3.3. DIMENSION OF A COMPONENT UNDER STATIC LOADING 161
3.3.5 BEAM UNDER STATIC BENDING MOMENT
e limit state function of a component and its reliability calculation under bending moment for
strength issue and deflection issue have been discussed in detail in Section 4.9 of Volume 1 [1].
After the limit state function of a component under static bending loading is established, we
can run the dimension design with the required reliability. Now we will use examples to show
how to conduct component dimension design under bending moment.
Example 3.13
A square cantilever beam with a length L D 20
00
˙ 1=16
00
is subjected to a lateral force F which
follows a normal distribution with a mean
F
D 3:675 (klb) and a standard deviation
F
D
0:52 (klb). e yield strength S
y
of the beam’s material follows a normal distribution with a
mean
S
y
D 34:5 (ksi) and a standard deviation
S
y
D 3:12 (ksi). Use the modified Monte
Carlo method to design the side height d of the square beam with the required reliability 0.99
when the side height d has a tolerance ˙0:010.
Solution:
In this example, there is no dimension-dependent parameter.
(1) e limit state function.
e normal stress of the square beam caused by the lateral force F is
D
FL
.
d=2
/
I
D
FL
.
d=2
/
d
4
=12
D
6FL
d
3
: (a)
e limit state function of the beam is
g
S
y
; F; L; d
D S
y
6FL
d
3
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(b)
In the limit state function, there are four normally distributed variables. e mean and standard
deviation of the length L can be calculated per Equation (1.1). e standard deviation
d
of the
side height, d can be determined per Equation (1.1). eir distribution parameters in the limit
state function (b) are listed in Table 3.30.
(2) Use the modified Monte Carlo method to determine the mean
d
of the height d .
Following the modified Monte Carlo method procedure discussed in Section 3.2.5 and the
flowchart shown in Figure 3.5, we can make MATLAB program for the limits state function (b).
e iterative results for the limit state function of this problem are listed in Table 3.31.
162 3. THE DIMENSION OF A COMPONENT WITH REQUIRED RELIABILITY
Table 3.30: Distribution parameters for Example 3.13
S
y
(ksi) F (klb) L (in) d (in)
μ
S
y
σ
S
y
μ
F
σ
F
μ
L
σ
L
μ
d
σ
d
34.5 3.12
3.675 0.52 20 0.015625
μ
d
0.0025
Table 3.31: e iterative results of Example 3.13 for the strength issue
Iterative #
μ
d
*
R
*
∆R
*
1 2.530692 0.929195 -0.0608
2 2.531692 0.930219 -0.05978
… … … …
116 2.645692 0.99001 1.03E-05
117 2.646692 0.990198 0.000198
According to the result obtained from the program, the mean of the side height of the
square beam with a reliability 0.99 is
d
D 2:647
00
: (e)
erefore, the side height d of the square beam with the required reliability 0.99 under the
specified loading is
d D 2:647 ˙ 0:005
00
:
Example 3.14
A circular simple support beam with a length L D 24
00
˙ 1=16
00
is subjected to a lateral force
F D 4:5 ˙ 0:5 (klb) in the middle of the beam. e yield strength S
y
of the beam’s material
follows a normal distribution with a mean
S
y
D 34:5 (ksi) and a standard deviation
S
y
D
3:12 (ksi). e Young’s modulus E of the beam material follows a normal distribution with
a mean
E
D 2:76 10
4
(ksi) and a standard deviation
E
D 6:89 10
2
(ksi). e allowable
deflection of the beam is 0.030
00
. Use the modified H-L method to design the diameter d of the
beam with the required reliability 0.99 when the diameter d has a tolerance ˙0:005.
Solution:
In this example, there is no dimension-dependent parameter.
(1) Limit state functions.
3.3. DIMENSION OF A COMPONENT UNDER STATIC LOADING 163
e maximum stress and the maximum deflection of this beam will be in the middle of
the beam. e normal stress in the middle section of the beam caused by the bending moment
is
D
.
FL=4
/
.
d=2
/
I
D
FLd=8
d
4
=64
D
6FL
d
3
: (a)
e limit state function of the beam for the strength in this problem is
g
S
y
; F; L; d
D S
y
6FL
d
3
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure
(b)
e beam deflection in the middle section of the beam in this example is
ı D
FL
3
48EI
D
FL
3
48E
.
d
4
=64
/
D
4FL
3
3Ed
3
: (c)
e limit state function of the beam for the deflection issue in this problem is
g
.
E; F; L; d
/
D 0:030
4FL
3
3Ed
3
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(d)
In these two limit state functions (b) and (d), there are five normally distributed random vari-
ables. e mean and the standard deviation of the force F can be calculated per Equation (1.2).
e mean and the standard deviation of the length L can be calculated per Equation (1.1). e
standard deviation of the beam diameter d can be calculated per Equation (1.1). eir distribu-
tion parameters in these two limit state functions are listed in Table 3.32.
Table 3.32: Distribution parameters for the limit state functions (b) and (d)
S
y
(ksi) E (ksi) F (klb) L (in) d (in)
μ
S
y
σ
S
y
μ
E
σ
E
μ
F
σ
F
μ
L
σ
L
μ
d
σ
d
34.5 3.12 2.76×10
4
6.89×10
2
4.5 0.125 24 0.015625
μ
d
0.00125
(2) Use the modified H-L method to determine the dimension.
Following the modified H-L method procedure discussed in Section 3.2.3 and the
flowchart shown in Figure 3.2, we can compile MATLAB programs for the limit state func-
tions (b) and (d).
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