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