Answer #2 and use AREA METHOD only. Find the Shear Moment Diagram of the following problems usingONLY AREA METHOD.30 kN50 kNtoBA1.1 m3 m2 m2000 lbM = 4800 lb-ftCAD3 ftRA6 ft3 ftRo2.800 lb/ftAD.CA2 ftRA4 ft2 ftRc3.

Given DataP=2000lbMx=4800lb-ftTo find SFD and BMD **As question demanding for solving Question no 2…

Answered: 2000 lb M = 4800 lb-ft В C A D 3 ft RA…

Engineering

Mechanical Engineering

2000 lb M = 4800 lb-ft В C A D 3 ft RA 6 ft 3 ft Ro 2.

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter11: Columns

Section: Chapter Questions

Problem 11.5.7P: Solve the preceding problem for a column with e = 0.20 in,, L = 12 ft, I = 2L7in4, and E = 30 ×...

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Concept explainers

Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and after loading.

Question

Answer #2 and use AREA METHOD only.

Find the Shear Moment Diagram of the following problems using
ONLY AREA METHOD.
30 kN
50 kN
to
B
A
1.
1 m
3 m
2 m
2000 lb
M = 4800 lb-ft
C
A
D
3 ft
RA
6 ft
3 ft
Ro
2.
800 lb/ft
A
D.
CA
2 ft
RA
4 ft
2 ft
Rc
3.

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PRIOR ANSWERS: Reaction Forces: By = 76.809 Cy = 15.94 Shear Forces: V (× = 10.5- ft (i.e., just to the left of support B)) = -36.75 kips V ( x = 10.5+ ft (i.e., just to the right of support B)) = 40.059 kips V (x = 25.5 ft) = -12.441 kips V (x = 26.5- ft (i.e.. just to the left of support C))= = -15.941 kips Bending moment: M (x = 10.5 ft (i.e., at support B)) = -192.9375 kips-ft M (x = 25.5 ft) = 14.1975 kips-ft