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Unformatted text preview: ENGINEERING MECHANICS  STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 81* A steel bar with a rectangular cross section is used to transmit
four axial loads as shown in Fig. P8—1.
(a) Determine the axial
forces transmitted
by cross sections in
intervals AB, BC, and
CD of the bar.
(b) Draw an axial force
diagram for the bar. SOLUTION For overall equilibrium of the bar: +—+ EFX = 35 + P  20 + 18 = 0 P = 37 kip = 37 kip —% A ioad diagram for the bar, freebody diagrams for parts of the bar to
the left of sections in intervals AB, BC, and CD of the bar, and an axial force diagram for the bar are shown below. 35 37 2° )8
H aov <—°c u Lea—>5 133 3: 37 'Illl'l'l'll'. iiicsf From the freebody diagrams: +—+ EFX = FAB — 35 = 0 FAB = 35 kip = 35 kip (T) Ans.
+—+ EFX = FBC  35 + 37 ='0 FBC = 2 kip = 2 kip (C) Ans.
+—+ ZFX = FCD  35 + 37  20 = 0 FCD = 18 kip = 18 kip (T) Ans. ENGINEERING MECHANICS  STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES The motor shown in Fig. P86 supplies a torque of 500 Nm to shaft
BCDE. The torques removed at gears C, D, and E are 100 N'm, 150
N'm. and 250 Nm, respectively.
(a) Determine the torques transmitted by transverse cross sections in intervals BC, CD, and DE of the shaft.
(b) Draw a torque diagram for the shaft. SOLUTION 5 load diagram for the shaft, freebody diagrams for parts of the shaft
to the left of sections in intervals BC. CD, and DE of the shaft, and a torque diagram for the shaft are shown below. 5°C! From the freebody diagrams: +C2Mx TBC — 500:0 +(j53Mx '_ +100=0 +C2Mx 100+150=0 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 810 For the steel shaft shown
in Fig. P810, (a) Determine the maximum
torque transmitted by
any transverse cross
section of the shaft. (b) Draw a torque diagram
for the shaft. I5 kNl'n 6 kNm SOLUTION
For overall equilibrium of the shaft: +CXM‘=T8156+14=0 T=15kNm=15kNm A load diagram for the shaft, freebody diagrams for parts of the Shaft
to the left of sections in intervals AB, BC, CD, and DE of the shaft, and a torque diagram for the shaft are shown below.
8 IS I. E_[(GH B .
C The 'IlIII/II/l/l/l/l/lﬂ uuuvzzngnn From the freebody diagrams: + C XMX = TAB  8 kN'm = 8 kNm —C— + C XMX TBC  7 kN~m = 7 kNm Q— + C 2Mx TCD 8 kN'm = 8 kNm —C— + C >2Mx TDE  i4 kNm = 14 kNm —C— Ans. ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 812* A bar is loaded and supported
as shown in Fig. P812.
(a) Determine the maximum
axial load transmitted
by any transverse
cross sectiOn of the bar.
(b) Draw an axial force
diagram for the bar. SOLUTION
For overall equilibrium of the bar: + T 2Fy = R  2(40) + 2(50) .U
 2(10)  10 = 0 ﬁg.P842
R = 10 RN = 10 RN T
A load diagram for the bar. freebody diagrams for parts of the bar above sections in intervals AB, BC, and CD of the bar. and an axial force
diagram for the bar are shown below. From the free—body diagrams: + T 2F = FAB  l0  2(10) = 0 FAB = —30 kN = 30 kN (C) + T 2F FBC  10  2(10) 2(50) = 0 FBC = 70 kN = 70 kN (T) + T [F FCD + 10 + 2(10)  2(50) + 2(40) = CD = 10 kN = 10 kN (C) BC = 70 kN = 70 kN (T) Ans. ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Five BOOmm diameter pulleys
are keyed to a steel shaft as shown in Fig. P8—14. The
pulleys carry belts that are
used to drive machinery in a
factory. Belt tensions for
normal operating conditions
are indicated on the figure.
Determine the maximum torque
transmitted by any transverse cross section of the shaft.
Fig. P844 SOLUTION A load diagram for the shaft, freebody diagrams for parts of the shaft
toathe left of sections in intervals AB, BC, CD, and DE of the shaft, and a torque diagram for the Shaft are shown below.
(OcalinL '7ZC>)int From the freebody diagrams: + C 251x = TAB — 600 = o  = 600 um —C + C 2Mx TBC  600 + 480 = o = 120 N'm 120 Nm C TCD  600 + 480  720 = o 840 Nm 840 Nm C—  600 + 480 — 720 + 720 = o  120 Nm = 120 Nm C— = 840 Mm = 840 Nm C— Ans. ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8—17 Determine the internal resisting forces and moment transmitted by section aa in the curved bar shown in Fig. P8l7. SOLUTION From a freebody diagram for
the part of the curved bar
to the right of section aa: II
D P  750 sin 30° + /’ 2F ,
X ,375 lb = 375 lb 1’ v — 750 cos 30° = o 649.5 lb 3 650 lb “\ Ans. M  750 cos 30° (30 cos 30°)  750 sin 30° (30  30 sin 30°) = 0 22,500 in.lb = 22.5 in.kip C 707 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 820* Determine the internal resisting forces and moment transmitted by section aa in bar ABC of the threebar frame
P8—20. shown in Fig. IOOmm IOOmm IOOmm 100mm
Fig. P820 SOLUTION
Link BD is a twoforce member:
therefore, the line of action
of force 3 is known and the
free—body diagram for member
ABC can be drawn as shown. + C EMA = 3(200)  3(400) = 0 )oo B = 6.00 kN = 6.00 kN T From a free—body diagram
for the part of member ABC to the right of section aa: 30.96° + v/ 2F = p  6 sin 30.960
+ 3 sin 30.96° = 0 1.5433 kN a 1.543 kN v/ v  6 cos 30.96°
+ 3 cos 30.960 = o v = 2.573 kN z 2.57 kN ‘3 Ans. + C 2M2 = M + 6(0.100)  3(0.300) = 0 M = 0.300 kN'm = 0.300 kN'm C ENGINEERING MECHANICS  STATICS, 2nd. 830 A threebar frame is loaded and supported as shown in Fig. P830.
Determine the internal
resisting forces and
moment transmitted by (a) Section aa in bar BD.
(b) Section bb in bar ABC. SOLUTION .a From a freebody diagram
for the complete frame: E(960)  900(1140) = 0 +CEMA= E 1068.8 N = 1068.8 N T + —% SE A = 0 A = 0 X X X + T SE = A
y y y = 168.8 N = 168.8 N ¢ + 1068.2  900 = 0 From a free—body diagram
for bar BD: + C EMD = By(600)  900(360) = 0 B = 540 N = 540 N ¢ Y From a freebody diagram
for bar ABC: + C EM = B (500) — 540(300)
C x + 168.8(480) = 0 . = 162.0 N = 162.0 N + X Ed. W. F. 722 RILEY AND L. D. STURGES ENGINEERING MECHANICS  STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 8—30 (Continued) (a) From a freebody diagram
for the part of bar BD to
the left of section aa: + 4 BF“ = P  162.0 = 0 P = 162.0 N = 162.0 N 9 V  540 = 0 = 540 N = 540 N T ’ + C ZM’ = M + 540(0.300) = o M = 162.0 Nm = 162.0 N'm D (b) From a freebody diagram
for the part of bar ABC
below section bb:  1M_ 0
¢  tan 300  59.04 0 + /’ 2F“ = p + 540 sin 59.04 + 162.0 cos 59.04°
 168.8 sin 59.04° = o
= ~401.6 N = 402 N v/
v + 540 cos 59.o4°
 162.0 sin 59.04°
— 168.8 cos 59.04° = o
v = 52.0 N = 52.0 N \N
C + 2M: = M + 162.0(0.250)
— 540(o.150)
+ 163.3(o.330) = o = —15.20 Nm = 15.20 Nm D Ans. ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 832* A threebar frame is loaded and supported as shown in Fig. P832.
Determine the internal
resisting forces and
moment transmitted by (a) Section an in bar CGE.
(b) Section bb in bar DEF. SOLUTION From a freebody diagram
of the pulley: '+——92F
X From freebody diagrams
of members DEF and CGE: D quanta. E Anonn. F Ey(400)  500(800) = 0 500(500)  Ex(600)  1000(400) = 0 726 ENGINEERING MECHANICS  STATICS. 2nd. Ed. H. F. RILEY AND L. D. STURGES 8—32 (Continued) (a) From a freebody diagram
for the part of bar CGE
below section 38: aw _ 0
200 — 06.31 ¢ = tan = p  500 cos 56.31°
250 cos 56.31°
1000 sin 56.31° = 0 1248 N = 1248 N ‘\ Ans. — v  500 sin 56.31°  250 sin 56.31° + 1000 cos 56.31°= 0 69.3 N = 69.3 N v/ M + 500(0.200) + 250(0.300)  1000(0.200) = 25.0 Nm = 25.0 Nm C (b) From a freebody diagram
of the part of member DEF
to the right of section bb: ZoomE 400mm. F
+ 6— 2F = P + 250 + 500 = 0
—750 N = 750 N *
V — 1000 + 500 = 0
500 N = 500 N ¢
M + 1000(0.200)
500(0.600) = 0 100 N'm = 100 Nm C ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 837 A beam is loaded and supported
as shown in Fig. P837. Using
the coordinate axes shown, write
equations for the shear V and
bending moment M for any section 'w»"
of the beam in the interval L3“*F————'°"
0 < x < 10 ft. Fm P&37 SOLUTION 5 75035
4Soogt.n, I 300 lLHt " From a freebody diagram for the complete beam:
+ C 2MB = 4500  A(10)
+ 300(10)(5)  750(2) = 0 ,a A = 1800 lb = 1800 lb T the interval 0 S x S 10 ft:
= 1800  300x = 300x + 1800 lb 4500 + 1800x  300(x)(x/2) = 150x2 + 1800x  4500 ft'lb A beam is loaded and supported
as shown in Fig. P838. Using
the coordinate axes shown, write
equations for the shear V and
bending moment M for any section
of the beam in the interval 0 < x < 3 m. SOLUTION From a free—body diagram
for the complete beam: + T EFy = VO  2(3)  3 = 0 o = 9 kN = 9 kN T + c 2M0 = —M0  2(3)(1.5) — 3(4) Mo = —21 kN'm = 21 kN'm C
For the interval 0 S x S‘3 m:
V = 9 — 2x = 2x + 9 kN —21 + 9x  2(x)(x/2) = ‘x2 + 9x — 21 kNm ENGINEERING MECHANICS  STATICS. 2nd. Ed. W. F. RILEY AND L; D. STURGES 840 A beam is loaded and
supported as shown in
Fig P840. Using the
coordinate axes shown,
write equations for the
shear V and bending
moment M for any section
of the beam in the
interval 0 < x < 4 m. SOLUTION From a freebody diagram
for the complete beam: +g 2MB = Ay(8) + 30(4)(6) + 40(2) = 0 ; 100 kN = 100 kN T For the interval 0 < x < 4 m:
V = 100  30x = 30x + 100 kN M 100x — 30(x)(x/2) = 15x2 + 100x kN'm A beam is loaded and
supported as shown in Fig P841. Using the
coordinate axes shown.
write equations for the shear V and bending
moment M for any section
of the beam in the
interval 4 ft < x < 12 ft. SOLUTION From a freebody diagram
for the complete beam: + C 2MB = A(16) + 750(8)(8)  3000(5) = 0 the interval 4 ft < x < 12 ft:
’ = 1875  750(x  4) = 750x + 4875 lb 1875x — 750(x — 4)(x — 4)/2 = 375x2 + 4875x  6000 ftlb
7H3‘7 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 8—51* Draw complete shear and moment diagrams for the beam shown in Fig. P851. SOLUTION From a freebody diagram'
for the complete beam: + C 2MB = A(13.5) + 5000(9.5) + 3000(4.5) = o A = 6185 lb = 6185 lb T + C XMA = B(13.5)  5000(4)  8000(9) = 0 B = 6815 lb = 6815 lb T Load, shear, and moment diagrams for the beam are shown below: 5000 IL 8090 lb ENGINEERING MECHANICS  STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 855* Draw complete shear
and moment diagrams
for the beam shown
in Fig. P855. SOLUTION From a freebody diagram
for the complete beam: + C EMA = 3(23)  2500(8)  800(10)(18) = o
B = 7130 lb + C EM = A(23) + 2500(15) + 800(10)(5) = 0 B
A = 3370 lb Load, shear, and moment diagrams for the beam are shown below: 25“ ’5 800 mm: ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 856* Draw complete shear
and moment diagrams
for the beam shown
in Fig. P856. SOLUTION From a freebody diagram
for the complete beam: ,4 + C EN = 3(8)  30(4)(2>  40(6) = o A
B = 60 kN + C 2MB = A(8) + 30(4)(6) + 40(2) = 0 A = 100 kN Load, shear, and moment diagrams for the beam are shown below: ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 867 Draw complete shear lOOOlb 2501b,“ 375Ib/n and moment diagrams for the beam shown in Fig. P867. I ' 6n——*—6n——*—4n
Fig. P867 SOLUTION ‘4§t L§t
H Zooogtqb From a freebody diagram
for the complete beam: + C EMA = 1000(4)  2000  250(12)(10) + B(20)  %(375)(4)[20 + a] : 0 B = 2250 lb H
O + ChZM = 1000(24)  A(20)  2000 + 250(12)(10)  %(375)(4)[] B
A = 2500 lb Load, shear, and moment diagrams for the beam are shown below: )Ooo“, 395$)
250 )L/ﬁ; a ...
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