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For the beam shown, draw the reactions at supports A and B in the positive direction, and also draw the shear and bending momCalculate the support reaction Ay lb ENTER 3 tries remaining. 1 point(s) possible Calculate the support reaction Ax lb ENTER

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solution Given, W = 480 lb eblet L = 13 ft a = 2 ft P 480 eblft 9 OB A R 2ft 13 ft 2ft Ay By calculation of End reactions I Iо - (Aso is so se) - - By (480 + 6 2-otago) Ay о Ay 12 бо + 3 3 Аз + 3 = 2 оо EMA=0 + манза) - - 33x480)х(23) 2x489) ж (13 +2{640 } 106080 { $2560} - (By {6408x5 640 + 53040 +7520 - (By x17) = 61200 By x 1750 6) 200 - By x17 By = 61200 17 | By = 3600Answer i Ay = 3600 lb Answer 27 Acc = 0 lb Answer 37 By 3600 lb. Shear force rce and • Calculation of of Shear Bending momentBending g Homent Max = +(3600xx) - (5xxxwx Daltres My = 3600** - (£x*X240) **** Mn = 3600 x 240 x 23 6 Mx= 3600 x 40x3Sheare force Sfx = 3600 - (* *2*480) – [4801x-2)] SF2 = 3600 - 480 - [ 480 [480 x 960) si SFX = 3600-480 - 480x + 960 Sfx= 40Bending moment ( 3600 x 2) Ma = 2)->(*12 43)(2 – 3 ) [480*(x-2)]-(232)] 480(x-3) - 48060-13 Ma 3600 x 480 x (2-2) x (X-2 2Mac = 36002 - 480x - 480x4 3 2. Ma = 36002 - 480 (x² - p=42+4) 2480 480 x 1920? - {2408x24x+4)} Ma = 3600 x-4802 + 640 -{2402Taking section 3 - 480 lblft 3 ولنا s! B R Ri aft Bft aft Ay = 360006Range Section 3 15 Pt <a < 17 ftDUE TO LESS TIME I AM UNABLE TO COMPLETE IT

YOU CAN UPLOAD IT AGAIN AND I WILL SOLVE REMIANING PART

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