Question

Solve for the slope at B the moment-area theorem using E=29000ksi G I = 600 in D B C 3ft 3 ft 6 note

Answer 1

= 2,9x107 lb/inz 29000 ksi' = 29 x107 600 in Р + Hр 3' 3' 6' RA Ro . PEM EMA= + Px 6 – Rox/2 =O Ro= Peb es From symmetry » RA=P 2 EF HEO Hoio Р fara L 2 l 2 P an SFD » A Р 2 -P/2 M= 2 x 6 = 3P lb-ft 3 lb-ft A B

MB Area of SFD between A AB ļ x3 11 Mo 2 1 P lb-ft = 18 lb-fn similarity, Mc = 2 x 6 = 3.P lb ft = 36p-in 36xP 18*P El M EI 11 diagram (D Oec OBC slope at ③ w.rt.. tangent drawn from point ©. = Oo-de ORC = De Area of the diagram b/w ® & ©. .

inch conversion factor Orc = Og *(3x12) 2 972P 972 P 2,9x107x 600 Magnitude 100 = P45.586 *108) radian 59) radion positive sign shows point ® is above the poing from where tangent is drawn.) sence of Rotation *counter-clock coise