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l l
l l’l‘oblem 6.5 (a) Let the dimension is z 0.1 m. Deter
mine the axial forces in the members, and show that in
this case this truss is equivalent to the one in Problem 6.4.
(b) Let the dimension h = 0.5 m. Determine the axial
forces in the members. Compare the results to (a), and
observe the dramatic effect of this simple Change in de
sign on the maximum tensile and compressive forces to
which the members are subjected. Solution: To get the force components we use equations'of the
form TPQ = TPQGPQ : Tpri + Tpgyj where P and Q take
on the designations A, B, C, and D as needed. Equilibrium yields At joint A: Eli‘s : TABX + TACX = 0, andZFy :TABY+TACY—1kN=0. — ‘TABX + TBCX + TBDX = 0. ’TABY + TBCY + TBDY = 0 —TBCX — TACX + TCDX : 0» —TBCY  TACY + TCDY + CY 3 0 ‘TCDX * TBDX + Dx = 0, = ’TCDY  TBDY + DY = 0
Solve simultaneously to get
TAB = TED : 2.43m,
= —2.78 kN,
= 0, TOD = —2.88 kN. Note that with appropriate changes in the designation of points, the
forces here are the same as those in Problem 6.4. This can be explained
by noting from the unit vectors that AB and BC are parallel. Also
note that in this conﬁguration, B C carries no load. This geometry is
the same as in Problem 6.4 except for the joint at B and member BC
which carries no load. Remember member BC in this geometryiwe
wiil encounter things like it again, will give it a special name, and will
learn to recognize it on sight. (b) For this part of the problem, we set h. r 0.5 In. The unit vectors
change because h is involved in the coordinates of point B. The new
unit vectors are eAB = —O.986i+ 0.164j, BAG E —0.864i — 0.504j, 630 : 0i * lj. 9313 = D.768i — 0.64%,
and egg 3 —0.8321 + 0.555} We get the force components as above, and the equilibrium forces at the joints
remain the same. Solving the equilibrium equations simultaneously for this
situation yields ' TAB : 1.35 kN,
TAG = —1.54 kN,
THC = —1.33,
TED = 1.74 kN,
and TOD = mine kN. These numbers differ signiﬁcantly from (a). Most signiﬁcantly, member BB is
new carrying a compressive load and this has reduced the loads in all members
except member BD. “Sharing the load" among more members seems to have
worked in this case. Problem 6.2.2 The Warren truss supporting the walk
way is designed to support vertical 50kN loads at B,
D, F, and H. If the truss is subjected to these loads,
what are the resulting axial forces in members BC, CD, and CE? Solution: Assume vertical loads at A and I Find the external
loads at A and I, then use the method of goints to work through the
structure to the members needed. SinIT x
[Y Ay+Iy“4{50)=0(kN}
AB 2—180.3kN
“350 ~95O —1550 —2}. 50 +241 :0
()()()<) y 9:33.69,
Solving A1, = 100 kN 1,, : IDOkN BCcosH+BD—ABCOSH =0 —50—ABsin9—BCsin0:0 Solving, BC : 90.1 kN (T)
B1) = —225 lcN (0} a m 33.69“
A0 = 150 kN (T) AB 0036’ +AC = 0
BC 2 90.1 kN (T) ABSin0+Ay =0 ZFw: CE—ACl—C'DcosﬂiBCcosBZO Solving, AB =—180.3kN C
( ) 21:3,: CDsin€+BCsinﬂ=0 A0 = 150 kN (T) 50“ng CE = 300 kN (T)
y' CD = —90.1kN(C)
Hence BC 2 90.1 N (T) CD : —90.1kN(C)
CE : 300 kN (T) Problem 6.31 For the truss in Problem 6.30, use the
method of sections to determine the axial forces in mem bers BO, CF, and FG; " I 501mm“: , a Solving BC = 300 kN (T) .. .
CF : 7141.4 kN (C') :; FG z —200 kN (C) —BO—CFcos45—FG:0 g
—C’F 511145“ — 100 : o ' —(1)FG’ — 2(100) = o E
g n, Problem 6.32 Use the method of sections to determine
the axial forces in members AB, BC, and CE. Solution: First, determine the forces at the supports 1 B D Ay+Gy—3F=u —1{F) _ 2(2F) + 39y : 0 Solving Aw : 0 Gy = 1‘67F
Ay z 1.33F
Method of Sections: CE+AB=0 o +ZMB: (—1)Ay + (1)03 = 0 Solving, we get AB : 71.331? (0)
CE = 1.33F (T)
BC = —0.33F (0) J
.l
.'l
i
.1 Problem 6.45 Use the method of sections to determine
the axial force in member EF. Solution: The included angle at the apex BAG is 12
a =tan‘1 (—) c 36.37“.
16 The interior angles BCA. DEG, FGE, HIG are!)’ = 90° — o: =
5313". The length of the member ED is LED : 8 tan a x 6 ft.
The interior angle DEF is x3 =tan*1(~4—)= 33.69“. LED The complete structure as a free body: The moment about H is M H =
—1o(12) — 10(16) + 1(12) 2 o, from which I z % 2 23.33 kip.
The sum of forces: XE?! : Hy + I : 0) from which Hy z —I : w23.33 kip. EFT, :H..+20:o, from which H}, = —2() Rip. Make the cut through EC, EF, and
DE. Consider the upper section only. Denote the axial force in a
memberjoiningI I, K by IK. The section as afree body: The sum of
the moments about E is ME = —10(4) — 10(8) + DF(LED} : D,
from which DF : % : 20 kip. The sum of forces: EFL. : —EFsin,6—EGsin'y—DF=0, ZR“ = —EFcosﬁ+EGcosry+2U: D, from which the two simultaneous equations: 0.5547EF + 0.8EG :
“20, and 0.8320EF — 0.6EG = 20. Solve: EF = 4.0 kip (T) 10kip_ $0.10? l
i
‘5
! Problem 6.69 The mass m : 12 kg. Determine the forces on member ODE. Solution: The equations of equiiibrium for the entire truss are: ZFX :AX+CX =0,
ZFY sz —mg=0,
and 2 MA = 0.40): — 0.7mg : 0.
From these equations we get
Ax : 7206.0 N, AY = 117.7 N,
and 0:1: 2 206.0 N. Member ABD: The equations are ZFX :AX+BX+DX+T=0, ZFY =Ay+By+Dy=0, and EMA = 0.2131, + 0.2BX + 0.2Dy + DADX + 0.1T = 0. Member CDE: The equations are
ZFX :meDX +EX =0.
ZFY = "DY +EY = 0,
and 2 MD = 0.4133, — 0.2EX : 0.
Member BE: The equations are
“BX+PX”‘EX =0, “BY+PY;'EY:0: and 2 ME = 0.43,, = n. The Pulley: The equations are
V 7T i PX z 0. The Weight: The equation is
2 FY = T — mg : 0. Soiving the equations simultaneously. we get
BX = 117.7 N, BY = G,DX = —29.4 N, DY : —117.7 N,
Ex=w117 N,Ey=0 ,T=117.7N, PX = —117.7N, Py : 7117.7N Problem 6.80 The weight W : 60 kip. What is the
magnitude of the force the members exert on each other
at D? Solution: Assume that a tong half will carry half the weight, and
denote the vertical reaction to the weight at A by R. The complete
structure as a free body: The sum of the forces: 217g =R—W=0,
fromwhichﬂzw TongHay’ACD: Elemem AC: The sum of the moments about A: (1) (2) (3) (4) (5) ZMA=30y+3Cszo The sum of the forces:
R
25;, = §+Cy+Ay :0,and ZFw=cm+Amzo
Element CD: The sum of the forces:
2F :DmiP703:O,and W
ZEJzoy—Oy—Em20‘
The sum of the moments:
3
2MB :20: ~30y —3P+§W20 Element AB: The sum of the forces: ZFy=—Ay+%—By:0,and ZF$=—A$—Bm :0. Element BD: The sum of the forces: W
ZFy=By—Dy—3:O,and ZFm=Bw—DE+P:U. These are ten equations in ten unknowns. These have the soiutiou
R : 60 kip. Cheek, A1: = —30 kip, Ag 2 0, Bx = 30 kip,
By = 30 kip, CI = 30 kip, Cy : 73D kip, m r 110 kip.
Dy r 0, and P = 80 kip. The magnitude of the force the
members exert on each other at D is D = 110 kip. L E 4
i
i
‘ Problem 6.81 Figure 3 is a diagram of the bones and
biceps muscle of a person’s arm supporting a mass. Ten
sion in the biceps muscle holds the forearm in the hor
izontal position, as illustrated in the simple mechanical
model in Fig. b. The weight of the forearm is 9 N, and
the mass m = 2 kg. (a) Determine the tension in the biceps muscle AB.
(1:) Determine the magnitude of the force exerted on the upper arm by the forearm at the elbow joint 0. Solution: Make a cut through AB and BC just above the elbow
joint C. The angle formed by the biceps muscie with respect to the
forearm is on = tan—1 (%) : 802°. The weight of the mass is
W : 2(9‘81}: 19.62 N. The section as a free body: The sum of the moments about 0 is 2 Mo = ~50Tsin a + 150(9) + 350W : o, from which T 2 166.76 N is the tension exerted by the biceps muscle
AB. The sum of the forces on the section is ZFX =C$+TCGSD£:O,
from which CE = —28.33 N.
23F}; :Cy+Tsincv—9—W=G, from which Cy 2 —135.72. The magnitude of the force exerted by
the forearm on the upper arm at joint 0 is F:1/C§+CE=138.65N Problem 6.90 For the bolt cutters in Problem 6.89,
determine the magnitude of the force the members exert
on each other at the pin connection B and the axial force
in the two—force member OD. Solution: From the solution to 6.107, we know BX : —695 N,
and By : —435 N. We also know that 0;; : 695 N, and Cy :
535 N, from which the axial load in member CD can be caiculated. The load in CD is given by Top : 1/0} + 03, : 877N Problem 6.91 The device is designed to exert a large
force on the horizontal bar at A for a stamping operation.
If the hydraulic cylinder DE exerts an axial force of
800 N and or : 80°, what horizontal force is exerted on
the horizontal bar at A? Solution: Deﬁne the acy coordinate system with origin at C. The
projection of the point D on the coordinate system is Ry = 2505ina : 246.2 mm,
and R3 = 250 (:0st: = 43.4 mm. The angie formed by member DE with the positive :1: axis is 9 = 180 — tan—1 = 145.38°. The components of the force produced by DE are Fm = Fcosﬁ = —658.3 N, and
F‘ = F sinG : 454.5 N. The angie of the element AB with the
positive 3: axis is ,6 = 180 —— 90 — a : 10°, and the components
of the force for this member are Pm = P cos ,6 and P.” = P sin ,9,
where P is to be determined. The angle of the arm BC with the {:05“
itive :1: axis is 'y = 90 + a 2 170°. The projection of point B is
Lm = 250 6057 : 7246.2 mm, and Ly = 2503in‘y = 43.4 mm.
Sum the moments about 0: 2 MC : RmFy 7 Rye. + mpg — Lyra... = 0. Substitute and solve: P : 2126.36 N, and Pa: = P c0553 3 2094N
is the horizontal force exerted at A. Problem 6.104 Determine the force exerted on the
bolt by the bolt cutters and the magnitude of the force
the members exert on each other at the pin connection A. Solution: Element AB: The moment about A is EMA : 71013 w 54F m o, where F 2 90 N. From which B r 7486 N. The sum of the forces:
Zr}, =A+B—F=U, from which A = 576 N Element BC: The moment about U: 2M0 = —lﬁB—8Fc :0, from which the cutting force is For : 972 N Problem 6.105 The 600113 weight of the scoop acts at
a point 1 ft 6 in. to the right of the vertical line CE. The
line ADE is horizontal. The hydrauiic actuator AB can
be treated as a twoforce member. Determine the axial
force in the hydraulic actuator AB and the forces exerted
on the scoop at C and E. SOIHﬁOl‘l: The free body diagrams are shown at the right. Place the
coordinate origin at A with the m axis horizontai. The coordinates (in
ft) of the points necessary to write the needed unit vectors are A (0, 0).
B (6, 2), C {8.5, L5), and D (5, G). The unit vectors needed for this
problem are my, = w0.9491 — 0.3163,
uBc = 0.981i — 0.196j,
and LIED : 70.447i H 0.894j.
The scoop: The equilibrium equations for the seoop are
ZFX = —TCB“BCX + Ex 3 0,
2 FY = TCBUBGY + Ev — 500 : 0,
and 2M0 :1.5EX w 1.5(6001b) = 0.
Solving, we get
EX = 6001b,
Ey z 4801b,
and T03 : 611.9 lb.
Joint B: The equilibrium equations for the scoop are
2 F): = TBAuBAX + TBDﬂBDX + TGBUBCX : 0,
and 2 FY S TBABBAY +TBD'uBDY + TCBUBCY 3 0.
Solving, we get
TBA = 835 lb,
and TED : W429 lb. ._ lft6" 600 lb ...
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This note was uploaded on 10/12/2009 for the course MAE 210306210 taught by Professor chatterjee during the Fall '08 term at UCLA.
 Fall '08
 CHATTERJEE

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