Download - Assignment #0 Solution - Spring 2015
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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
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MAE 311: Machines and Mechanisms I Spring 2015
Assignment #0 Not Collected
Copyright 2015 Phillip M. Cormier and Jobaidur R. Khan
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
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MAE 311: Machines and Mechanisms I Spring 2015
Assignment #0 Not Collected
Copyright 2015 Phillip M. Cormier and Jobaidur R. Khan
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
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MAE 311: Machines and Mechanisms I Spring 2015
Assignment #0 Not Collected
Copyright 2015 Phillip M. Cormier and Jobaidur R. Khan
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
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MAE 311: Machines and Mechanisms I Spring 2015
Assignment #0 Not Collected
Copyright 2015 Phillip M. Cormier and Jobaidur R. Khan
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
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MAE 311: Machines and Mechanisms I Spring 2015
Assignment #0 Not Collected
Copyright 2015 Phillip M. Cormier and Jobaidur R. Khan
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