Piper's Pit BBS/FTP: ibm 0020 - 0029/ibm0020-0029/ibm0028.tar/ibm0028/STATTUT.ZIP/S05.CHN

Turbo Pascal Chain module | 1987-10-26 | 36.7 KB | 285 lines

DATA AREA FOR BUILDEXE FOLLOWS
GBIGTURBO - Large Code Model. Copyright (c) 1985 by TurboPower Software.
"All Rights Reserved. Version 1.08D
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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(AB)
F(AB)
F(KL)
F(KL)
F(KL)
F(KL)
F(AB)
F(AB)
F(KL)
F(KL)
F(KL)
F(KL)
F(AB)
F(KL)
F(KL)
=The frame shown consists of two vertical members AKC and BLD
<supported by pins and brackets at C and D and pin-connected
>to the links AB and KL.  Two horizontal loads (~P and ~Q) are
applied at A and K as shown.
>\We propose to find the reactions at the supports C and D and
4the forces exerted on member BLD by links AB and KL.
iSelect the magnitude and sense of each load (from 5 to 100 kN), specifying its sense, and the dimensions
(b (from 1 to 6 m) and c (from 0 to 6 m).
D\For the value of c you have chosen, member BLD becomes a two-force
Imember. While the method of analysis that we shall follow is still valid
$it is not the most efficient to use.
:For later reference, we shall determine the angle ~
 that
$member KL forms with the horizontal:
          ~     ~[======]  m `
     tan (~
          ~     ~[======]  m
VALUES
\     ~
Member/Frame/
?Shall we draw the free-body diagram of the entire frame (F) or
'of an individual member (M)?  ~[======]
YOUR ANSWER [M or F]
6\Do we know the direction of the reaction at C? ~[===]
6This means the reaction at C should be represented by
~[===]  unknowns.
YOUR ANSWER
DThese 2 unknowns represent the components Cx and Cy of the reaction
7at C. We shall show them both on the free-body diagram.
6Do we know the direction of the reaction at D?  ~[===]
FThis means the reaction at D should be represented by ~[===] unknowns.
YOUR ANSWER
;These 2 unknowns represent the components Dx and Dy of the
Greaction at D. We shall show them both on the figure and thus complete
#the free-body diagram of the frame.
J&We are now ready to write the equilibrium equations for the entire frame.
BHow many _independent_ equilibrium equations can we write?  ~[===]
YOUR ANSWER
$How many unknowns do we have? ~[===]
YOUR ANSWER
D\This means that we cannot determine _all_ of the unknowns from the
>equilibrium conditions for the entire frame. We may, however,
determine _some_ of them.
DIn order to write an equation involving only one unknown, should we
Kconsider the components or the moments of the various forces? ~[==========]
;About what point do you wish to compute the moments of the
forces? ~[=]
YOUR ANSWER [C or D]
We write the equation\
~MC = 0:  ~[=] Dy (~[===] m) ~[=] (
 kN) (~[===] m) ~[=] (
 kN) (~[===] m) = 0
~MD = 0:  ~[=] Cy (~[===] m) ~[=] (
 kN) (~[===] m) ~[=] (
 kN) (~[===] m) = 0
SIGNS AND VALUES
Solving this equation for
, we obtain
 kN.
<Does the sign obtained indicate that this force is directed
upward or downward? ~[========]
Upward/Downward/
YOUR ANSWER [U or D]
CCan we write _another_ equation involving only one unknown?  ~[===]
6Will it involve moments or components?  ~[===========]
Moments about what point?  ~[=]
A/B/C/D/
YOUR ANSWER [A,B, or D]
YOUR ANSWER [A,B, or C]
#Components in what direction?  ~[=]
YOUR ANSWER [X or Y]
We write the equation
~Fy = 0:
 ~[=]  ~[======] kN = 0
SIGN AND VALUE
Solving this equation for
, we obtain
=\Does the sign obtained indicate that this force is directed
upward or downward? ~[========]
Upward/Downward/
YOUR ANSWER [U or D]
CWe have used 2 equilibrium equations associated with the free-body
Dof the entire frame and found 2 of the unknown reaction components,
namely Cy and Dy.
;\How many more unknown reaction components are left? ~[===]
YOUR ANSWER
These components are Cx and Dx.
J\Can you write an equation involving only _one_ of these unknowns? ~[====]
ECan you write an equation involving _both_ of these unknowns? ~[====]
<Let us write the most obvious equation involving Cx and Dx :
     ~
~Fx = 0:  Cx + Dx +
 kN +
 kN = 0
\We therefore have
     Cx + Dx =
Cx+Dx
 kN
A\Can we write another equation involving Cx and Dx, so that both
Eequations can be solved simultaneously for these two unknowns? ~[===]
Actually
:, we could write an equation involving moments about some
<appropriate point, but this equation _would not contain any
@information we do not already have_, since only 3 _independent_
5equations can be obtained from any 2-dimensional FBD.
So, we must try something else.
>In order to proceed with the solution of the problem, we must
@dismember the frame and sketch the free-body diagram of each of
)its members, namely AKC, BLD, AB, and KL.
3\Is any of these members a two-force member? ~[===]
1/2/3/4/
A\Please list all two-force members in order by number. ~[=] ~[=]`
$     1. AKC   2. BLD   3. AB   4. KL
YOUR ANSWERS [1, 2, 3, or 4]
:Let us first consider the two-force members AB and KL and
the forces exerted on them.
0Do we know the direction of these forces? ~[===]
#Do we know their magnitude? ~[====]
Do we know their sense? ~[====]
=While we know neither the magnitude nor the sense of the two
Gforces acting on AB, we know they must be equal and opposite. We shall
Marbitrarily assume that they _pull_ on AB, that is, the member is in tension.
@Since the forces exerted on each other by two members are equal
Band opposite (see Sec. 6.10 of _Vector Mechanics for Engineers_),
=we may represent as shown the forces exerted by AB and KL on
#the two vertical multiforce members
VSince we have no further use for the two-force members AB and KL, we shall erase them.
BWe now may use either member AKC or BLD as a free body. Since the
Dproblem calls for the determination of the forces exerted on member
EBLD (in addition to the reactions at both supports), we shall choose
BLD as a free body.
EWriting that the sum of the y components of the forces must be zero,
Gwe conclude that Dy = 0. But we found earlier a different value for Dy.
R\This inconsistency is due to the fact that the frame is _improperly constrained_
*and will collapse under the given loading.
HWe recall that we found Dy = 0 from the free-body diagram of the entire
Lframe. We check that the equilibrium equation (~
Fy = 0) written for _this_
free body is satisfied.
6But this leaves only _two_ more independent equations
D(out of three) to solve for the three unknowns Dx, F(AB), and F(KL).
KThe frame is statically indeterminate. It is also _improperly constrained_
-and would collapse under a different loading.
GLet us remember that we have already determined the reaction component
JDy from the free-body diagram of the entire frame. How many of the forces
3shown on this free-body diagram are unknown? ~[===]
VALUE
,The unknown forces are Dx, F(AB), and F(KL).
BHow many independent equilibrium equations are available from the
'free-body diagram of member BLD? ~[===]
VALUE
?This means we can write as many equations as we have unknowns.
4So, we should be able to determine all the unknowns.
~LLet us try to write an equation with only one unknown. Should this equation
W~;involve the components or the moments of the forces acting
~ on the free body? ~[===========]
}"Components in what direction? ~[=]
YOUR ANSWER [X or Y]
We write the equation\
~Fy = 0: ~[=] F(KL) sin
 ~[=] ~[======] kN = 0\
SIGNS AND VALUES
z1Solving this equation for F(KL), we find F(KL) =
<zRDoes the sign indicate that F(KL) is directed toward or away from L? ~[==========]
Toward/Away from/
YOUR ANSWER [T or A]
2x2The force exerted by KL on member BLD is therefore
     F(KL) =
 kN ~
F(KL)
33333
vIThe relatively large value obtained for F(KL) indicates a poor design of
vJthe frame. Point L should not have been selected so close to the midpoint
of member BD.
uCLet us try to write another equation with only one unknown. Should
uCthis equation involve the components or the moments of the various
gu/forces acting on the free body?  ~[===========]
\Moments about what point? ~[=]
YOUR ANSWER [B or D]
(t9Referring to the free-body diagram, we write the equation
cos/sin/
 /+/-/
~MB = 0: ~[=] Dx (~[======] m) ~[=] F(KL) (~[===] ~
) (~[======] m) = 0\
 /+/-/
~MD = 0: ~[=] (F(AB) ~[======] m) ~[=] F(KL) (~[===] ~
) (~[======] m) = 0\
Oq-SIGNS, VALUES, AND S for SINE OR C for COSINE
p:Substituting the numerical values of ~
 and F(KL), we have
p6     + F(AB) (6 m) - (~[======] kN) (cos ~[======] ) (
 m) = 0
o6     + F(AB) (6 m) + (~[======] kN) (cos ~[======] ) (
 m) = 0
n3     + Dx (6 m) + (~[======] kN) (cos ~[======] ) (
 m) = 0
n3     + Dx (6 m) - (~[======] kN) (cos ~[======] ) (
 m) = 0
NUMERICAL VALUES
Solving for Dx, we find Dx =
l#Solving for F(AB), we find F(AB) =
l@\Does the sign obtained indicate that this force is directed to
?l"the right or to the left? ~[=====]
Left/Right/
YOUR ANSWER [L or R]
YOUR ANSWER [L or R]
F(AB)
dj5The horizontal component of the reaction at D is thus
     Dx =
Yi,The force exerted at B on member BLD is thus
     F(AB) =
e,force F(AB) exerted on member BLD at point B
e!component Dx of the reaction at D
We still have to determine the
.  Equating to zero
eAthe sum of the x components of the forces shown on the free-body
diagram, we write
cos/sin/
F(AB) ~[=] Dx
Dx ~[=] F(AB)
~Fx = 0: ~[=]
 ~[=] F(KL) ~[===] ~
b$SIGNS AND S for SINE or C for COSINE
- F(AB)
+ Dx
 F(AB)
~Fx = 0:
 F(KL) cos ~
a<\Substituting the numerical values obtained earlier we have
- F(AB)
 ~[======]
+ Dx
 ~[======]
 ~[======] cos ~[======] = 0
SIGNS AND VALUES
_DWe still have to determine the horizontal component of the reaction
^Gat the support C of the frame. Recalling the equation obtained earlier
^8from the free-body diagram of the entire frame, we write
     Cx + Dx =
]FSubstituting the value found for Dx with the appropriate sign, we have
     Cx ~[=] ~[======]  =
VALUE
Solving for Cx, we find Cx =
[8\Does the sign obtained indicate that Cx is directed to
O["the right or to the left? ~[=====]
Left/Right/
YOUR ANSWER [L or R]
 kN ~
Y5The horizontal component of the reaction at C is thus
   Cx =
X?\We have now determined all the forces and reactions requested.
SETUPJUMPTABLE FOLLOWS
FARCALLHANDLER FOLLOWS