HW 4 Solutions

HW 4 Solutions - Problem 5.4 {a} Draw the free—hotly...

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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 free-hotly diagram of the beam. {le Determine the reactions at the supports. Solution: (a) The FED 4b] The equilibrium equations 2MB : —t-l-II| 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 free-body 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 1-440 ft-lb I 1440 a- 11 l 36 1h Ll 6E] lb.-"ft{ ——12ft—— son-'IhH-IIHIJ"l"l--l--l..l..l--1'1--,.__.._ 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: [marl-air 3: V3“) : R, + B. Mg“) = (R, +BJI + C — 33. The reactions are 1+4m—u- ‘n—fildTl—I-l-I— 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 Wat-TN. 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]|5|M3(x]|. The mtirmml moment magnitude Boom in the Illiid interval: ...
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HW 4 Solutions - Problem 5.4 {a} Draw the free—hotly...

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