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Ce315 Ce31s4 p2.4 de Guzman

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0% found this document useful (0 votes)
59 views3 pages

Ce315 Ce31s4 p2.4 de Guzman

Uploaded by

Christian
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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‘Submitted by: De Guzman, Christian M. Ison, Vince Isaac Peter P, Recrea, Margie T. Sumbilla, Lalaine A. Virtucio, Lian Nica D. Arlegui,Quiapo, Manila Sructural Theory CE315 CE31S4 Plate No. 2.4 Shear and Moment Diagram Rate Submitted to: 2011684 Engr. Jamaica Alyanna D. Asor 1510851 1810980 2010282 1811855 Date Submitted: ‘September 22, 2022 Scanned with CamScanner PROBLEM NO. 1 SKETCH THE SHEAR AND MOMENT Dinca! = AND COMPLETE «THE §= VALUES. IN THe DiAcRAM. je BOWING FOR REACTIONS SMe = 02+ (2 + Ba (e)-4 BX) (4x9) =0 Ras tN /m x-07t Ba=O Fy Ott -4 (26) 44 + By =0 B= Bin SOWING FoRx IF SPRNOREL 1 “—S fa = = 1 e41x4e Vat =O Mas 0 +12 =12 SS ra Var= OF 421 Me 124% (Xi) = 2.3839 Vous 1-$)e)=8 Mee 13.3833 - + xa) + LmuA) +144)-0 Ve+ -8+8°0 = $0(9)- -4 «133353 Scanned with CamScanner PROBLEM NO. 2 SVErOA THE SHEAR AND = MOMENT DIAGRAM] = ANO COMPLETE THE VALUES IN THE DIAGRAM. 50 Ibi SOLVING FOR REACTIONS 2 Mn = Ot 50(20)(10) ~ Re (20) + 200 =O Re = SiO Ib /py FXO 7t An =O FyOtt ~50 (20) + Slo t Av =O Av = 49010 /pr VaL>O Vae= 0+ 440 = 440 Na+ Hap - 50 (28) = -510 Vee = -510 + 510 =O ¥ too , ie -lo00, . 510 zx a0 * x X29.B X= 10.2 4Ma=O Ho= 04 4 (490) (4.8) = 240) Ma = 2401 - + (210) (10.2) > ~200 Mer = 295 +0 = - 200 Nee = -260 + 200 =O Scanned with CamScanner

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