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Unformatted text preview: ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES ' 812* A bar is loaded and supported as shown in Fig. P8—12. (a) Determine the maximum
axial load transmitted
by any transverse
cross section of the bar.
Draw an axial force
diagram for the bar. SOLUTION For overall equilibrium of the bar:
+ T ZFY : R  2(40) + 2(50)  2(10) — 10 = 0
R = 10 kN = 10 kN T Fig. P812 A load diagram for the bar. free—body 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 freebody diagrams: = FAB — 10  2(10) = 0 FAB = 30 kN i 30 kN (C) —FBC  10  2(10) 2(50) = 0 FBC = 70 kN = :9 kN (Tl FCD + 10 + 2(10) 2(50) + 2(40) = CD = 10 kN = 10 kN (C) _ _ “*2
BC _ 70 kN  70 RN (T) _ Ans. ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Five GOOmm 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. 1‘ SOLUTION ll 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.
LooNm. 720 Nm ll 1  From the freebody diagrams:
l + C ZMX = TAB  600 = 0 L + C zMx TBC  600 + 480 = 0 = 120 Nm 120 N'm C— l ‘ + C XMX TCD — 600 + 480 — 720 = 0 = 840 Nm 840 N;m C—  600 + 480  720 + 720 = 0 ' 120 N'm = 120 N'm C DE TCD = 840 Nm = 340 N'm C— I Ans. ESGINEERING MECHANICS — STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 8—15* Determine the internal resisting forces and moment transmitted by
section aa in the bracket shown in Fig. PH15. SOLUTION
From a free—body diagram for the part of the bracket to
the right of section aa: 2F = 300  P = 0 X P = 300 lb = 300 lb + = V + 500 = 0 500 lb 2 500 lb ¢ M + 500{12)  300(8) = 0 M —3600 inlb = 3500 in.°lb C ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 817 Determine the internal resisting
forces and moment transmitted by
section aa in the curved bar shown in Fig. P817. SOLUTION From a freebody diagram for
the part of the curved bar
to the right of section aa: + x7 ZFX, —P  750 sin 30° = 0 P 375 lb = 375 1b /” v — 750 cos 30° 0 649.5 lb a 650 lb “\ Ans. M — 750 cos 30° (30 cos 30°)
 750 sin 30° (30  30 sin 30°) 22,500 in.lb = 22.5 in.'kip C 8—36* A beam is loaded and
supported as shown in
Fig. P8—36. 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 < 6 m. SOLUTION From a freebody diagram
for the complete beam: I!
Q + C 2MB = 18  A{8} + 5(6){5} A = 21 kN = 21 kN T  S 6: If\
y interval 0 V a 21  5x = —5x + 21 RN —18 + 21x — 5{x1{x/2i z ~2.5x2 + 21x — 18 kN'm '73) ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 844 A beam is loaded and supported as shown in Fig P844. 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: + C 2MB = 9  AllO} + 18(6){9} + 36(41 = 0 112.5 kN = 112.5 kN T For the interval 0 < x < 4 m:
V = 112.5  18(x + 2} = 18x + ?6.5 kN M 9 + 112.5x  18(x + lex + 2)/2
—9x2 + ?6.5x — 45 kNm 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: + C EMA = 8(8) — 30(4)(2) — 40(6) = 0
B = 60 kN + c 2MB = A(8) + 30(4)(6) + 40(2) =
A = 100 kN Load, shear, and moment diagrams for the beam are shown below: 40 RM 30 Km»... ...
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 Spring '08
 McVay

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