Solution to Problem 415 | Shear and Moment Diagrams

Tags:

Problem 415
Cantilever beam loaded as shown in Fig. P-415.

Click here to read or hide the general instruction

Write shear and moment equations for the beams in the following problems. In each problem, let x be the distance measured from left end of the beam. Also, draw shear and moment diagrams, specifying values at all change of loading positions and at points of zero shear. Neglect the mass of the beam in each problem.

Solution 415

Click here to show or hide the solution

Segment AB:
$V_{AB} = -20x \, \text{kN}$
$M_{AB} = -20x(x/2)$
$M_{AB} = -10x^2 \, \text{kN}\cdot\text{m}$

Segment BC:
$V_{BC} = -20(3)$
$V_{BC} = -60 \, \text{kN}$
$M_{BC} = -20(3)(x - 1.5)$
$M_{BC} = -60(x - 1.5) \, \text{kN}\cdot\text{m}$

Segment CD:
$V_{CD} = -20(3) + 40$

$V_{CD} = -20 \, \text{kN}$

$M_{CD} = -20(3)(x - 1.5) + 40(x - 5)$

$M_{CD} = -60(x - 1.5) + 40(x - 5) \, \text{kN}\cdot\text{m}$

To draw the Shear Diagram

V_AB = -20x for segment AB is linear; at x = 0, V = 0; at x = 3 m, V = -60 kN.
V_BC = -60 kN is uniformly distributed along segment BC.
Shear is uniform along segment CD at -20 kN.

To draw the Moment Diagram

M_AB = -10x² for segment AB is second degree curve; at x = 0, M_AB = 0; at x = 3 m, M_AB = -90 kN·m.
M_BC = -60(x - 1.5) for segment BC is linear; at x = 3 m, MBC = -90 kN·m; at x = 5 m, M_BC = -210 kN·m.
M_CD = -60(x - 1.5) + 40(x - 5) for segment CD is also linear; at x = 5 m, M_CD = -210 kN·m, at x = 7 m, M_CD = -250 kN·m.

Submitted by Romel Verterra on January 29, 2009 - 10:18pm

58630 reads

MATHalino.com

Subjects

Solution to Problem 415 | Shear and Moment Diagrams

Recent Updates

Strength of Materials

Our Online Reviewers

MATHalino.com

Subjects

You are here

Solution to Problem 415 | Shear and Moment Diagrams

Recent Updates

Strength of Materials

Our Online Reviewers