Piper's Pit BBS/FTP: ibm 0020 - 0029/ibm0020-0029/ibm0028.tar/ibm0028/STATTUT.ZIP/S04.CHN

Turbo Pascal Chain module | 1987-10-26 | 28.0 KB | 206 lines

DATA AREA FOR BUILDEXE FOLLOWS
GBIGTURBO - Large Code Model. Copyright (c) 1985 by TurboPower Software.
"All Rights Reserved. Version 1.08D
0123456789ABCDEFLL
module:
 procedure:
 called from
module:  0
1Out of BIGTURBO stack space or no modules loaded.
3Attempting to call routine in uninitialized module.
3Invalid FarIncoming pointer or corrupt module code.
Program error - code
 in FarOutgoing.
Error while calling module:
, procedure:
 from module
F(BD)
F(AD)
F(AC)
 6 ft
 6 ft
 6 ft
F(CD)
F(DF)
F(AC)
 6 ft
 6 ft
F(CE)
F(CF)
F(DF)
 6 ft
 6 ft
F(CE)
F(EF)
F(FG)
 6 ft
CThe cantilever truss supports three loads as shown.  We propose to
Bdetermine the force in a member of our choice for a given loading
of the truss.
 6 ft
 6 ft
 6 ft
7\Enter the magnitude of each load (from 0 to 12 kips).\
MAGNITUDES
 kips
 kips
 kips
B\Select the member for which you want to find the force. (Members
@AB, EF, EG, and FG have been omitted since the force in each of
=these members may best be found by the method of joints) ~[=]
9`   1. AC   2. AD   3. BD   4. CD   5. CE   6. CF   7. DF
1/2/3/4/5/6/7/
YOUR SELECTION [1 through 7]
&We will solve for the force in member
>To do this, we shall take a section, that is, we shall draw a
>line through 3 members of the truss to isolate the portion of
(truss located to the right of that line.
Looking for F(
C&Select by number the members you wish to cut: ~[===] ~[===] ~[===]
N`1. AB  2. AC  3. AD  4. BD  5. CD  6. CE  7. CF  8. DF  9. EF  10. EG  11. FG
MEMBERS TO CUT [1 through 11]
/\Your section does not cut through the truss.
7Select a section which isolates a portion of the truss.
2\Your section does not pass through the member in
$which the force is to be determined.
0We shall remove the members which have been cut
>and draw the free-body diagram of the portion of truss to the
right of the section.
?\We wish to write an equation which contains _only the desired
force F(
:)._  To eliminate the other two unknowns, should we write
;an equation involving components or moments? ~[==========]\
3In what direction should we take the components to
$eliminate the extra unknowns?  ~[=]
YOUR ANSWER [X or Y]
5About what point should we compute the moments? ~[=]
A/C/D/E/G/
[A, C, D, E or G]
C/D/E/F/G/
[C, D, E, F or G]
C/E/F/G/
[C, E, F or G]
E/F/G/
[E, F or G]
YOUR ANSWER
;To facilitate the computation of the moment of F(CF) about
AG, we shall move that force along its line of action until it is
applied at point C.\
;To facilitate the computation of the moment of F(DF) about
AC, we shall move that force along its line of action until it is
applied at point D.\
    We write&
"We write the equilibrium equation&
 /+/-
~MD = 0:  ~[=] F(AC)( ~[===] ft) ~[=] (
)( ~[===] ft)`
~ ~ ~         ~[=] (
)( ~[===] ft) ~[=] (
)( ~[===] ft) = 0
SIGNS AND VALUES
sin/cos/tan/
Fy = 0:  F(AD) ~[===] ~
S for Sine OR C for COSINE
~MA = 0:  ~[=] F(BD)( ~[===] ft) ~[=] (
)( ~[===] ft)`
~ ~ ~         ~[=] (
)( ~[===] ft) ~[=] (
)( ~[===] ft) = 0
SIGNS AND VALUES
 /+/-
~MG = 0:  ~[=] F(CD)( ~[===] ft) ~[=] (
)( ~[===] ft)`
~ ~ ~         ~[=] (
)( ~[===] ft) ~[=] (
)( ~[===] ft) = 0
SIGNS AND VALUES
 /+/-
~MF = 0:  ~[=] F(CE)( ~[===] ft) ~[=] (
)( ~[===] ft)`
~ ~ ~         ~[=] (
)( ~[===] ft) = 0
SIGNS AND VALUES
~MF = 0:  ~[=] F(CE)( ~[===] ft) ~[=] (
)( ~[===] ft) = 0
SIGNS AND VALUES
 /+/-
~MG = 0: ~[=] F(CF) sin ~
( ~[===] ft) ~[=] (
)( ~[===] ft)
~[=] (
)( ~[===] ft) = 0
SIGNS AND VALUES
 /+/-
~MC = 0:  ~[=] F(DF) cos ~
 ( ~[===] ft) ~[=] (
)( ~[===] ft)`
~ ~ ~         ~[=] (
)( ~[===] ft) ~[=] (
)( ~[===] ft) = 0
SIGNS AND VALUES
~MC = 0:  ~[=] F(DF) cos ~
 ( ~[===] ft) ~[=] (
)( ~[===] ft)`
~ ~ ~         ~[=] (
)( ~[===] ft) = 0
SIGNS AND VALUES
!\We observe from the diagram that
tan/cos/sin/
"     ~[===] ~
  = (8 ft) / (6/ft)\
TRIGONOMETRIC FUNCTION
F(AC)
F(AD)
F(BD)
F(CD)
F(CE)
F(CF) sin ~
F(DF) cos ~
/It follows that ~
 = 53.1~
 and sin ~
 = 0.800.
(\Substituting into the equation, we have
&    F(AD)(0.800) -
 kips = 0           F(AD) =
 kips\
%We shall now solve this equation for

 kips

 = ~[======]
ANSWER
tan/cos/sin/
"We observe from triangle EFG that\
      ~[===] ~
 = (4 ft) / (6 ft)
"We observe from triangle CDG that\
!     ~[===] ~
 = (8 ft) / (12 ft)
TRIGONOMETRIC FUNCTION
~0It follows that ~
 = 33.7~
 and sin ~
 = 0.555.\
}"We have, therefore F(CF)(0.555) =
 kips, or F(CF) =
 kips\
}0It follows that ~
 = 33.7~
 and cos ~
 = 0.832.\
{"We have, therefore F(DF)(0.832) =
 kips, or F(DF) =
 kips\
The force in member
 is F(
 kips&
[zEWhat does the sign indicate?  Tension or compression? ~[===========]&
ly2By the way, did you know that you could have used
3y5a different section to find the force in this member?
x/&Do you wish to try a different section? ~[===]
x*Do you wish to try another example? ~[===]
-x*Do you wish to try another example? ~[===]
SETUPJUMPTABLE FOLLOWS
FARCALLHANDLER FOLLOWS