Answered: 5 kN 2 m C E. A 6 kN 8 kN 2 m 2 m 2 m

Engineering

Civil Engineering

5 kN 2 m C E. A 6 kN 8 kN 2 m 2 m 2 m

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Drawing Framing Plans

A framing plan is a plan that is a blueprint to display the locations, materials (like timber, steel), spacings, sizes, framing members, and the number of structural components required to construct the house. Drawing the joists and double framing in such a position so that they can be occupied in the completed building. There is no requirement for joist dimensions at each position. Perpendicular to the joist, the drawing of bridges is made and noted as “3-2” x 3 bridgings. Wall framing plans depict the position and methodology of framing openings and heights of the ceiling so that studs and posts can be cut out. As it is a vertical plane view, technically a wall framing plan is not a plan.

Question

Engineering Mechanics

DETERMINE MEMBER BD USING THE METHOD OF SECTIONS.

Note: for compression, include a negative sign ( - ) and for tension, type answer as is. Use four (4) decimal places during computation and two (2) decimal places for the final answer.

5 kN
2 m
C
E.
A
V6 kN
8 kN
- 2 m
2 m
2 m

Branch of science that deals with the stationary and moving bodies under the influence of forces.

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sorry，what are these： Step 3 Slope deflection equation In section AB MAB= MFAB+22EIL2θA+θB-3δLMAB= -12.5+4EI5θB MBA= 12.5+4EI52θB In section BC MBC= 0+2EI52θB+θC MCB= 2EI5θB+2θC In section CD MCD= -16.67+4EI52θC MDC= 16.67+4EI5θC arrow_forward Step 4 Equilibrium equations Equilibrium equations MBA+MBC= 0 -(i)MCB+MCD= 0 -(ii) Now, From i12.5+8EIθB5+4EIθB5+2EIθC5= 0EIθB125+EIθC25 = -12.5 -(iii) From ii2EIθB5+4EIθC5+-16.67+8EIθC5=02EIθB5+EIθC125= 16.67 iv Solving equations (iii) and (iv), we get EIθB= -6.548EIθC = 8.037