Piper's Pit BBS/FTP: ibm 0020 - 0029/ibm0020-0029/ibm0028.tar/ibm0028/STATTUT.ZIP/S06.CHN

Turbo Pascal Chain module | 1987-10-26 | 46.4 KB | 360 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
 lb/ft
lb/ft
lb/ft
LabwP
GSelect the dimensions and loading of the beam within the limits shown.\
!   Length of beam    : 0 to 24 ft
!   Distance a        : 0 to 22 ft
?   Distance b        : Greater than distance a, less than 24 ft
$   Distributed load  : 0 to 25 lb/ft
"   Concentrated load : 0 to 250 lb
VALUES
,Dimension b must be greater than dimension a
"Dimension L must be greater than b
Loading must be non-zero
8.  Please enter new values that meet the criteria below:
1OK. This is the beam and loading you have chosen.
\_REACTIONS_
<\We note that for this loading the reactions at A and B are
?vertical.  We first compute the reaction at B using the entire
beam as a free body.
2\About what point do you wish to sum moments? ~[=]
YOUR CHOICE [A or B]
>\Before writing ~
~MA = 0, we will compute the resultant of
Athe distributed load and show the resultant at the center of the
,rectangle representing the distributed load.
%\The value of the resultant is wa = (
 /+/-/
 /+/-/
7Choosing counterclockwise moments as positive, we write
~MA = 0: ~[=] (
 lb) ( ~[======]  ft)  ~[=] (
 lb) ( ~[======]  ft )`
)~ ~  ~       ~[=] RB ( ~[======]  ft) = 0
VALUES
%&We shall solve this equation for RB.
     RB =
     RB = ~[======] lb
YOUR ANSWER
6To compute the reaction at A, we sum the y components.
~Fy = 0:   RA - P - wa + RB = 0
C&~ ~ ~          RA - ~[======] lb - ~[======] lb + ~[======] lb = 0
VALUES
&&We can now calculate the value of RA.
     RA = ~[======] lb
VALUE OF RA
     RA =
V (lb)
CONSTRUCTION OF SHEAR DIAGRAM
HStarting at the left support we will compute the shear at the following
points:
B    Just to the right of the support at A    (this is equal to RA)
)    At the end D of the distributed load
7    Just to the left of point E where load P is applied
8    Just to the right of point E where load P is applied
B    Just to the left of the support at B     (this is equal to RB)
6_Shear just to the right of A:_  (This is equal to RA)
\     VA = RA = + ~[======] lb
VALUE
     VA =
D_Shear at the end of distributed load:_  We consider the portion of
/the beam to the left of point D as a free body.
~Fy = 0:  RA - (Distributed load between A and D) - VD = 0
+\     ~[======] lb - ~[======] lb - VD = 0\
VALUES
Solving, we find
     VD =
?Between A and D the loading is uniformly distributed, therefore
\     dV/dx = -q = constant
>\The slope of the V diagram is constant, which means that the
7diagram is a straight line for the portion from A to D.
C_Shear just to the left of point E:_  Since the beam is not loaded
'between D and E, the shear is constant.
?\The shear diagram in portion DE is a horizontal straight line
and we have:
\ Shear to the left of E =
>_Shear just to the right of point E:_  Subtracting the load P
3from the shear just to the left of point E, we have
\Shear to the right of E = (
 lb) - ~[======]  lb
VALUE
Solving, we have VE =
<_Shear to the left of support at B:_  Since the beam is not
9loaded between E and B, the shear is constant. The shear
?diagram from E to B is a horizontal straight line, and we have
 VB =
-\_Construction of the bending moment diagram_
    M (lb
=\Since dM/dx = V, the maximum moment will occur at the point
where V = ~[======]
ANSWER
2Which of the following statements is correct? ~[=]
2\1. Maximum bending moment occurs between A and D.
'\2. Maximum bending moment occurs at E.
YOUR ANSWER [1 or 2]
7We locate F by determining the distance xm.  Since the
triangles are similar, we have
%&        xm        (~[=======] - xm)`
!     ~[=======]        ~[=======]
VALUES
<Since these quotients are equal, they are also equal to the
<quotient obtained by adding, respectively, their numerators
and denominators, we find:
\        xm

`


        xm


Solving for xm, we have
\     xm = ~[======]
VALUES
We find
     xm =
;Since the area under a shear diagram between two points is
;equal to the change in bending moment between the same two
>points, we will compute the area of each portion of the shear
diagram.
)Complete the following area calculations:
     A1 = (0.5) (
 lb) (~[======]  ft)`
     A2 = (0.5) (
 lb) (~[======]  ft)`
     A3 = (
 lb) (~[======]  ft)`
     A4 = (
 lb) (~[======]  ft)
VALUES
@The shear diagram crosses the horizontal axis under the load P.
:The maximum bending moment occurs at point E where load P
is applied.
4Compute the area of each part of the shear diagram.\
 A1 = 0.5 (
 lb) (~[======]  ft)`
 A3 = (
 lb) (~[======]  ft)`
 A4 = (
 lb) (~[======]  ft) `
VALUES
_BENDING MOMENT DIAGRAM_
;&We recall that the moment at each end of the beam is zero
and plot these two points.
;&Since the area of the shear diagram between two points is
?equal to the change in the bending moment between the same two
points, we write
_For points A and F:_
\     MF - MA = A1 \
'     MF - ~[======] = ~[======]  lb~
_For points A and D:_
\     MD - MA = A1 \
'     MD - ~[======] = ~[======]  lb~
VALUES
     MF =
_For points F and D:_
\     MD - MF = A2 \
'     MD - ~[======] = ~[======]  lb~
VALUES
     MD =
_For points D and E:_
\     ME - MD = A3 \
}'     ME - ~[======] = ~[======]  lb~
VALUES
     ME =
|B_For points E and B:_  (Since we already know MB = 0, we use this
to check our calculations)
     MB - ME = A4&
4{'     MB - ~[======] = ~[======]  lb~
VALUES
     MB =
z#MB = 0, the computations check out.
y9The moment diagram is now drawn by connecting the points
you have determined.
_From A to F and then to D:_
_From A to D:_
x@\Since the V diagram is an oblique straight line, the M diagram
tx7is a parabola with its vertex at the maximum moment MF.
#v@\Since the V diagram is an oblique straight line, the M diagram
u;is a parabola.  As the shear decreases, the slope of the M
diagram decreases.
_From D to E:_
tB\Since the V diagram is a horizontal straight line, the M diagram
s<is an oblique straight line joining the points representing
MD and ME.
_From E to B:_
r=\Again, the V diagram is a horizontal straight line, and the
r&M diagram is an oblique straight line.
r=You have completed the construction of the shear and bending
moment diagrams.
|kAThe edge ~A of a half section of pipe of weight ~W and radius ~r
rests on a horizontal floor.\
k=In order to keep the edge ~B of the section a fraction of an
jAinch off the floor, a hook attached to a cable has been inserted
|j?under the edge and a force ~P is being applied to the cable in
6j5a direction forming an angle ~
 with the horizontal.\
qc@After selecting the values of ~W, ~r, ~
, and of the coefficient
-cC of static friction ~
s between the pipe section and the floor, we
bD propose to determine the required force ~P and the reaction at ~A.\
bBWe also propose to check whether the pipe section can actually be
Db"maintained in the position shown.\
a@\Select the values of ~W (from 10 to 1000 lb), ~r (from 2 to 30
a>in.), ~
 (from 0.00 to 90.0~
), and ~
s (from 0.00 to 0.900).\
fffff
VALUES
#]:We choose as a free body the pipe section and the portion
\.of cable BD which is in contact with the pipe.
\D&The forces applied to that free body are the weight ~W of the pipe
Z\.section, the force ~P, and the reaction at ~A.
\@&How many components should we use to represent the reaction at
[    A? ~[===]
YOUR ANSWER
0[=These components are the normal component N and the friction
force F.
T?Can we express the friction force ~F in terms of ~N?  In other
words, can we write
     F = (
) N ?  ~[===]
That's right!
S%Wrong!  The correct answer is _NO_.
pS6The product of the coefficient of static friction ~
3S@and the normal component N represents the _maximum value_ Fm of
R@the static-friction force F.  This value is reached when motion
is impending.
RA\Since we do not know whether motion is impending in the present
>R>situation, we cannot replace F by Fm.  However, after we have
Q=determined F _from the equilibrium equations_ we shall check
Q4that the value obtained for F is not larger than Fm.
KQCWe shall write the equations expressing the equilibrium conditions
of the free body.
PC&Can we write an equation involving only one of the unknown forces
~P, ~F, or ~N? ~[===]
HPA&Will this equation involve components or moments ? ~[==========]
O9About which of the points shown in the free-body diagram
O$should we compute the moments ? ~[=]
A/B/C/D/E/
YOUR ANSWER [A, B, C, D, or E]
We write therefore\
/+/-/
cos/sin/
~MA = 0:  ~[=] (P ~[===] ~
) (DE) ~[=] (P ~[===] ~
) (AE)
~[=] W(AC) = 0   (1)\
FM$SIGNS AND S for SINE OR C for COSINE
L5We shall now determine the distances AC, DE, and AE:\
      AC = ~[======] in  `
K,      DE = ( ~[======] in) cos ~[======] ~
KD      AE = AC + CE = ~[======] in + ( ~[======] in) sin ~[======] ~
VALUES
J&Carrying out the computations, we find
     DE =
      AE =
AC =
AE =
DE =
H=      + (P cos ~
)(DE) + (P sin ~
)(AE) - W(AC) = 0     (1) \
qHMSubstituting these values and the selected data into Equation (1), we write \
3GP + (P cos ~[======] ~
) ( ~[======] in) + (P sin ~[======] ~
) ( ~[======] in) `
F& - ( ~[======] lb)( ~[======] in)  = 0
VALUES
     + (P cos
) + (P sin
 lb)(
(E     in) = 0
D%We shall solve this equation for ~P.
 P =
 P = ~[======] lb
ANSWER
We now write the equation \
~Fx = 0:  F - P cos(~
) = 0 \
BJSolving this equation for ~F and substituting the numerical data, we have\
A&   F = ( ~[=====] lb) cos ~[======] ~
VALUES
We find F =
We now write the equation `
~Fy = 0:   N - W + P sin ~
 = 0 \
?JSolving this equation for ~N and substituting the numerical data, we have\
>7   N = ~[======] lb - ( ~[=======] lb) sin ~[======] ~
VALUES
We find N =
<!Is our solution complete? ~[===]
That's right!
Wrong!
;HOur solution is _incomplete_ because we still have to check whether the
;Jfriction force ~F that we have computed can actually be developed between
A;.the floor and the edge ~A of the pipe section.
:HIn other words, we have to make sure that the value found for ~F is not
:Glarger than the maximum value ~Fm of the static-friction force at that
d::point  (see Sec. 8.2 of _Vector Mechanics for Engineers_.)
9NThe maximum value of the static-friction force is obtained by multiplying the
9Gnormal component ~N of the reaction at ~A by the coefficient of static
friction ~
8$     Fm = ~[======] ( ~[======] lb)
VALUES
We find Fm =
6EAfter comparing the values obtained for ~F and ~Fm, indicate whether
6: our solution is valid (V) or not valid (N)  ~[=========]
Valid/Not valid/
YOUR ANSWER
5;We thus conclude that the pipe section _can_ be maintained
|5Cin the position shown and that the values found for ~P, ~F, and ~N
are correct.
5GWe suggest that you try the same problem again, with the same data for
40~W, ~r, and ~
s, but with an angle ~
 less than
Do you want to do so? ~[===]
4;We thus conclude that the pipe section _can_ be maintained
3Cin the position shown and that the values found for ~P, ~F, and ~N
are correct.\
@3EAfter comparing the values obtained for ~F and ~Fm, indicate whether
27our solution is valid (V) or not valid (N) ~[=========]
Valid/Not Valid/
YOUR ANSWER
24 We thus conclude that the pipe section _cannot_ be
1?maintained in the position shown and that the values found for
P, F, and N are _incorrect_.
(1CThe pipe section will slide to the left with an accelerated motion.
0BThus, the pipe section is not in equilibrium and _the equilibrium
equations cannot be used_.
^0DThe problem becomes a problem of _Dynamics_, which should be solved
0Fby the method given in Chapter 16 of _Vector Mechanics for Engineers_
/W, taking into account the fact that ~F is now equal to the kinetic-friction force ~F~k.
0/GWe suggest that you try the same problem again, with the same data for
.2~W, ~r, and ~
s, but with an angle ~
 larger than
Do you want to do so? ~[===]
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