ME 352 - Machine Design I
Name of Student:____________________________
Fall Semester 2016
Lab Section Number:________________________
Homework No. 9 (30 points). Due at the beginning of lecture on Monday, November 7th.
Solve Problem 13.42, see page 611.
1
Solution to Homework 9.
The force F, acting at point C on link 2, is perpendicular to link AC as shown in Figure 1.
Figure 1. Link BD is Pinned to the Ground at Pin D and to Link AC at Pin B.
Part (i). The slenderness ratio at the point of tangency between the Euler column formula and the
Johnson parabolic formula.
The slenderness ratio of link 3, at the point of tangency, can be written as
2C E
S yc
 Sr D  
(1)
From the pinned-pinned end conditions, the end condition constant for link 3 is
C=1
(2)
Therefore, the slenderness ratio of link 3 at the point of tangency is
 Sr D  
2  1  206  109
 98.4
420  106
(3)
Part (ii). The critical load and the critical unit load acting on link 3.
In order to determine the critical load on link 3, first determine if this link is an Euler column
or a Johnson column. The Euler and Johnson criterion, respectively, are valid when
Sr   Sr  D
Sr   Sr  D
and
(4)
The slenderness ratio is defined as
Sr 
2
L
k
(5a)
where the radius of gyration is
k
I
A
(5b)
The second moment of area of link 3 (see Appendix A, Table 4, page 871) is
I

64
D4 

64
(0.005) 4  3.068 1011 m 4
(6a)
and the cross-sectional area of link 3 (see Appendix A, Table 4, page 871) is
A

4
D2 

4
(0.005) 2  1.963 105 m 2
(6b)
Substituting Equations (6) into Equation (5b), the radius of gyration of link 3 is
k
I
3.068 1011

 1.25 103 m
5
A
1.963  10
(7)
Then substituting Equation (7) into Equation (5a), the slenderness ratio of link 3 is
Sr 
0.15
 120
1.25 103
(8)
Comparing Equation (8) with Equation (3) gives
Sr   Sr D
i.e.,
120  98.4
(9)
Therefore, link 3 is an Euler column.
A plot of the critical unit load on link 3 against the slenderness ratio is shown in Figure 2.
Figure 2. A Plot of the Critical Unit Load against the Slenderness Ratio.
3
From the Euler column formula, the critical load on link 3 can be written as
C  2 E 
PCR  A 

2
 Sr 
(10)
Substituting the known values into this equation, the critical load on link 3 is
1   2  206  109 
PCR  1.963  105  
  2,772 N
1202


Therefore, the critical unit load on link 3 is
(11)
PCR
2,772

 141.2  106 N/m 2
5
A 1.963  10
(12)
Part (iii). The force F in order for the factor of safety to guard against buckling of link BD to be
N  1.
The free body diagram of link 2 is shown in Figure 3.
Figure 3. The Free Body Diagram of Link 2.
The sum of the moments about pin A is
M
A
0
(13a)
Equation (13a) can be written as
( R2 X FY  R2Y FX )  ( R22 X F32Y  R22Y F32 X )  0
(13b)
or as
( R2 cos  2 )( F sin  F )  ( R2 sin  2 )( F cos  F )  ( R22 cos  22 )( F32 sin 32 )  ( R22 sin  22 )( F32 cos 32 )  0
or as
4
R2 F sin ( F   2 )  R22 F32 sin (32   22 )  0
(13c)
Rearranging this equation, the force can be written as
F
R22 F32 sin ( 22  32 )
R2 sin ( F   2 )
(14)
The free body diagram of link 3 is shown in Figure 4.
Figure 4. Link 3 is in Compression.
Note that link 3 is a two-force member and the member is in compression.
The internal reaction force at pin B can be written as
FB  F23   F32
i.e., the magnitude of the internal reaction force FB is equal to the magnitude of the internal
reaction force F32 . Therefore, Equation (14) can be written as
F
R22 FB sin ( 22  32 )
R2 sin ( F   2 )
(15)
The factor of safety guarding against buckling for link 3 is defined as
N
PCR
FB
(16)
Rearranging Equation (16), the compressive load on link 3 is
FB 
PCR 2,772

 2,772 N
N
1
5
(17)
Substituting  2   22  120, 32  60,  F  210, R2  0.2 m, R22  0.15 m, and Equation (17)
into Equation (15), the force is
F
0.15  2, 772 sin (120  60)
 1,800.6 N
0.2 sin (210  120)
(18)
Part (iv). If the force is specified as F  3,000 N then determine: (a) the critical load on link BD for
the factor of safety to guard against buckling N  1 ; and (b) the slenderness ratio of the column.
Rearranging Equation (14), the internal reaction force can be written as
F32 
R2 F sin ( F   2 )
R22 sin ( 22  32 )
(19a)
Substituting F  3,000 N and the given data into this equation gives
FB  F32 
0.2  3, 000 sin (210  120)
 4, 618.8 N
0.15 sin (120  60)
(19b)
Rearranging Equation (16), the critical load can be written as
PCR  N  FB
(20a)
Substituting Equation (19b) and the factor of safety N = 1 into Equation (20a), the critical load is
PCR  1  4, 618.8  4, 618.8 N
(20b)
I. Assume that the column is an Euler column.
Rearranging Equation (10), the slenderness ratio of link 3 can be written as
 Sr Euler 
C2 E A
PCR
(21)
Substituting Equations (2), (5) and (20), into Equation (21), the slenderness ratio of link 3 is
 Sr Euler 
1  2  206  109  1.963  105
 92.97
4, 618.8
(22)
II. Assume that the column is a Johnson column.
The Johnson parabolic equation is
PCR
1  S y Sr 
 Sy 


A
C E  2 
2
(23)
Rearranging this equation, the slenderness ratio can be written as
 Sr  Johnson 
2
Sy
6
PCR

 Sy  A


C E

(24a)
or as
 Sr  Johnson 
2
420  106
4, 618.8 

6
 1 206  109  92.3
 420  10 
5 
1.963 10 

(24b)
From Equation (3), the slenderness ratio of link 3 at the point of tangency is
 Sr D  98.4
Note that
 Sr  Johnson   Sr D
(25)
i.e., 92.3  98.4
(26)
which is the CORRECT ANSWER.
Therefore, the column must be an Johnson column; i.e., the correct value for the slenderness
ratio is
S r   S r  Johnson  92.3
Note this answer is Equation (24b).
CHECK: Since
 Sr Euler   Sr D
i.e., 92.97  98.4
(27)
(28)
which is NOT POSSIBLE then the column must be an Johnson column. Therefore, the correct
value for the slenderness ratio is
S r   S r  Johnson  92.3
7
(29)