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Unformatted text preview: 3,113 =. M 71.10
Determine the internal forces and moment at A for each loading. §QLQTIQN: r—— 4 m
(a) Draw the F30 and determine the reactions at the left and (b) right ends of the beam. SRN kNImJ l f r L, A
ln—J I : m—Jv
4m r—l I
L, Ry
EMR = 0 = (3,000 N) (2 m) — L, (4 m) —> L, = 4,000 N T 2F, = 0 = L, — 8.000 N + R, = 4.000 N — 8,000N + R=y —> R, = 4.000 N T 2F, = 0 = L,( —> L, = 0
Cut the beam through point A and draw the FBD: V
A p MA
1, A
(F
Elm—0
LY
EF,=0=L,—VA=4.000NVA £F,=0=PA
ANS \_/A=4000N~L EA.=_Q EMA = 0 =  L,('1m)+ MA = (4.000 N)('1m)+ MA
ANS MA = 4,900 Nm (1:) Draw the F80 and determine the reactions at the left and right ends of the beam. — FERN/m
Lx I I I I
m
1m—{p‘
4m
L, Ry
ZMR = 0 = (8.000 Nim) (4 m) (2 m) — L, (4 m) —> L, = 4.000 N T 2F, = 0 = L,— 8.000 N + R, =4,000 N — 8.000N + R=y —> R, = 4.000 N T IF, = 0 = L, —> L, = Cut the beam through point A and draw the FBD: EkN/m p. P Mn
l n
101—4
L y EMA = 0 = ~ (4,000 N} (1 m) +[(2,000 Nlm) (1 m)] (0.5m) + MA
ANS MA = 3,000 N—m 2F, = 0 = L,— (2.000 Him) (1 m)  VA = 4.000 N  2.000 N  VA 2F, = 0 = PA
ANS \_/A=2,000 NJ. EAEQ 358 PROBLEM72.1§ (9991,) E O 0 N IOUN BOON
BUUN v' .
WN 400Nm if 9 “lEUONﬁm
' —1600Nm a h" (a : " .L.>! 3_ The beam in Problem 72.13 will safely support shear
j forces and bending moments of magnitudes 2 kN and
; _: ' 6 kNrn. respectively. Based on this criterion. can it
' safely be subjected to the loads F = 1 kN, C = 1.6 kNm? W:
' i' 3; Draw the FBD for the beam and determine the reactions
: at points A and B. Drawing the shear force and IR N bending moment diagrams:
t» 4 m '+' 4 m — 8 m '"E'1\r.":'.: .—_ 1:" lkN EMA: 0 =  (1000 N) (16 m)+1600 Nm + e, (0 m)
ey=1300NT EF,=0=—1DODN+BY+Ay=1DODN+1BDON+Ay
A,=300Ni . We see from the sear force and bending moment diagrams that, while
the shear force is within the given limit, the magnitude of the maximum
bending moment exceed the given limit. . ANS Beam gangs}; safely sggggrt the lgagg.
_. 379 P 7 .
The bar 30 is rigidly ﬁxed to the beam ABC at 8. Draw the
shear force and bending moment diagrams for the beam ABC. sgLUTIQu: ' some The angle made by the cable with the horizontal is:
(9: tan" a =14.04°
3ft Summing moments about point A:
EMA = o =  (600 lb) (10 ft) + T (sin 14.04“) (5 ft) + T (cos 14.04”) (2 n)
T = 1546 lb Summing horizontal forces on the bar:
2F, = o = A,‘ — T (cos 14.04“) = Ax  (1546 lb) (cos 14.04") .:
Ag=1500 lb.» 1800131410 Summing vertical forces on the bar:
22Fy = o   600 lb + (1546 lb) (sin 14.04°) + A,
AV 2 225 lb
180 OF ‘t llo mm Draw the shear force and bending moment diagrams for
beam ABC. i—Em—ﬂhem—ﬂ—Bm—ﬁ Cy 2m M:
We consider only the F80 of bar ABC. The effect of the BIDsection 1333N “(N
of the beam is reduced to an equivalent force and moment at point B. (3,)50 = 4 RN 4 (MB)ED =  (4 kN} (2 m) =  8 kNm 1333M Summing moments on bar ABC about point A: V
EMA = 0 =  (4.000 N) (2 n1)  (8.000 Nm) + C), (6 m)
EE>E> 7N c, = 2667 N T
10,667N—m Summing vertical forces on bar ABC:
2F,=0=A,+C, 4.000N=A,+2667N—4,000N 2667Nm AY=1333NT M ...
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 Spring '06
 Keil

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