Internal Forces in Frames and Beams: Shear and Bending Moments, Study notes of Statics

Download Internal Forces in Frames and Beams: Shear and Bending Moments and more Study notes Statics in PDF only on Docsity! Chapter 7: Internal forces in Frames and Beams In Chapter 6, we considered internal forces in trusses. We saw that all the members are 2-force members that carry only tension or compression. In this chapter, we will consider internal forces in Frames and Beams. Recall that these structures have atleast one multi-force member. Multi-force members can carry additional types of internal forces such as shear and bending moment in addition to tension/compression. For example, consider the cantilever beam shown with an end load. We can find the external forces using the FBD of the entire beam. However we may also want to find out the internal forces (and moments) at different points of the beam. This will help us decide if the beam can support the applied load or not. To do this, we imagine two (or more) sub-parts of the beam as shown. Tuesday, November 17, 2009 10:53 PM CE297-FA09-Ch7 Page 1 Read example 7.1 Radius of pulleys = 200 mm Exercise 7.17 & 7.18 Find the internal forces (& moments) at J & K. CE297-FA09-Ch7 Page 2 for region @_ Feo. oan at 240 bb Broblb an 240 300 V Fer ogy i nage O: WP], 2A +0 => _120-v-0 He . > =a M eu ‘o ZMy=027 120% 4M =0 “ reer aor —% for region @) FRDs: 29 at a r2blo we 300 120 mM My v 240 Ub Bobb x guo 2 Boo me for ‘a! Ai i ‘ | Vb ae sega! am aaj » v 7a e 1b 2h, =05 -l204+240-V20 “oa m 1g00 ey © ZMy=0 > oa -240(n-10) +m =0 | 8 * AM = =2400 + 120% | aot — CE297-FA09-Ch7 Page 5 | | mM, it, 7 AR 240 Up Br0lb Ho Le <———_—> fbr nt in region @: ——~ * d ZBy 20 27 4V+300-120=0 120 Uo ae EN) 120 T Ve A600 7 SS] Ww 55. a > -™ +300(55-%) ~120(P0-2)=0 3— 2 nt = [0-H] for oregon, @ FRDs: me " ait 7b “ { momy > 240 Uy arobb 2H0 300 ~~ he Foxe ony "a" ine nopion GB): Ww aR-0 > “ * oe ZMy20 a> —-M—120(70-4)-0 > [a= Bio DH] Matin Avante Valu of Shar: |$0 bb Momunt : 1800 Loin CE297-FA09-Ch7 Page 6 7.6 Load vs. Shear vs. Bending moment Drawing Shear force and Bending moment diagrams for a beam can be simplified by using relationships between Load vs. Shear and Shear vs. Bending Moment. These relationships can be derived simply from statics as follows. Consider a small ∆x length of any beam carrying a distributed load. ( )curve loadunder area−=−=− ∫ D C x x CD dxwVV ( )curveshear under area==− ∫ D C x x CD dxVMM Read examples 7.4, 7.5, 7.6 and 7.7. Sunday, November 22, 2009 8:09 PM CE297-FA09-Ch7 Page 7