52k ft 707k ft A B 2klf 1245k 114k B f 136k B xft Mmax Page 2 UNIVERSITY OF

52k ft 707k ft a b 2klf 1245k 114k b f 136k b xft

This preview shows page 1 - 4 out of 4 pages.

70.7k-ft AB2klfAB2klfAB2klf-12.45k 1.14k BfBfBf13.6k B xft Mmax
Background image
Page 2 UNIVERSITY OF CALIFORNIA, BERKELEY Department of Civil Engineering CE 120 - Instructor: J. P. Moehle - Spring 2015Problem 2 For the beam shown, calculate the reactions, and draw the shear and bending moment diagrams. Sketch the deflected shape considering the boundary conditions and considering flexural deformations only. Note that there are hinges at C and D. Solution: Free Body Diagram: From 3 + ∑Fx=0; P2=0kFrom 2 + ∑Fx=0; P1=0k+ Mc=0; 5*4.5+V2*9=0V2=-2.5k+ ∑Fy=0; V1-5-5-V2=0V1=7.5k From3 + MF = 0; V2*12-0.5*6*3*8+R4*6=0R4=17k + ∑Fy=0; V2-9+R4+R5=0 R5=5.5k From-1 + ∑Fx=0; R1=0k+ MA=0; V1*15+9*2*4.5-R3*9=0R3=21.5k + ∑Fy=0; -V1-18+R2+R3=0 R2=4k 2klf3klf5 k5 k2klf3klf5 k5 kR3 R1 R4 R2 V1V1 P1 Vtt2V2 P2 R5 1 2 3 3klf
Background image
Page 3 UNIVERSITY OF CALIFORNIA, BERKELEY Department of Civil Engineering CE 120 - Instructor: J. P. Moehle - Spring 2015Problem 3 For the frames shown, determine the reactions and draw the axial force, shear, and bending moment diagrams. Sketch the deflected shapes considering the boundary conditions and considering flexural deformations only. Solution: (a) 2klf4k(a)12’14’7’2klf2klf4k(a)R3 R1 R2 + ∑Fx=0; R1= -4k=4k+ Ma=0; 4*7+24*6-R3*12 =0R3=14.33k + ∑Fy=0; R2+R3-24=0R2=9.66kDeflected Shape
Background image
Page 4 UNIVERSITY OF CALIFORNIA, BERKELEY Department of Civil Engineering CE 120 - Instructor: J. P. Moehle - Spring 2015(b) R3 R1 R2 + ∑Fy=0; R2= 0k+ Ma=0; 5+R3*20=0R3=-0.25k=25k + Fx=0; R2+R3=0R2=0.25k Deflected Shape
Background image

You've reached the end of your free preview.

Want to read all 4 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture