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Unformatted text preview: Problem 5.4 {a} Draw the free—hotly diagram of the
beam. [13} Determine the tension in the rope and the
reactions at B. Solution: Let T he the tension in the rope. T
. . . . . gr,” 600 1b
The equilibrium equations are: 3Ou I

 B.
EPA : —T5i1130‘— (600 lb] sinSEJ=+B,t :0 I I '1‘
EPA, : Tcos 3El° — [GUI] lb] EDS 30’ +1?)f = [I
5 ﬁ 9 [‘1
EM}; : moo lb]r053]°t9 ft] — Teosﬁﬂ’tlJ ft] = U Solving Yyields Problem 5.13 (at Draw the freehotly diagram of the
beam. {le Determine the reactions at the supports. Solution: (a) The FED
4b] The equilibrium equations 2MB : —tlII mm m] +At6 m] = o 2F; : —.4 +31 = 0 EF. :40 RN +3. = 0 Solving we ﬁnd Problem 10.24 (a) Determine the shear force and
bending moment as functions of .t. {b} Show that the equations for V and M as functions
of .1' satisfy the equation V = dedx. Strategy: For part (on. cut the beam at an arbitrage
position I and draw the freebody diagram of the part
of the beam to the right of the plane. Solution: Cut the beam at arbitrary I and examme the section to
the right Problem 10.25 Draw the shear force and bending
moment diagrams for the beam in Problem 10.24. Solution: The diagrams 1440 ftlb I
1440 a 11 l 36 1h Ll 6E] lb."ft{ ——12ft—— son'IhHIIHIJ"l"lll..l..l1'1,.__.._ le Problem 10.30 The beam in Problem 1025) will safely
5111]me shear fumes and bending moments of magni—
tudes 2 LIN and bi kN—m. respectively. On the basis
of this criterion. can it safely be subjected In the loads
F = 1 IN. C =1.6kN—m’_' 301111101]: From the solution tn Problem 10.29, the shear and the
momean in the intervals are Internal 1‘: V1=R,, M1:x]=R,..t.
Internal 2: Van] : Ry: Mgr) : R,.\' + C: [marlair 3: V3“) : R, +
B. Mg“) = (R, +BJI + C — 33. The reactions are 1+4m—u ‘n—ﬁldTl—IlI— 3m — B:()q16F—C). MIH :dey =F_B_ Mguj =(R,+B]x——C— SB=F.r+C—:lﬁF—C] The maximum shears in each lumen'31 have the magmt'ude rank: = F" _ 16F " EC Ihe maximum magnitude occurs at .t = S: M3tSJI = SF = 3 kN m
and it exceeds the 3:11 limit by 2.5 kN 111. NO IVMJI = I‘r’ztrll S WatTN. so that the largest shear fur a farce F :1 kl? is V3§1’]=R,. +
B = F — 8 +8 = F = 1 kN, which can be safely supported. The
maximum mument magnitudes in each inten'al have the rank:
M1t.t] 5 M1{.t]5M3(x]. The mtirmml moment magnitude
Boom in the Illiid interval: ...
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 Spring '08
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