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Beam: Shear and Moment Diagrams

A beam is supported as shown in the figure; it is intended to resist concentrated vertical load F1, located L1 from the left end, and distributed vertical load Fd2, which acts over a length L2 from the right end of the beam. The beam is sectioned into units with endpoints A, B, C, and D. The assignment in this problem is to draw the Shear and Moment diagrams for this beam. The X direction is along the beam.

L1

L2 F1

B C

Fd2

D

A

L

The first step is to draw a free-body diagram (FBD) of the overall beam to calculate the reaction forces at A and D. To determine the reaction forces, the distributed vertical load may be replaced with a concentrated vertical load F2, which acts at a length L2/2 from the right end of the beam.

L1 Ax Ay

We write three equations of static equilibrium: Fx 0 Ax

L2 /2 F1 F2

L

Dy

F M

y

0 Ay D y F1 F2 0 D y L F1 L1 F2 L L2 / 2 F2 L L2 / 2

L

Az

The reaction force solutions are: Dy

F L1 1

Ax 0 Ay F1 F2 D y

To determine the shear forces in each subsection of the beam, free-body diagrams of each subsection are drawn, representing the forces on the left and right sides. For this problem, the shear forces are: v AB x Ay v BC x Ay F1 vCD x Ay F1 Fd 2 x The shear forces in the first two subsections AB and BC are constant. The shear forces in CD starts from the BC value and decreases linearly with slope Fd2, ending at Dy. The end reaction +Dy then closes the shear diagram back to zero. Note vCD above assumes that x starts at zero; hence the axes must be shifted horizontally to fit in with the overall shear diagram. In order to determine the moment diagram given the shear diagram, one integrates the shear functions in each subsection. The difference between the moment subsection endpoint values in any case equals the area under the shear curve in that subsection. m AB x m A Ay x mBC x mB Ay F1 x mCD x mC Ay F1 Fd 2 x 2 / 2 Again in the moment equations, each x range starts at zero at the left of the subsection; hence the axes must be shifted horizontally to fit in with the overall moment diagram. Note, due to the beam support conditions in this problem, the moment must be zero at the left and right ends of the beam. In the moment equations, the constants of integration are: mA 0 mB Ay L1 mC mB Ay F1 L L1 L2 0 F1 10,000 User sets: F1 and Fd2 (N), plus L1 and L2 (m): 0 Fd 2 2,000 0 L1 L / 2 0 L2 L / 2 Computer sets: Visualize: L = 10 m Beam with loading conditions, Shear and moment diagrams

Numerical Display: Shear forces vA, vB, vC, and vD, moment magnitudes mA, mB, mC, and mD, plus reaction forces Ax, Ay, and Dy.

User Feels: Any of the above shear forces, moment magnitudes, or reaction forces (Y force only with Joystick user feels appropriate force or moment magnitude as they move along shear/moment diagram). Examples: There are two distinct shear/moment diagrams, depending on whether Ay F1 0 or

Ay F1 0 ; the above shear and moment equations apply equally to both cases, but the resulting diagrams have different characteristics. Example 1: Ay F1 0 When the user enters F1 = 5,000, L1 = 2.5, Fd2 = 1,000, and L2 = 3 (N and m), the results are: vA = 4,200, vB = -800, vC = -800, and vD = -3,800 (N); mA = 0, mB = 10,500, mC = 6,900, and mD = 0 (Nm); Ax = 0, Ay = 4,200, and Dy = 3,800 The associated shear/moment diagrams (N, Nm) for Example 1 are:

4000 2000

Shear

0 -2000

0

1

2

3

4

5 X

6

7

8

9

10

15000

M om ent

10000

5000

0 0

1

2

3

4

5 X

6

7

8

9

10

Example 2: Ay F1 0 When the user enters F1 = 1,000, L1 = 2, Fd2 = 2,000, and L2 = 4.5 (N and m), the results are: vA = 2,825, vB = 1,825, vC = 1,825, and vD = -7,175 (N); mA = 0, mB = 5,650, mC = 12,038, and mD = 0 (Nm); Ax = 0, Ay = 2,825, and Dy = 7,175 The associated shear/moment diagrams (N, Nm) for Example 2 are:

2000 0

Shear

-2000 -4000 -6000 0 1 2 3 4 5 X 6 7 8 9 10

15000

M om ent

10000

5000

0 0

1

2

3

4

5 X

6

7

8

9

10

In Example 1, the constant shear Ay F1 in BC is negative, while in Example 2, the constant shear Ay F1 in BC is positive. Thus, in Example 1, the linearly-changing moment in BC has negative slope, while in Example 2, this slope is positive. Further, though the linearly-changing shear in CD has negative slope Fd2 in both examples, the moment parabola in CD is strictly decreasing to zero in Example 1, while this parabola increases before decreasing to zero moment in Example 2. Note further that at point B, there is a discontinuity in shear, which translates to a change in moment slope at point B, but there is no discontinuity in shear at point C, hence moment slopes match at point C; this is true in both examples.

Comprehension Assignment: Once you get the `feel' for this simulation, run the program several times to collect and plot data: for a constant applied distributed load Fd2 = 1000 (N/m) and L2 = 5 m, and a fixed value of L1, vary applied load F1 over its allowable range and determine the resulting maximum shear vMAX, maximum moment mMAX, plus reaction forces Ax, Ay, and Bx. Plot vMAX, mMAX, Ax, Ay, and Bx vs. F1. Repeat these plots for various values of L1 over its allowable range. Discuss the trends you see do the results make sense physically?

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