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Unformatted text preview: PROBLEM 5.71 Owl m 0”] m 0.125 m 22L2$ N In *5.71 Beam AB supports a xmifomﬂy distn'buted load of’Z kN/m and have concentrated
loads P and Q” It has been experimentally determined that the normal stress due to
bending on the bottom edge of the beam is —569 MPa at A and —299 MPa at C
Draw the shear and bending—moment diagrams for the beam and determine the
magnitudes of the loads P and Q. SOLUTION 4 I = r‘a(\8)(scl‘ Gmsux/o‘m 'I s = ET : 3.883on mg: 3.888x166m2
A+ A M. = 6"
MA=(3.888XIO‘°)(‘5é.°1)=22125 Nm M c Mg: 56’c
M5 = (aﬁsleo‘CX—Z‘H): ne.25 N~ m 9 2M. = 0
221.23 — (ammo) — 0.2 P  0.325Q = o 0.11” + 0.325 G. — I3/.ZS m +DML = 0
H625 ‘(o.orl(2ool O.I P  0.22312 = O.IP+ 0.22561 r 10625 (2) O Soﬁvina (l) Avml (23 simuj'i‘avxeousjj
P= 500 N: 4:
Q: 250 N ‘ gear/hon "Force a)" A
PA— 400 500 — 250 = 0
RA : I350 N'M VA = “50 N V9: 250 MA: 422.25 Nam
M‘f llé.2§ Nm MD 2 ‘ N‘v" 5.97 Determine the largest permissible uniformly distributed load w for the shown, knowing that the allowable normal stress is +12 ksi in tension and  19,5 ksi
in compression” PROBLEM 5.97 SOLUTION Qeaol‘ious: Q +C36w=0 B=C=lgaw Shear: VA : o
Vg‘: 0— SW = ' 8W
VB+= —8w+1$w= IOw
Va": 10w  20w =—H>W
Va ‘: ’.!C>w + L8 w = 8w
V9: 8w — 3w = O AV‘EQS T A 1‘» B "(&)(2}é§8wl r—32 w
8 +«» E (#)(Io)(to w) : 50 w
Benohnﬁ mome'l‘S? MA = 0
Mg : O ‘ 1’ ‘32W
M53—32W'f50vl 1' I8w Cen‘f‘woi'cl and momen‘l‘ o‘P [me/4‘42; D._7£_I 09492 O.7l_lmq>_ 2,5294 n,"
Bendinﬁ momen+ 19ml}: M f  (VI/j
Ten$:'on nit 8 duel C  (l23( ‘ﬂ_‘3706) 2  RI'P {h
COMP. ad 8 va. c ‘ (415w 14343 3 = — 17. €144 tar," in <\
Tension d E (12Y— Limits): H.212 kipl in <1 Compression at} E  (17.5 )(23906 )3 44.6 ‘0"... iv; AWwaMe foul w B F c: — 32w : — 27, 4m W: 0.87% top /."n
E l8 W: [7. 2’2 W: 0.754 /fn
Smal’les'l’ w . 0379 lap A». r to.“ («P IH. «I 5.99 Beams AB, BC, and CD have the cross section shown and are pinconnected at B
PROBLEM 599 and C Knowing that the allowable normal stress is +110 MRI in tension and e 150
MPa in compression, determine (a) the largest'permissible value of’w if beam BC is not
to be overstressed, (b) the corresponding maximum distance a for which the cantilever
beams AB and CD are not overstressed 125 mm r290 mm—vI i SOLUTION T MB 2 ML 3 O
g___i Vs =  Vc. = (%)(7.z.)w= as w Area. 8 +0 E 0‘? shear Amara“
$808: w) = 6.42 w ME: O+6J42Hu r 648mi Cen‘l’roicl anal momew+ o‘F {ner‘l’iq 3‘10625 344.82 140625 HC.‘!3 9.043;); 3.5l6xlo‘
53:25:: 7.073 xzo‘ 3.5tasrb‘
= I2L'I3 mm _ 534250
4375‘ I = EAJ‘ + 23:" r we. “0‘ m: Leea‘l‘iou .Yé'm) I/ (ﬁr3x6? aﬂéo (Io"m3) 4”?
LO‘H’OM Bending momen‘l‘ Anti“: Mt  SI/y
Tension «+2; i,  (How/ofW—37Azmo‘é) s: 1422403 Mu.
Wail E3 — (lSoxlo")(z$8éx107‘») =.—‘ 38. Sat/03 Mm
Tagabu A MD: —(.Hoxlo‘) (253.5160 242.4910" Nu
Comp.¢+ A41):  (Isarlo‘X—Szwxlo‘fl:ﬂamxlo’ N = 1.420103 N/M
= 1.48: kN/m « Alpavagje In“, W 6.48%! '‘ iéaxlcs W {Shear 3:} A LV‘; = (a.+'3.é) W Area A +1: 13 of? shear disarm iaG/A + VB) 1‘ %a(a.+7.2)w Bending momewl A (415.: D) M, a(a+"7.2.) W
7inta+22mthsnaﬁ =  3,:u x103 7; o.‘ 4. 3.6 a. e— 9‘83? = o a.’= L635 m 4 5.37 and 5.38 Draw the‘shear and bendingmoment diagams for the beam and
loading shown and determine the maximum normal stress due to bending 300 20 mm SOLUTION c Al D chc boob, EFGH 30mm Nah «wa ME = 0 Jun
7 4» king.
QEME = O
0.: H(o2).C‘+o?(O.4))(3oo) = O
H = 213.33 N
.1sz : o vg4o~—300+2:333=0
vE = I24 €7 N Skew: E +0 F v = 124.57 um
F b G V = 86.67 Nm
6+0 H V =2I3.83 Nm
BanalM3 mud a F
E Mr :0
D M; (O.2)ClZé.€7) = o V Mp r Nm
was? Bending! Momen+ on" G 7 M; .. DEMG=O
x (a  M; + (0.2)(2I3..33) = 0
M5 = Nm 28.3%
2€.G7 Frye bed] A BC 0 E
+9? Mta = o —o_e A +(o.4 )(3ao) +(O.2)(3oo) (0.1)(I2‘c.c1) 2 o
A: 257.7? N
4' M, = O ~(O.2)(3oo\ —» (04)(300) (oz)(12(..57)
+ 0.629 = O
D = 468.8? N A 3 MB Bending Momen‘f 03‘ 8
ﬂ 92MB=O Muir/I“: 5!5C Nm
—Co.2)(257.78) + We '—‘ 0
“5733" M3 = 51.56 MM 5 = gbb" = 2"(20‘X30)z
Boo = 3 “03 rams = 3KIO‘ M3
A B c Bending meme/21‘s.! C _
1 +92 r)? = 0) Normal) shes:
—(o.4 257.72 +(0.2)(3oo)
1577" V + Me, = 0 Stﬂa— = III“: ><lo‘ Pa.
M; r 42.“ Mm 3““) : I7.I°r MP0. < MD Benchnj momen‘} a} D
C D 5 +3 EH, , o _ T ‘ M9 ~(O.2)(1l3.33)= 0
V Mo = — 25.33 Nm 5.97 Determipe the largest permissible uniformly distributed load w for the PROBLEM 5.97 shown, knowmg that the allowable normal stress is +12 ksi in tension and ~19,5 ksi
in compression“ SOLUTION Wax/Hons: B+C3éw=0 B=C=l§,w VA: 0
Vgr O~ 8w = #8“!
VB+= 8w+13~w—‘ 10w
V5: jaw—20w=10W
Vc+=410w+ 13w =8w
VD: 8w—8w‘: 0 Areas? A 1‘. B "(%)(2)&8wl r—32 w
B +0 E (MIon w) :50 w
Bending M0m9w+s 3 MA = 0
MB: r ‘32“)
ME ~‘ —35Lw +50 w = I8w Cen+v~o£cl anal momen+ J‘inerh‘a Bendinrj MOM€A+ firm}: TenJon «f 8 wt C ~ (my 3906) = — 22.;37 mp. {n C9mp of} B and C  («FLSY— 1.4393] 2 " 17. 749 tarp in 4
Tension ~+ E —(12Y— 1.4393): 17.212 id,» in <\
Compression a} E (~H.5 )(2.3q06 )2 44,6 klr. in Mooth Lao! w B as C — 32w =  2?. 449 w: 0.27% k7, A».
E W 7 I7. w t O. /[M Smﬂes‘r w = 0.879 lap/n. r 10m up /H. «a CEN 302 — Mechanics of Materials
Section Exam #2A 30w 710K352. Name ................................................................................................. .. Problem 1. (30 points} The concrete post in Figure (a) is reinforced axially with four symmetrically placed bars, each with a cross
sectional area of lSOOmmz. The moduli of elasticity are 150GPa for steel and 15 GPa for concrete. (a) Compute the stress in each material when theﬁl 000 kN load is applied. (10 points), (b) If the initial length of the column is L = 10 ﬁ, how much does the column shorten when the load is applied?
(.5 points) (0) If the allowable stress in the post is IOOMPa for steel and 4MPa for concrete, compute the maximum safe
axial load P that may be applied. (5 points) Cw Wje (d) If instead of using steel and concrete we only used concrete determine the stress inWand the ﬁnal length
of the column (after the application of the load).(5 points) (e) What are the longitudinal and lateral strains in this case? What are the new cross—sectional dimensions
(average value)? Assume that the Poisson’s Ratio is v = 0.33. (5 points) PCorb n. at” o Mk; E4343 M A}? 31:23:31“ 2
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, ,wwmmwl M Problem 4 (15 points) A) Derive the expressions for the shear force and bending moment for each segment of the beam (1) and (3)
(7.5 for each). Express shear and moment functions in terms of X. For segment 1, x should be taken as shown
below (starting at the left support), While x in the segment 2 should be taken from the right support. 400 lb/ft 200 lb/ft / 50% 0‘7 7L 200% 2..
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Mi—f ISD Q Problem 2 (35 Points)
The box beam shown in the ﬁgure below supports the concentrated loads 8001b and 20001b. Calculate the shear and moment diagrams (10 points each). Compute the maximum tensile and maximum compressive stress due to
the bending moment (10 points). Show the stress distribution in the section shown below (5 points). 6in
8001b 20001b <—————» [V] ~600‘ [M1 Section — Draw Stress
‘ Distribution Co wright
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 Force, ........., allowable normal stress

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