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Transcript
```© 2008 by R.C. Hibbeler. Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6-45. The beam is subjected to a moment M. Determine
the percentage of this moment that is resisted by the
stresses acting on both the top and bottom boards, A and B,
of the beam.
Section Property:
B
25mm '
/ = — <0.2)(0.2')-— (0.15)(0.15 3 ) =91.14583(10-') m4
"»
: • 25 mm
150m m
Bending Stress: Applying the flexure formula
Afv
£"
M(O.l)
= 1097.143 M
91.14583(11)-*)
M(0.075)
Resultant Force and Moment:
= 822.857 M
For hoard A or B
F = 821857M(0.025)(0.2)
I
+ -(I097.I43M - 822.857Af>( 0.0251(0.2)
= 4 800 M
M' = F(O.I76I9) = 4.80A/IO.I7619) =0.8457 M
— | = 0.8457( 100%) = 84.6 *
Ans
6-46. Determine the moment M that should be applied to
the beam in order to create a compressive stress at point D
of <TD = 30 MPa. Also sketch the stress distribution acting
over the cross section and compute the maximum stress
developed in the beam.
M
25 mm
25nu
Section Property:
/ = 1(0.2)(0.2')-— (0.15)(O.I5 5 ) =91.14583(10'') m'
Bending Stress: Applying ihe flexure formula
Mv
c=T
M0.075)
M = 35458 N m = 36.5 kN m
Me
/
36458(0.11
•=40.0 MPa
91.I4583(10-»)
Ans
266
25 mm
copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6-57. Determine the resultant force the bending stresses
produce on the top board A of the beam if M = 1 kip • ft.
Section Properties:
IA
0.75(10)(1.5) + 7.5(l)( 12) + 14.2S(6)( 1.5)
10(1.5)
= 6.375 in.
.5') + 10(1.5)<6.375-0.75) 2
+ -M6)( 1.5') + 61 1.5)( 14.25 -6.375)'
= 1 196.4375 in'
Bending Stress: Applying Ihe flexure formula
<To =
Myc
/
1000(12X6.375-1.5)
= 48.90 psi
1196.4375
MyD
~7~ :
1000( 12)16.375)
1196.4375
63.94 psi
The Resultant Force: For lop board A
F =-(63.94 + 48.90) (10)(1.5) = 846 Ib
Ans
6-58. The control level is used on a riding lawn mower.
Determine the maximum bending stress in the lever at
section a-a if a force of 20 Ib is applied to the handle. The
lever is supported by a pin at A and a wire at B. Section a-a
is square,0.25 in. by 0.25 in.
20 Ib
20(2) - M - 0 ;
M = 40 Ib in
Me
40(0.125)
/ ~ £(0.25X0.25')
15.4 ksi
272
copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6-93. The beam is subjected to the loading shown. Determine its required cross-sectional dimension a, if the allowable
bending stress for the material is cranow = 150 MPa.
60 kN
40 kN/m
2 !*
'
-2m-
-1m-
Support Reactions: As shown on FBD.
Section Properties:
4«V««frotoil
. ^ £vA ^ X h) «+ H r«) ( f")
"
"
r
1
I \ /• 1 V 5
1V
/ = — 10) -a + 0 - 0 — a — a\2
U >/
2
b A 12
I
^
6 )
5
37 .
Internal Moment: As shown on the moment diagram
Allowable Bending Stress: The maximum moment is
A/,,,, = 60.0 kN m as indicated on the moment diagram
Applying the flexure formula
150(10')
60.0(IO')(a-ia)
a=O.I599 m = 160mm
Ans
6-94. The wing spar ABD of a light plane is made from
2014-T6 aluminum and has a cross-sectional area of 1.27 in ,
a depth of 3 in., and a moment of inertia about its neutral
axis of 2.68 in4. Determine the absolute maximum bending
stress in the spar if the anticipated loading is to be as shown.
Assume A, B, and C are pins. Connection is made along the
central longitudinal axis of the spar.
801b/in.
D
2ft
3ft-
10 li/.
AM
Note thai 2S.8 ksi < ar = 60 ksi
OK
291
-6ft-
copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6-98. The wood beam is subjected to the uniform load
of w = 200 Ib/ft. If the allowable bending stress for the
material is <ranow = 1.40 ksi, determine the required dimension b of its cross section. Assume the support at A is a pin
and R is a roller.
.irrrrrm
V(lb)
Allowable Bending Stress: The maximum moment is
M mll = 5688.89 Ib ft as indicated on moment diagram.
Applying the flexure formula
M,.,c
l.40( 10')
2844.44(12X0.75*)
b = 4.02 in.
Ans
6-99. The wood beam has a rectangular cross section in
the proportion shown. Determine its required dimension b
if the allowable bending stress is cral)ow = 10 MPa.
500 N/m
2m
r
— 2m
I
,
fm
7#>
Allowable Bending Stress: The maximum moment is
Wm<1 = 562.5 N m as indicated on the moment diagram.
Applying the flexure formula
;
^N\
-zso
MCA/-*-;
10( I0 6 ) =
562.5(0.75t)
b = 0.05313m = 53.1 mm
*
L,
Ans
294
copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6-120. The composite beam is made of 6061-T6 aluminum
(A) and C83400 red brass (£'). If the height h = 40mm,
determine the maximum moment that can be applied to the
beam if the allowable bending stress for the aluminum is
(<railow)ai = 128 MPa and for the brass (cr aUow ) br = 35 MPa.
k
50 mm
]
150mm
Section Properties: For transformed section.
Allowable Bending Stress: Applying the flexure formula
Assume failure of red brass
Eb, lOl.O(lO')
btl, = nbii = 0.68218(0.15) = 0.10233m
Me
(<T-"°")br = £7
. _ lyA
35( 10')
_ 0.025(0.10233)(0.05) + (0.07)(0.15)(0.04)
0.10233(0.051 + 0.15(0.04)
M(0.09 -0.049289)
7.45799( lO-«)
= 6412N m = « . 4 l k N - m (controls!)
Ans
Assume failure of aluminium
= 0.049289 m
Me
4M-|5(ft«mj)(a051) +0.10233(0.051(0.049289-0.025)2
7.45799(10-')]
•|-J2(0.15)(0.041)-HO.I5(0.04)(0.07-0.049289)!
= 7.45799(10-') m4
6-121. A wood beam is reinforced with steel straps at
its lop and bollom as shown. Determine the maximum
bending stress developed in the wood and steel if the beam
is subjected to a bending moment of M = 5 kN • m. Sketch
the stress distribution acting over the cross section. Take
£w = 11 GPa, EA = 200 GPa.
20 mm
4- — M = 5 k N - m
20 mm
7 = —(3.63636X0.34)' - — (3.43636)(0.3)' = 4.17848(10"')m'
Maximum stress in steel:
_nMc, _ 18.I82(S)(10')(0.17)
'"' "" ~ 7
4.17848(10-')
3.70 MPa
Ans
Maximum stress in wood:
7
= 0.179 MPa
4.17848(10-')'
An
» st ) = n(<rw)-m, = 18.182(0.179) = 3.26 MPa
1\
F
I rt
3U Mf.
309
```
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