7-18. If the force P = 800 lb. determine the maximum shear stress in the beam al the critical section.The supports at A and B only exert vertical reactions on the beam. Support Reactions: As shown on FBD. Internal Shear Force: As shown on Shear diagram, V.u = 800 lb. Section Properties: . I>A 2.75(2.J)(3) + 0.75(6)(1.5> 200 lb,ft -3ft - 6 ft — —4— 3 ft 6 in. »T fan. 2.5(3)1-6(1.5) = 1.6591 in. *xXtj*iuoA goo tk 1.5 in. I /w - ^(3) ( 25') + 3(2.5)(2.75- 1.6591)1 + ^(6)(l.5') +6(1.5)( 1.6591 -0.75)2 = 21.9574 in4 fi... " fA'» 1.17045(2.3409)(3)» 8.2198 ui' Mazimum Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula p* 'l^c |*/« ' Moo It 00 N, 1 N -t * a 00 xao /f 21.9574(3) 357
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© 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 allcopyright 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.
7-17. Determine the largest end forces P that the membercan support if the allowable shear stress is Ta||ow = 10 ksi.The supports at A and B only exert vertical reactions on thebeam.
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on Shear diagram, V>u = P.
Section Properties:
. IjA 2.75(2.5)(3) + 0.75(6)(1.5)
IA 2.5(3)+ 6(1.5): 1.6591 in. 3 in. Lj
"̂ 1.5 in.1.5 in.
i(3)(2.5')+3(2.5)(2.75-l.6591)212
^(6)(l.5')+S<1.5)(1.6591-0.75)J
= 21.9574 in*
S... » ?*' " I- 17045(2.3409X3) - 8.2198 in1
Allowable Shear Stress: Maximum shear stress occursat the point where the neutral axis passes through the section.Applying the shear formula
I r-4-ni i
'' 3y« T ift * J>e T J/f 1P+kOO P-K>00
*„,,
10 =•
ItP (8.2198)21.9574(3)
P * 80.1 kip
0
-F
(, 00
3 N-J-b
P
' aoo
xfjt;
A n s
7-18. If the force P = 800 lb. determine the maximumshear stress in the beam al the critical section.The supportsat A and B only exert vertical reactions on the beam.
Support Reactions: As shown on FBD.
Internal Shear Force: As shown on Shear diagram,V.u = 800 lb.
Section Properties:
. I>A 2.75(2.J)(3) + 0.75(6)(1.5>
200 lb,ft
- 3 f t - 6 ft — —4— 3 ft
6 in.
»Tfan.
2.5(3)1-6(1.5)= 1.6591 in.
*xXtj*iuoA goo tk 1.5 in.
I/w - ^(3) ( 25') + 3(2.5)(2.75- 1.6591)1
+ ̂ (6)(l.5') +6(1.5)( 1.6591 -0.75)2
= 21.9574 in4
fi... " fA'» 1.17045(2.3409)(3)» 8.2198 ui'
Mazimum Shear Stress: Maximum shear stress occursat the point where the neutral axis passes through the section.Applying the shear formula
p* 'l^c |*/« 'Moo It
00
N,1 N-t
* a
00
xao
/ f 21.9574(3)
357
© 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 allcopyright 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.
7-27. Determine the shear stress at points B and Clocated on the web of the fiberglass beam.
in|—3ft- -3ft- - 6 f t
Support Reactions: As shown on FED.
Internal Shear Force: The shear force at section a- a,V... = 428.57 Ib.
Section Properties:
/», =i(4)(7.53)--!-(3.5)(V) =77.625 in'
Qc =y'A'= 3.375(4)(0.75)= 10.125 in
Shear Stress: Applying the shear formula
= V&
428.57(10.125)
lfo(tj*900lk to
r—r—i H
3K 7ft
876 -57 It bZI-4) 'i
ii>:=4235711
i=jLh
77.625(0.3)112 psi Ans
3;n.
o7*;i.
*7— 28. Determine Ihe maximum shear stress acting in thefiberglass beam at the critical section.
2001b/ft
,tITm> Support Reactions: As shown on FBD.
Internal Shear Force: As shown on shear diagram.K... = 878.57 Ib.
- 6 f t
Section Pri
2 f t- 6 f t
= ^(4)(7.5')-l(3.5)(6!) =77.625^
0.5 in.-
°-7-iin-
6 in.
cu.• 3.375(4)(0.75) + l.5(3)(0.5) = 12.37J in'
4 in. 0.75 in.
Maximum Shear Stress: Maximum shear stress occursat the point where the neutral axis passes through the section.Applying the shear formula
^T_.; -iit
878.57(12.375)77.625(0.5)
280 psi AIU
875-57*
-r<j<.)
361
© 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 allcopyright 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.
The beam is made from three polystyrene strips thatare glued together as shown. If the glue has a shear strengthof 80 kPa, determine the maximum load P that can beapplied without causing the glue to lose its bond.
Maxim urn shear is «1 the supports
30mm
40mm
60mm
40 mm
i'— 20mm
0.8 m*H—1 m -
-1(0.02X0.06)' + 2[ — (0.03)(0.04)' + (0.03X0.04X0.05)2] = 6.68(10~')m4
VQ (3f/4K0.05K0.04X0.03)
6.68(IO-«X0.02)
P . 238 N Ans
7-46. The beam is made from four boards nailed togetheras shown. If the nails can each support a shear force of100 lb., determine their required spacings s' and s if thebeam is subjected to a shear of V = 700 lb.
Section Properties :
j,g*"
3.3548 in
'MA = dO)( l') + 10(1)(3.3548-0.5)!
yj(1.5>( 10') + 1.5(10){6-3.3548)2
= 337.43 in4
fit ->i'X'» 1.8548(3)( I) = 5.5645 in'
& «>»'<*' = 2.6452(10) (1.5) = 39.6774 in'
Shear Flow : The allowable shear flow at points A and B LS100 100
0, » — md ?j => —~ . respectively.
100 _ 700(5.5645)
* " 337.43"
_VQ,
"'- —100 700(39.6774)
"7 337.43
J = 8.66 in. Aiu Ans
371
© 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 allcopyright 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.
11-11. Select the lightest-weight steel wide-flange beamhaving the shortest height from Appendix D that will safelysupport the loading shown, where w — 0 and P = 10 kip.The allowable bending stress is callow = 24 ksi, and theallowable shear stress is Ta]]ow = 14 ksi.
Bending Stress: From the moment diagram. AfmJI = 60.0 kip fL
Assume blinding controls the design. Applying the flexure formula.
IIIP
— 8 ft -• 6 f t -
60.0(12)30.0 in'
Three choices of wide flange section having the weight 26 Ib/f t can bemade They arc W12 x 26, W14x26. and Wl6x26. However, the
shi>riL-stis the w 12x26.
Select W|2x26 (S, * 33.4 in'. d= 12.22 in.. ru. = 0.230 in.)
VShear Sirest: Provide a shear stress check using r = — for the
'.«WI2 x26 wide-flange section. From the shear diagram, Vmlx = 10.0 kip.
10.0
Sft ift
100
Hence.
0.230(12.22)
= 3.56ksi < r,,,,,, = Uksi (O.K!)
Use W12x26 Aiu -60-0
11-12. Determine the minimum width of the beam tothe nearest j in. that will safely support the loading ofP = 8 kip. The allowable bending stress is trMcm = 24 ksiand the allowable shear stress is raUow = 15 ksi.
- 6 f t - - 6 f t -
Beam design: Assume moment controls.48.0(12><3)Me
—; 24 =
= 4in.
Check shew:
T . H -It ~
= 0.5ksi < ISksi OK
580
© 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 allcopyright 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.
11-17. The steel cantilevered T-beam Is made from twoplates welded together as shown. Determine the maximumloads P that can be safely supported on the beam if theallowable bending stress is tranow = 170 MPa and the
150 mm
~y *
mm
150 mm
15 mm
r
1
Section properties:
0.0075(0.15X0.015) + 0.09(0.15)(0.015)
0.15(0.015) + 0.15(0.015)= 004g75m
/= -L(0.15)(0.015)5 + 0.1S(0.015)(0.0487S-0.0075)112
-I- —(0.015)(0.15)3 + 0.015(0.15)(0.09-0.04875)2= ll.9180r.ur6) m*
/ 11.9180UO-6)
c (0.165-0.04875)
-,., .(0.165-0.04875)fi«,.i = /A = ( — )(0.165-0.04875)(0.015) = 0.101355( UT5) m3
Maximum load: Assume failure due to bending moment
W... = <T.HO. S; 6P = 170(106)(0.10252)(10"3)
P - 2904.7 N - 2.90 kN Am
Check shear:
_ »4.,C=... _ 2(2904.7)(0.101353)(10-3)3.29MPa < r.iio, = 95 MPa
11-18. Draw the shear and moment diagrams for theW 12 X 14 beam and check if the beam will safely supportthe loading. The allowable bending stress is <ral|OW = 22 ksiand the allowable shear stress is Tallow = 12 ksi.
Bending Stress: From the moment diagram. Afmi( = 50.0 kip ft
Applying the flexure formula with 5 = 14.9 in3 for a wide-flangesecuon W12x 14,
1.5 kip/ft
50 kip-ft
V
- 3 f t - 12ft
<*„., = 550.0(12)
14.940.27 ksi > (TlM3. = 22 ksi (No Good!) Cf
Ir ------ \ f,0 L'P
Shear Stress: From the shear diagram. Vmtl = 13.17 kip. UsingV
r= — where d= 11.91 in. and f,. = 0.20 in. for WI2x 14 wide
fiance section.
K_
" Vt t.f
t.d13.17
0.20(11.91)= 5.53 ksi < rlllo. = 12 ksi (O.K.')
Hence, the wide flange section W 1 2 x l 4 fails due lo thebending stress and wi l l not safely support the loading. Ans
/3/7
-50-0
77?
583
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