 

2.Pass: Data Entry + Calculation (results in parts approximated)

11) Left Support: Free , Okay

12) Single Load: Force, Value of Load -33218, Okay

13) Compute: w = -0.21778,   psi=  0.001747,   M=  0,   Q=  33218

14) Shaft Section: just enter half of the length to get intermediate results:

Length 75, Young’s M. 210000, Diameter 80, Distrib. Load no, Okay

15) Compute: w= -0.092295,   psi=  0.001526,   M=  2491350,   Q=  33218

16) Shaft Section: go now 75 ahead, thus, x is now 150: Length 75, Young’s M. 210000, Diameter 80, Distrib. Load no, Okay

17) Compute: w= 0,   psi= 0.000862,   M= 4982700,   Q= 33218

Q is correct, because the support is not yet entered!

18) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support, Okay

19) Compute: w= 0,   psi= 0.000862,   M= 4982700,   Q= -24378.7

Damit wird die LagerForce am 1.Lager (=Sprung in der Q-Linie!): Q17) + Q19) = 33218+24379=57597

20) Shaft Section: Length 240, Young’s M. 210000, Diameter 80, Distrib. Load no, Okay

21) Compute: w= 0,   psi= -0.000308,   M= -868198,   Q= -24378.7

22) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support, Okay

23) Compute: w=  0.,   psi= -0.000308,   M= -868198,   Q  = -6454.86

Thus becomes the support force for the 2nd support (=leap in the shear diagram!): Q21) + Q23) = 24379-6455=17924

24) Shaft Section: Length 200, Young’s M. 210000, Diameter 80, Distrib. Load no, Okay

25) Compute: w= 0,   psi=  0.000409,   M= -2159170,   Q  = -6454.86

Oh, we did not want to go so far, but we wanted to calculate the results for half of the length between the 2nd and the 3rd support. No problem. Just go back on the  x axis:

26) Shaft Section: Length -100, Young’s M. 210000, Diameter 80, Distrib. Load no, Okay

27) Compute: w= -0.017925,   psi= -0.0000255,   M= -1513683.9,   Q  = -6454.86

and again to the right:

28) Shaft Section: Length 100, Young’s M. 210000, Diameter 80, Distrib. Load no, Okay. This must deliver the same results as for 25)...

29) Compute: w= 0,   psi=  0.000409,   M= -2159170,   Q  = -6454.86

30) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support, Okay

31) Compute: w= 0.,   psi=  0.000409,   M= -2159170,   Q  =  33218

Thus becomes the support force for the 2nd support (=leap in the shear diagram!): Q29) +Q31) = 6455+33218=39673

32) Shaft Section: Length 65, Young’s M. 210000, Diameter 80, Distrib. Load no, Okay

33) Inter.Cond./Right Sup.: Right Support, Okay, Free, M  0, Q  33218, Okay

Warning: Do not enter the force via „Single Load“! Enter F in this step. Quite similar you would enter a bending moment of e.g. 30000 at the right support: Right Support, Free, M=30000, Q=0.

34) Compute: w= 0.033816,   psi= 0.000576,   M= 0,   Q= 33218

Hint: You could have ommited the step 33); You could have calculated the results directly after step 32). This was only done here for training purpose.

 

Example 2: Continuous Beam, two times statically over-determined. q0 = 10 N/mm, EI= 2.1x1013 Nmm2 , l= 6000 mm

Literature: Gross/Schnell: Formel- und Aufgabensammlung zur Technischen Mechanik II (S.97) (in German language)

 

1.Pass:

1) Left Support: Clamping

2) Beam Section: Length 3000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 10, qs 0, Okay

3) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support, Okay

4) Beam Section: Length 3000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 10, qs 0, Okay

5) Inter.Cond./Right Sup.: Right Support, Okay, Joint Support, M 0

 

2.Pass (Rechenergebnisse t.w. gerundet):

6) Left Support: Clamping

7) Compute: w= 0,   psi= 0,   M= -6428571,   Q= 13928.6

8) Beam Section: Length 3000, Young’s M. 210000, Moment of Inertia 100000000,Distrib. Load yes, Okay, q 10, qs 0, Okay

9)  Compute: w= 0,   psi= 0.000076531,   M= -9642857,   Q= -16071.4

10) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support, Okay

11) Compute: w= 0.,   psi= 0.000076531,   M= -9642857,   Q= 18214.3

Thus becomes the support force for the 2nd support (=leap in the shear diagram!): Q9) + Q11)= 16071+18214=34285

12) Beam Section: Length 3000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 10, qs  0, Okay

13) Compute: w= 0,   psi= -0.000306,   M= 0,   Q= -11785.7

14) Inter.Cond./Right Sup.: Right Support, Okay, Joint Support, M  0

15) Compute: w= 0,   psi= -0.000306,   M= 0,   Q= -11785.7

(this step must deliver the same results as step 13); thus, you may omit the right support in the 2nd pass).

 

Example 3: Gerber Carrier = Beam with pin-joints so that the beam is always statically determined. Invented by Heinrich Gottfried Gerber, a German engineer. The most famous Gerber carrier? The Forth Bridge over the Firth of Forth in the east of Scotland. Data for this example: q0 = 10 N/mm, EI= 2.1x1013 Nmm2 , a= 3000 mm

Literature: Gross/Schnell: Formel- und Aufgabensammlung zur Technischen Mechanik I (S.117) (in German language)

 

1.Pass:

1) Left Support: Clamping

2) Beam Section: Length 3000, Young’s M. 210000, Moment of Inertia 100000000,Distrib. Load no

3) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Gerber Joint, Okay

4) Beam Section: Length 3000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 10, qs 0, Okay

5) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support, Okay

6) Beam Section: Length 3000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 10, qs 0, Okay

7) Inter.Cond./Right Sup.: Right Support, Okay, Free, M  0, Q  5000, Okay

 

2.Pass:

8) Left Support: Clamping

9) Compute: w= 0.,   psi= 0.,   M=  15000000,   Q= -5000

10) Beam Section: Length 3000, Young’s M. 210000, Moment of Inertia 100000000,Distrib. Load no

11) Compute: w= -2.14286,   psi= -0.001071,   M=  0,   Q= -5000

12) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Gerber Joint, Okay

13) Compute: w= -2.14286,   psi= -0.000179,   M= 0,   Q  = -5000

14) Beam Section: Length 3000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 10, qs 0, Okay

15) Compute: w= 0,   psi= 0.003036,   M= -60000000,   Q= -35000

16) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support, Okay

17) Compute: w  = 0,   psi= 0.003036,   M  = -60000000,   Q  = 35000

Support force for the 2nd support: Q15) + Q17)  = 35000+35000=70000

18) Beam Section: Length 3000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 10, qs 0, Okay

19) Compute: w= 16.0714,   psi= 0.00625,   M= 0,   Q= 5000.

20) Inter.Cond./Right Sup.: Right Support, Okay, Free, M  0, Q  5000, Okay

20) Compute: w= 16.0714,   psi= 0.00625,   M= 0,   Q= 5000

 

Example 4: Gerber Carrier. q0 = 50 N/mm, EI= 2.1x1013 Nmm2 , a= 6000 mm, l= 10000 mm

 

Literature: Gross/Schnell: Formel- und Aufgabensammlung zur Technischen Mechanik I (S.116) (in German language)

 

1.Pass:

1) Left Support: Clamping

2) Beam Section: Length 6000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 0, qs  0.005, Okay

3) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Gerber Joint, Okay

4) Beam Section: Length 4000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 30, qs  0.005, Okay

5) Inter.Cond./Right Sup.: Right Support, Okay, Joint Support,  M=0

 

2.Pass:

6) Left Support: Clamping

7) Compute: w= 0,   psi= 0,   M=  -800000000,   Q= 163333

8) Beam Section: Length 6000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 0, qs  0.005, Okay

9) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Gerber Joint, Okay

10) Beam Section: Length 4000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 30, qs  0.005, Okay

11) Compute: w=  0,   psi= -0.11045,   M=  0,   Q= -86666.7

Example 5: Carrier. Moment = 100000 Nmm, E= 206000 N/mm2, l= 500 mm, d= 20 mm

Literature: Marguerre: Technische Mechanik Zweiter Teil (S.77) (in German language)

 

1.Pass:

1) Left Support: Joint Support

2) Single Load: Bend.Moment  -100000

3) Shaft Section: Length 500, Young’s M. 206000, Diameter 20, Distrib. Load no

4) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support

5) Shaft Section: Length 500, Young’s M. 206000, Diameter 20, Distrib. Load no

6) Inter.Cond./Right Sup.: Right Support, Okay, Joint Support,  M  100000

 

2.Pass:

7) Left Support: Joint Support

8) Single Load: Moment  -100000

9) Compute: w=  0,   psi=  0.007726,   M=  100000,   Q= -300

10) Shaft Section: Length 500, Young’s M. 206000, Diameter 20, Distrib. Load no

11) Compute: w=  0,   psi= 0,   M= -50000,   Q= -300

12) Inter.Cond./Right Sup.: Intermed.Condition, Okay, Joint Support

13) Compute: w  =  0,   psi= 0,   M= -50000,   Q= 300

Support force for the support B:  300+300=600

14) Shaft Section: Length 500, Young’s M. 206000, Diameter 20, Distrib. Load no

17) Compute: w=  0,   psi= -0.007726,   M=  100000,   Q=  300

 

Example 6: Carrier, F= 16667 N, EI= 2.1x1013 Nmm2 , a= 500 mm, q0= 50 Nmm, c=171818 N/mm

Literature: Göldner: Übungsaufgaben aus der Technischen Mechanik (S.52) (in German language)

 

1.Pass:

1) Left Support: Clamping

2) Beam Section: Length 1000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 50, qs  -0.05, Okay

3) Spring Support: c  171818

4) Beam Section: Length 500, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load no

5) Inter.Cond./Right Sup.: Right Support, Okay, Free, M  0, Q=16667

 

2.Pass:

6) Left Support: Clamping

7) Compute: w=  0,   psi=  0,   M= -8333413.6,   Q=  16666.58

8) Beam Section: Length 1000, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load yes, Okay, q 50, qs  -0.05, Okay

9) Compute: w=  0.145505,   psi=  0.000298,   M= -8333500,   Q=  -8333.42

10) Spring Support: c  171818

11) Compute: w=  0.145505,   psi=  0.000298,   M= -8333500,   Q=  16667

Thus becomes the spring force: Q9) + Q11) 8333+16667=25000

12) Beam Section: Length 500, Young’s M. 210000, Moment of Inertia 100000000, Distrib. Load no

13) Compute: w=  0.327387,   psi=  0.000397,   M= 0,   Q=  16667