e 265 p cable a b c d 8 ft 6 ft 6 ft 6 ft solution

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) E 265 P Cable A B C D 8 ft 6 ft 6 ft 6 ft Solution 4.3-11 Beam with a cable E P Free-body diagram of section AC P Cable P A 6 ft 4P __ 9 B 6 ft C 6 ft D 8 ft P 4P __ 9 4P __ 9 A 6 ft 4P __ 5 B 3P __ 5 6 ft C M N V MC M 0 UNITS: P in lb M in lb-ft 4P 4P (6 ft) (12 ft) 0 5 9 8P M lb-ft 15 Numerical value of M equals 640 lb-ft. 8P lb-ft 15 1200 lb 640 lb-ft and P Problem 4.3-12 A simply supported beam AB supports a trapezoidally distributed load (see figure). The intensity of the load varies linearly from 50 kN/m at support A to 30 kN/m at support B. Calculate the shear force V and bending moment M at the midpoint of the beam. 50 kN/m 30 kN/m A B 3m...
View Full Document

This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Tech.

Ask a homework question - tutors are online