MAE 311: Machines Machines and and Mechanisms Mechanisms I
Spring 2014
Name:
Exam #1
Print your name at the top each page. You will turn in both this exam sheet and your equation sheet. Point distribution is shown for each problem (this may be helpful in allocating time for each question. Any instance of academic dishonesty will result in a zero for the exam, and has the potential for further disciplinary action (including expulsion from the university).
By printing your name you
acknowledge you will not discuss this exam with anyone prior to 5:00 PM on March 14 th, 2014.
Problem 1 (15 Points Total ) A carbon steel component (elastic modulus of 30 106 psi and a Poisson’s ration of 0.3) has the stress ∙
state listed below (all other stresses are zero): σxx = 300 psi, σyy = -200 psi, τxy = 400 psi, τxz = -200 psi Do the following: a) Determine all strains. strains. (12 Points) b) Briefly describe what Poisson’s ratio captures. (This can be done in 1-2 sentences). (3 Points) Sol.
a)
= 300 psi, yy = – 200 200 psi, zz = 0 psi, xy = 400 psi, yz = 0 psi, zx = – 200 200 psi
xx
xx E1 E E 1 E yy E E zz E E E1 0 0 xy 0 yz 0 0 0 0 0 zx 0
0
0
0 x
0
0
0
0
0
1 G
0
0
1 G
0
0
y 0 z 0 xy 0 yz 1 G zx
For the material, = 0.3, E 30 10 psi ps i , G
E
6
xx 30110 300.29102 300.29102 0.29 2 1 300.29102 3010 yy 3010 zz 300.29102 300.29102 30110 0 0 xy 0 yz 0 0 0 0 0 zx 0 6
6
6
6
b)
6
6
6
21
30 10 psi 6
21 0.3
6
0
0
0
6
0
0
0
0
0
0
1 11.510 6
0
0
0
1 11.5106
0
0
0
1 11.5106
11.5 106 psi
300 1.2 10 5 9.667 10 6 200 0 1.0 10 6 5 400 3.478 10 0 0 200 1.739 10 5
Poisson’s ratio is the ratio between axial strain and lateral strain.
Copyright 2014 –Phillip M. Cormier and Jobaidur R. Khan
1
Problem 2 (15 Points Total ) For the component shown below in Figure 1, do the following: a) Determine the maximum stress in the component given it is made of carbon steel. (For carbon steel: Sy=190 MPa, elastic modulus of 206.8 GPa, and a Poisson’s ration of 0.3) (10
Points) b) Would your analysis process change if the material were made of gray cast iron? If so, what would change and why? (5 Points) (For gray cast iron: SUT=150 MPa, elastic modulus of 103.4 GPa, and a Poisson’s ration of 0.28)
Figure 1: Machine Component for Problem 2 Sol.
(a)
If the material is carbon steel, it is ductile. So, stress concentration is not considered here. The maximum stress will take place at minimum area, which is the center of the component. So, the minimum area is A = Width (with hole) × Thickness = (25 – 6) × 10 = 190 mm² 200 N P So, the maximum stress will be, max 1.05 MPa A 190 mm 2
(b)
If the material is gray cast iron, it is brittle. Because brittle materials do not yield locally, the stress concentration needs to be considered. We would take our nominal stress and multiply it by the stress concentration factor. Stress concentration factor (from the chart provided) 6 mm d The parameter, [So, the correlation is valid] 0.24 065 W 25 mm 2
3
4
d d d d 7.9735 9.2659 1.8145 2.9684 K t 3.0039 3.753 W W W W W 3.0039 3.753 0.24 7.9735 0.242 9.2659 0.243 d
5
1.8145 0.244 2.9684 0.245 2.4427 So, the maximum stress will be,
max K t o 2.4427 1.05MPa 2.56MPa
Copyright 2014 –Phillip M. Cormier and Jobaidur R. Khan
2
MAE 311: Machines and Mechanisms I
Spring 2014
Name:
Exam #1
Problem 3 (30 Points Total ) For the given state of stress (magnitudes and directions indicated in Figure 2): a) Create the stress tensor for this problem. (5 Points) b) Determine the principal stresses and order them appropriately. Clearly document your approach showing all work. (15 Points) c) Identify the principal direction for σ1. (10 Points)
Figure 2: Stress Element for Exam Problem 3
Sol.
(a)
= 900 psi, yy = 400 psi, zz = – 500 psi, xy = 600 psi, yz = 0 psi, zx = 0 psi
xx
xx xy zx xx xy zx 900 (b) xy yy yz xy yy yz 600 zx yz zz zx yz zz 0 For non-trivial solution, det pI 0
900 p p I 600 0
So,
600 400 0
0 500 0
900 p 600 0 det I p 500 p 0
600
0
400 p 0
900 p
600
0
600
400 p
0
0
0
500 p
600 400 p 0
0 500 p 0
900 p 400 p 600 600 0 0 500 p
500 p 3600 400 p 900 p p 2 3600 0
p 2 1300 p 0 500 p 500 p p p 1300 0 Copyright 2014 –Phillip M. Cormier and Jobaidur R. Khan
3
1,400 1,400
2
p 400, p
4 90,000
1400 1,600,000
2 2 Ordering the results following the convention on largest to smallest:
1332.5, 67.5
1 1300 psi, 2 0 psi, 3 500 psi (c)
With Eigen Values now known, we can determine the principal direction of 1,
p p 0 pI 0
nx n ny nz
Determine 1 using 1 = 1300 psi, 600 0 900 1300 x1 0 1 0 400 1300 0 1 I 600 y 500 1300 z 1 0 0 0 1 0 x 0 400 600 1 1.5 0 x1 0 1 0 1 1.5 0 1 0 600 900 0 y y 1 0 1800 z 0 0 0 1 z 1 0 0
1
For row 1 or 2, x
1.5 1 0 1 1.5 1 , y
x
y
For row 3, 1z 0
x1 1.5 x1 0.8321 So, y1 1 y1 0.5547 z 1 0 z 1 0 ˆ
ˆ
This will not be necessary
ˆ
Copyright 2014 –Phillip M. Cormier and Jobaidur R. Khan
4
MAE 311: Machines and Mechanisms I Name:
Spring 2014 Exam #1
Problem 4 (40 Points Total ) A machine component has been simplified to create the diagram shown in Figure 3. The loading in the problem is Fx = 0 lb, Fz = -100 lb and Fy = -200 lb. If all connections are assumed to be rigid, do the following: a) Draw the stress element at Point A, clearly labeling the source of all stresses. (15 Points for Labeling Element, 5 Points for Sources of Each Stress) b) Determine all stress magnitudes and create the stress tensor. (15 Points for Magnitudes, 5 Points for Putting them in the Tensor)
Figure 3: Component for Problem 4
Copyright 2014 –Phillip M. Cormier and Jobaidur R. Khan
5
Sol.
(a)
Point A will have a bending moment from Fy, which is My = 200 × 10 = 2,000 lb.in and it will cause axial tension. Fy will also have a torque (Ty = 200 × 5 = 1000 lb.in) and it will cause shear. Point A will have a bending shear from Fz.
Let’s build the stress on element A.
y
xz
from Fz x from My
xz
z
The normal stress from bending,
Ty c
(b)
J
Myc I
16Ty
d
3
32M y
d 3
32 2,000 lb.in
1.5 in 3
6,036 psi ,
16 1,000 lb.in
1,509 psi , 1.5 in 3 16Fz 16 100 lb.in FQ 4Fz 75.4 psi , The shear from bending, Fz 2 Ib 3A 3d 2 31.5 in So, shear in xz, xz Ty Fz 1,509 75.4 1,434 psi The shear from torsion,
Ty
x
x
from Ty
xx xy zx 6,036 psi The stress tensor is, xy yy yz 0 zx yz zz 1,434 psi
Copyright 2014 –Phillip M. Cormier and Jobaidur R. Khan
0 1,434 psi 0
0
0
0
6
