# HW 5 - 260 CHAPTER 4 Shear Forces and Bending Moments...

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Unformatted text preview: 260 CHAPTER 4 Shear Forces and Bending Moments Problem 4.3-3 Detennine the shear force V and bending moment M at the midpoint of the beam with overhangs (see figure). Note that one load acts downward and the other upward. Solution 4.3-3 Beam with overhangs 2:M A = 0: R B = P (1 + 2;) (downward) i P 'J A C c= I ~b-4:-Ll2~ R A v FREE-BODY DIAGRAM (C IS THE MIDPOINT) = 0: 2:F VERT 1 2bP RA=-[P(L+b+b)] V = R A - P = P L P (I + ~)- L - 2:M c = 0: M=P(1 +~)(~)-P(b+~) PL PL M = - + Pb - Pb - - = 0 - 2 2 Problem 4.3-4 Calculate the shear force V and bending moment M at a cross section located 0.5 m from the fixed support of the cantilever beam AS shown in the figure. Solution 4.3-4 Cantilever beam :4.0kN 1.5 kN/m 2:.F VERT = 0: V = 4.0 kN + (1.5 kN/m) (2.0 m) = 4.0 kN + 3.0 kN = 7.0 kN - 2:M D = 0: M = -(4.0 kN)(0.5 m) - (1.5 kN/m) (2.0 m)(2.5 m) = -2.0 kN . m - 7.5 kN . m FREE-BODY DIAGRAM OF SEGMENT DB = -9.5 kN' m - Point D is 0.5 m from support A. ! 4.0kN M SECTION 4.3 Shear Forces and Bending Moments 265 Problem 4.3-11 A beam ABCD with a vertical ann CE is supported as a simple beam at A and D (see figure). A cable passes over a small pulley that is attached to the arm at E. One end of the cable is attached to the beam at point B. What is the force P in the cable if the bending moment in the beam just to the left of point C is equal numerically to 640 lb-ft? (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.) Solution 4.3-11 Beam with a cable E FREE-BODY DIAGRAM OF SECTION AC N 4P 9 4P 4P M --(6 ft) + - (12 ft) = 0 5 9 8P M = --lb-ft 15 Numerical value of M equals 640lb-ft. 8P :. 640 lb-ft = -lb-ft 15 and P = 1200 lb - 4P 9 UNITS: Pin lb Min lb-ft Problem 4.3-12 A simply supported beam AB supports a trapezoidally 50 kN/m distributed load (see figure). The intensity of the load varies linearly j < 30 kN/m from 50 leN/m at support A to 30 leN/m at support B. I ! i ' i Calculate the shear force V and bending moment M at the midpoint of the beam. SECTION 4.5 Shear-Force and Bending-Moment Diagrams 273 Problem 4.5-8 A beam ABC is simply supported at A and B and has an overhang BC (see figure). The beam is loaded by two forces P and a clockwise couple of moment Pa that act through the arrangement shown. Draw the shear-force and bending-moment diagrams for beam ABC. Solution 4.5-8 Beam with overhang lP 'P ii, ~ C ">"=>', ~~= £1=(==a====Za=Z;AZ : : y!;=: Z.=Za=,_Z,::3I)Pa •.• ~P Jp .p :p ; B it lower E£=C ==a=======a ==:&:::;;=y;i "==a::::==::jY'1 C beam: Y J2P . ' . .( v o ---~-p-----'--Ll~;-=--==--=--=--=-.-=. F· MO---~-Pa Problem 4.5-9 Beam ABCD is simply supported at Band C and has q overhangs at each end (see figure). The span length is L and each overhang has length L/3. A uniform load of intensity q acts along the entire length of the beam....
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## This note was uploaded on 11/14/2011 for the course ECIV 310 taught by Professor Huckelbridge during the Spring '11 term at Case Western.

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HW 5 - 260 CHAPTER 4 Shear Forces and Bending Moments...

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