Assignment #0 Solution - Spring 2015

MAE 311: Machines and Mechanisms I Spring 2015

Assignment #0 Not Collected

Copyright 2015 Phillip M. Cormier and Jobaidur R. Khan

Problem 1

For the beam shown in Figure 1:

a) Determine the reactions at A and B.

b) Draw the Shear Force Diagram.

c) Draw the Bending Moment Diagram

d) Identify where the critical point is if the main beam has a constant cross section. What

assumptions or simplifications are you making?

Figure 1: Beam Diagram for Problem 1

Solution:

a) First resolve the loads to simple point loads on the main beam.

Distributed load: equivalent load = 100 lb/ft 2.5 ft = 250 lb. Point of action is geometric center

(mid-point), which is 1.25 ft from end.

Point load on bar: Assume L-shaped bar is rigid and transfers load to main bar (2.5 ft from left

end). This also applies a couple equivalent to 500 lb 0.5 ft = 250 ft.lb. The following free body

diagram can now be drawn.

Fy = 0 = RA + RB 500 lb 250 lb RA + RB = 750 lb

lb5.579Rft25.1ft8lb250ft5.2lb500lb.ft250ft5.2ft8R0M BBA So, RA = 750 lb 579.5 lb = 170.5 lb

1.25 ft

2.5 ft 2.5 ft

250 ft.lb

8 ft

y

x

250 lb 500 lb

RB RA




b) Redrawing the free body diagram with the new information and accounting for the distributed

load and couple (from L-bar).

Distributed load: equivalent load = 100 lb/ft 2.5 ft = 250 lb. Point of action is geometric center

c) Integrating the shear to get bending moment and add in couple from L-bar (F.d = 500 lb 5ft =

250 ft.lb)

d) Assuming that bending is the primary concern, the critical point is 2.5 ft from the end on the left.

2.5 ft 2.5 ft

250 ft.lb

8 ft

y

x

500 lb

579.5 lb 170.5 lb

100 ft/lb

x

V

170.5 lb

329.5 lb

250 lb

A = 426.25 ft.lb A = 312.5 ft.lb

A = 988.5 ft.lb

x

M (ft.lb)

426.25

312.25

676.25




Problem 2

For the beam shown in Figure 2:

a) Determine the reactions at A and B.

b) Draw the Shear Force Diagram.

c) Draw the Bending Moment Diagram

Figure 2: Beam Diagram for Problem 2

Solution:

a) First determine equivalent load and where it acts for distributed load. Equivalent load is the area

under the curve.

lb3.333dx440x165x5.12dxxpP8

4

2

x

x

eq

f

o

To determine the acting location of Peq,

ft24.6

P

dxx440x165x5.12

dxxp

dxxxp

xeq

8

4

23

x

x

x

x

Pf

o

f

o

eq

2 ft

5.5 ft

y

x

1000/3 lb 500 lb

RB RA 6.24 ft




Fy = 0 = RA + RB 500 lb 333.3 lb RA + RB = 833.3 lb

lb560Rft24.6lb3.333ft2lb500ft5.5R0M BBA So, RA = 833.3 lb 560 lb = 273.3 lb

b) Draw the free body diagram needed for shear. Equation for the shear force curve,

x440x5.82xdx440x165x5.12dxxpxV 236

252

c) Integrating the shear to get bending moment,

23324

2523

6

25 x220x5.27xdxx440x5.82xdxxVxM

x

M (ft.lb)

93.2

300.6

546.6

2 ft

4 ft

y

x

500 lb

560 lb 273.3 lb 5.5 ft

x

V (lb)

244.2

226.7

273.3

315.8




Problem 3

For the cross-section show in Figure 3:

a) Determine the area of the cross-section.

b) Determine the location of the centroid.

c) Determine the moment of inertia about its centroidal axis x'.

d) Determine the maximum stress due to bending at the critical point if this cross-section is used for

the beam in Problem 2.

Figure 3: Cross-section Dimensions for Problem 3

Solution:

a) The area of the cross-section consists of three rectangles.

A = AI + AII + AIII = 2in 0.2in + 0.2in 2in + 2in 0.2 in = 1.2 in

II

I

III

x

y

2 in

0.2 in 2 in




b) Due to symmetry, xI = 0.1in, xII = 1in, xIII = 2 0.1 = 1.9in,

yI = 1in, yII = 2 + 0.1 = 2.1in, yIII = 1in,

So, centroid,

in1in4.0in4.0in4.0

in4.0in9.1in4.0in1in4.0in1.0

AAA

AxAxAx

A

Ax

x222

222

IIIIII

IIIIIIIIIIII

III

Ii

i

III

Ii

ii

in30

41

in4.0in4.0in4.0

in4.0in1in4.0in1.2in4.0in1

AAA

AyAyAy

A

Ay

y222

222

IIIIII

IIIIIIIIIIII

III

Ii

i

III

Ii

ii

c) Determine Ix' for the combine area,

Step 1: find Ix' for each rectangle (about its own centroid)

For AI and AIII, 433

'x in15

2in2in2.0

12

1bh

12

1I

For AII, 433

'x in1500

2in2.0in2

12

1bh

12

1I

Step 2: find Ix for each rectangle about composite centroid using parallel axis theorem,

For AI and AIII, 442

2

242

'xIII'xI'xin188.0in

9000

1684in

30

411in4.0in

15

2AdIII

For AII, 442

2

242

'xII'xin216.0in

9000

1948in

30

411.2in4.0in

1500

2AdII

Ix' composite is sum of components (once determined for composite centroidal axis),

44

III'xII'xI'x'xin591.0in

3000

1772IIII

d) Neglecting shear, (at larger moment)

Psi200,15in

in12ft.lb6.546

I

My4

30001772

3041

ftin

x

x

Documents

Assignment #0 Solution - Spring 2015