fig. 6.1(a) fig. 6.1(b)solution part a) cross section (1): the net area moment of inertia for cross...
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ME 323: Mechanics of Materials Homework Set 6
Fall 2019 Due: Wednesday, October 9
Problem 6.1 (10 points): A design decision is to be taken based on whether a beam can support higher flexural stresses in an I-configuration (Fig 6.1(a)), or an H-configuration (Fig 6.1(b)), considering plane of bending is the ๐ฅ๐ฆ plane. Assuming that both cross sections are symmetric about the origins placed at the respective centroids:
a) Determine the second area moment of inertia of the two cross sections, ๐ผ$$% and ๐ผ$$& . b) Determine which cross section would experience a lower magnitude of maximum flexural
stress for the same loading conditions.
Fig. 6.1(a) Fig. 6.1(b)
Solution Part a) Cross section (1): The net area moment of inertia for cross section may be determined by subtracting the area moment of inertias of the two rectangular holes (of size ๐ ร 0.4๐) from the area moment of inertia of the larger rectangle (of size 1.4๐ ร ๐) about the ๐ง axis. Note that the holes also have their centroids at the ๐ง axis.
๐ผ$$% =112 ๐
(1.2๐)2 โ 2๐ผ4567 = 0.144๐8 โ 2 9112 ร 0.45๐ ร ๐
2;
=> ๐ผ$$% = 0.144๐8 โ 0.075๐8 = 0.069๐8
[1.1]
Cross section (2): The net area moment of inertia for cross section may be determined by adding the area moment of inertias of the web, and the two flange sections that have their centroids at the ๐ง axis.
โด ๐ผ$$& = ๐ผB7C + 2๐ผE6FGH7 =112(๐)(0.1๐)2 +
212(0.1๐)(๐)2 = 0.0168๐8 [1.2]
Part b) Since both cross sections are symmetric about the z axis, the distance from the centroid to the top, and bottom is same in each case, i.e.
๐ฆJFK,(%) = 0.6๐; ๐ฆJFK,(&) = 0.5๐ [1.3]
The loading conditions being same for both the configurations, ๐OPQ can be considered as the maximum bending moment
๐OPQ% =๐JFK๐ฆJFK,(%)
๐ผ$$%= 8.695๐S2๐JFK
[1.4]
๐JFK& =๐JFK๐ฆJFK,(&)
๐ผ$$&= 29.76๐S2๐JFK
[1.5]
Since, cross section (1) experiences a lower magnitude of flexural stress, it can support larger loads before elastic failure.
Problem 6.2 (10 points): A cantilevered beam AE of length ๐ฟ is subjected to a uniformly distributed downward load ๐V(Force/length) between A and B, a concentrated moment ๐W at C and a point force ๐acting upwards at D. The bending moment diagram corresponding to the applied load is shown in Fig 6.1(a). The cross section of the beam is shown in Fig 6.1(b).
Assuming ๐V =Z[\]^_
,๐W = ๐&๐V, ๐ =Za\]^
a) Determine the numerical values of ๐%, ๐&, and ๐2. b) Determine the second area moment of inertia of the cross section about its centroid ๐ผb in
terms of ๐8. c) Determine the location (๐ฅ) where the cross-section experiences zero flexural stress. d) Determine the (๐ฅ, ๐ฆ) coordinates where maximum tensile stress is experienced, and the
(๐ฅ, ๐ฆ) coordinates where maximum compressive stress is experienced. Also calculate the magnitude of the corresponding stresses.
Hint: Check both the positive, and negative extremes of bending moment.
Fig. 6.2 (a)
Fig. 6.2 (b)
Solution
FBD:
Equilibrium equations:
ฮฃ๐นe :๐V๐ฟ2 + ๐ โ ๐ธe = 0
=>๐ธe = ๐ โ๐V๐ฟ2 =
๐V
๐ฟ 9โ๐%2 + ๐2;
[2.1]
ฮฃ๐h :๐V๐ฟ2 9
3๐ฟ4 ; โ๐W โ ๐ 9
๐ฟ8; + ๐h = 0
=> ๐h = ๐&๐V +๐2๐V
8 โ38๐%๐V = ๐V 9โ
38๐% + ๐& +
๐28 ;
[2.2]
Part a)
The shear force and bending moment equations for each section can be written using the following two formulae:
๐(๐ฅ&) = ๐(๐ฅ%) + k ๐(๐)๐๐
K_
K[
[2.3]
Fig. 6.2 (c)
๐(๐ฅ&) = ๐(๐ฅ%) + k ๐(๐)๐๐
K_
K[
[2.4]
Section AB:
๐(๐ฅ) = ๐(0) + kโ๐V๐๐ฅK
V
= 0 โ ๐V๐ฅ; ๐ 9๐ฟ2; = โ
๐%๐V
2๐ฟ [2.5]
๐(๐ฅ) = ๐(0) + kโ๐V๐ฅ๐๐ฅK
V
= โ๐V๐ฅ&
2 ;๐ 9๐ฟ2; = โ
๐%๐V
8 [2.6]
Section BC:
๐(๐ฅ) = ๐ 9๐ฟ2; +
k0๐๐ฅK
^&
= โ๐%๐V
2๐ฟ ; ๐ 93๐ฟ4 ; = โ
๐%๐V
2๐ฟ [2.7]
๐(๐ฅ) = ๐ 9๐ฟ2; +
k๐(๐ฅ)๐๐ฅK
^&
= โ๐%๐V
8 โ๐%๐V
2๐ฟ 9๐ฅ โ๐ฟ2; =
๐%๐V
8 โ๐%๐V๐ฅ2๐ฟ ;
๐93๐ฟ4 ; = โ
๐%๐V
4
[2.8]
๐ 93๐ฟ4 ;
n
= ๐ 93๐ฟ4 ;
S
= โ๐%๐V
2๐ฟ ; ๐ 93๐ฟ4 ;
n
= ๐ 93๐ฟ4 ;
S
+๐W = ๐V 9โ๐%4 + ๐&;
[2.9]
Section CD:
๐(๐ฅ) = ๐ 93๐ฟ4 ;
n
+ k0๐๐ฅK
2^8
= โ๐%๐V
2๐ฟ ; ๐ 97๐ฟ8 ; = โ
๐%๐V
2๐ฟ [2.10]
๐(๐ฅ) = ๐ 93๐ฟ4 ;
n
+ k๐(๐ฅ)๐๐ฅK
2^8
= ๐V 9โ๐%4 + ๐&; โ
๐%๐V
2๐ฟ 9๐ฅ โ3๐ฟ4 ;
= ๐V 9๐%8 + ๐&; โ
๐%๐V๐ฅ2๐ฟ ;
๐97๐ฟ8 ; = ๐V 9
โ5๐%16 + ๐&;
[2.11]
๐ 97๐ฟ8 ;
n
= ๐ 97๐ฟ8 ;
S
+ ๐ =๐V
๐ฟ 9โ๐%2 + ๐2; ;
๐ 97๐ฟ8 ;
n
= ๐ 97๐ฟ8 ;
S
= ๐V 9โ5๐%16 + ๐&;
[2.12]
Section DE:
๐(๐ฅ) = ๐ 97๐ฟ8 ;
n
+ k0๐๐ฅK
o^p
=๐V
๐ฟ 9โ๐%2 + ๐2;; ๐(๐ฟ) = ๐ธe =
๐V
๐ฟ 9โ๐%2 + ๐2;
[2.13]
๐(๐ฅ) = ๐ 97๐ฟ8 ;
n
+ k๐(๐ฅ)๐๐ฅK
o^p
= ๐V 9โ5๐%16 + ๐&; +
๐V
๐ฟ 9โ๐%2 + ๐2; q๐ฅ โ
7๐ฟ8r
= ๐V 9๐%8 + ๐& โ
7๐28 ; +
๐V๐ฅ๐ฟ 9โ
๐%2 + ๐2; ;
๐(๐ฟ) = ๐h = ๐V 9โ3๐%8 + ๐& +
๐28 ;
[2.14]
Using the moment values at certain lengths from the equations above,
๐s^&t = โZ[\]
p= โ0.3๐V (from Fig. 6.2 (a))
โน ๐% = 2.4
[2.15]
๐so^ptn= ๐V sโ
vZ[%w+ ๐&t = โ0.7๐V (from Fig. 6.2 (a))
โน ๐& = 0.05
[2.16]
๐(๐ฟ) = ๐V sโ2Z[p+ ๐& +
Zapt = 0.2๐V (from Fig. 6.2 (a)) [2.17]
โน ๐2 = 8.4
Part b)
To find the centroid of the area, by considering the top surface to be ๐ฆ = 0 just for this purpose,
๐ด% = 0.3๐&;๐ด& = 0.35๐& [2.18]
๐ฆ% = โ0.15๐;๐ฆ& = โ0.53๐; [2.19]
The y-coordinate of the centroid, ๐ฆy can be found out by using,
๐ฆy =๐ด%๐ฆ% + ๐ด&๐ฆ&๐ด% + ๐ด&
= โ0.355๐ [2.20]
From the fig below, second area moment of inertia ๐ผ$
๐ผ% =๐โ2
12 + ๐ด%(๐ฆy โ ๐ฆ%)& =112๐
(0.3๐)2 + 0.3๐&(0.082๐)& = 0.015๐8 [2.21]
๐ผ& =๐โ2
36 + ๐ด&(๐ฆy โ ๐ฆ&)& =136๐
(0.7๐)2 + 0.35๐&(0.301๐)& = 0.0202๐8
๐ผb = ๐ผ% + ๐ผ& = 0.0355๐8
[2.22]
Part c)
The flexural stress can be denoted as ๐K = โ\e{
, hence the flexural stress is zero across the whole
cross-section when ๐ is 0. Here, ๐ = 0 at ๐ฅ = 0 and o^p< ๐ฅ < ๐ฟ. To find another ๐ฅ, the moment
in section DE is:
Fig. 6.2 (d)
๐(๐ฅ) = ๐V 9๐%8 + ๐& โ
7๐28 ; +
๐V๐ฅ๐ฟ 9โ
๐%2 + ๐2; = 0
โน ๐ฅ = 0.972๐ฟ
[2.23]
Hence, ๐K = 0 at ๐ฅ = 0 and ๐ฅ = 0.927๐ฟ.
Part d)
Maximum bending moment from the Fig. 6.2 (a) is ๐JFK = โ0.7๐V
๐}7G~๏ฟฝ67 =โ๐๐ฆ๐ก๐๐
๐ผ=0.7๐V(0.355๐)0.0355๐8 = 7๐V๐S2
๐๏ฟฝ5J๏ฟฝ๏ฟฝ7~~๏ฟฝ๏ฟฝ7 =โ๐๐ฆ๐๐๐ก๐ก๐๐
๐ผ=โ0.7๐V(๐ โ 0.355๐)
0.0355๐8 = โ12.72๐V๐S2
[4.11]
Problem 6.3 (10 points): A beam AF is loaded as shown in Fig. 6.3 (a) and its cross-section is shown in Fig. 6.3 (b). Find:
a) Draw the shear force diagram. b) Draw the bending moment diagram. c) Determine the maximum normal stress in the beam. d) Determine the maximum shear stress in the beam.
Take the dimension a = 2 cm in Fig. 6.3 (b)
Fig. 6.3 (a)
Fig. 6.3 (b)
Solution
FBD:
Equilibrium equations:
ฮฃ๐นe:๐ดe โ 1200 โ 1600 + ๐ธe โ 600 = 0
=> ๐ดe + ๐ธe = 3400๐
[2.1]
ฮฃ๐๏ฟฝ :โ 1200(1.5) โ 1200 โ 160097 +43; + ๐ธe
(8) โ 600(11) = 0
=>๐ธe = 2867๐
โน ๐ดe = 533๐
[2.1]
Fig. 6.3 (c)
Parts a), b)
Part c)
The maximum magnitude of bending moment from the fig. above is at E, ๐JFK = 2703๐๐. The maximum flexural stress would occur at the bottom and top fibers of the cross-section. Hence, the distance of these fibers from the centroid (๐ฆy) of the cross-section must be obtained.
Fig. 6.3 (d)
๐ฆy =๐ด%๐ฆ% โ ๐ด&๐ฆ&๐ด% โ ๐ด&
=15.75๐&(โ1.75๐) โ 6.25๐&(โ2.25๐)
9.5๐& = โ1.421๐ [3.1]
Therefore, the centroid is at 1.421๐from the top surface of the cross-section. Similarly, to find the second area moments of inertia,
๐ผ% =๐โ2
12 + ๐ด%(๐ฆy โ ๐ฆ%)& =1124.5๐
(3.5๐)2 + 15.75๐&(0.329๐)& = 17.785๐8 [3.2]
๐ผ& =๐โ2
12 + ๐ด&(๐ฆy โ ๐ฆ&)& =112(2.5๐)8 + 6.25๐&(0.829๐)& = 7.55๐^4
๐ผb = ๐ผ% โ ๐ผ& = 1.64๐10Sw๐8
[3.3]
๐ฆ}5๏ฟฝ = 1.421๐; ๐ฆC5}}5J = 2.079๐
๐}7G~๏ฟฝ67 = 46.84๐๐๐; ๐๏ฟฝ5J๏ฟฝ๏ฟฝ7~~๏ฟฝ๏ฟฝ7 = 68.53๐๐๐
[3.4]
Part d)
The maximum shear stress can be calculated using the following formula,
๐JFK =|๐|JFK๐
๐ผ๐ก=|๐|JFK๐ดโ๐ฆโ
๐ผb๐ก=1500 โ {(1.421๐)(4.5๐) โ 2.5๐(0.421๐)} โ {0.809๐}
10.235๐8 โ 4.5๐
โด ๐JFK = 351.65๐๐๐
[3.5]
Fig. 6.3 (e)
Fig. 6.3 (f)
Problem 6.4 (10 points): A wood beam is supported and loaded as shown in Fig. 6.4 (a). The weight ๐ค of the beam is to be considered, using ๐พ = 50๐๐ ๐๐ก2โ for the specific weight of the wood. The maximum shear stress in any cross-section (along the y-direction) must not exceed ๐F665B =160๐๐ ๐.
a) Calculate the maximum allowable value of the loading ๐ for the solid rectangular cross-section (A), with dimensions ๐ = 2.5๐๐ and โ = 6๐๐.
b) Calculate the maximum allowable value of the loading ๐ for the solid circular cross-section (B), with diameter ๐ = 6๐๐.
c) Calculate the maximum allowable value of the loading ๐ for the hollow circular cross-section (C), with outer diameter diameter ๐5 = 6๐๐ and inner diameter ๐๏ฟฝ = 3๐๐
d) Show the state of stress at three places (M, N, P) on the cross-section (A) as shown in Fig. 6.4 (b). The co-ordinates for these points are:
Points Co-ordinates
(x,y,z)
M ( 6ft , 0in , 0in )
N ( 6ft , 1.5in , 0in )
P ( 6ft , 3in , 0in )
Fig. 6.4 (a)
Solution
FBD:
Calculating unit loading ๐ค for each of the cross-sections A, B & C:
For (A): ๐ค = ๏ฟฝs&.v%&๐๐t s w
%&๐๐t๏ฟฝs50 6CE
E}at โ 5.21 6CE
E} [4.1]
For (B): ๐ค = ๏ฟฝ(๐)s w๏ฟฝG&โ%&
t&๏ฟฝ s50 6CE
E}at โ 9.82 6CE
E} [4.2]
Fig. 6.4 (b)
Fig. 6.4 (c)
For (C): ๐ค = (๐)s w๏ฟฝG&โ%&
t&โ (๐)s 2๏ฟฝG
&โ%&t&ยกs50 6CE
E}at โ 7.36 6CE
E}
[4.3]
Equilibrium Equations:
ฮฃ๐นe:๐ดe โ (8๐๐ก)๐ค โ ๐ โ 2๐ + ๐ทe = 0
=> ๐ดe + ๐ทe = 3๐ + (8๐๐ก)๐ค
[4.4]
ฮฃ๐๏ฟฝ :โ (8๐๐ก)๐ค(4๐๐ก) โ ๐(4๐๐ก) โ 2๐(7๐๐ก) + ๐ทe(8๐๐ก) = 0
=>๐ทe = 2.25๐ + (4๐๐ก)๐ค
๐ดe = 0.75๐ + (4๐๐ก)๐ค
[4.5]
Shear Force Diagram:
The maximum magnitude of shear stress is seen to |๐|JFK = ๐ทe = 2.25๐ + (4๐๐ก)๐ค and occurs at ๐ฅ = 8๐๐ก.
Part a)
Fig. 6.4 (d)
The maximum shear stress of a rectangular cross-section can by calculated using the reduced formula,
๐JFK =|๐|JFK๐
๐ผ๐ก =32|๐|JFK๐ด = 90๐๐ ๐
โน3
2
2.25๐๐๐๐๐๐ค + (4๐๐ก) s5.21๐๐๐๐๐ก t
(2.5๐๐)(6๐๐)= 90๐๐ ๐
โด ๐F665B โ 390.74๐๐๐
[4.6]
Part b)
๐JFK =|๐|JFK๐
๐ผ๐ก =34|๐|JFK๐ด = 90๐๐ ๐
โน43
2.25๐F665B + (4๐๐ก) 99.82๐๐๐๐๐ก ;
(๐) s6๐๐2 t& = 90๐๐ ๐
โด ๐F665B โ 830.77๐๐๐
[4.7]
Part c)
Calculating the ๐JFK at the neutral axis for this cross-section,
For (C): ๐JFK is expected to occur at the neutral axis. Calculating the maximum shear stress:
๐JFK =|๐|JFK๐
๐ผ๐ก = 90๐๐ ๐
๐ = ๐ดโ๐ฆโ = [๐ด% โ ๐ด&] ๏ฟฝ๐ด%๐ฆ% โ ๐ด&๐ฆ&๐ด% โ ๐ด&
๏ฟฝ = ๐ด%๐ฆ% โ ๐ด&๐ฆ&
โด ๐ = ยฅ๐296๐๐2;&ยฆ q9
43๐;96๐๐2;r โ ยฅ
๐293๐๐2;&ยฆ q9
43๐;93๐๐2;r = 15.75๐๐2
[4.8]
๐ผ =๐4 9
6๐๐2 ;
8
โ 93๐๐2 ;
8
ยก = 59.64๐๐8; ๐ก = ๐5 โ ๐๏ฟฝ = 3๐๐
โด ๐F665B โ 439.94๐๐๐
Part d)
The shear force ๐ and the bending moment ๐ at ๐ฅ = 6๐๐ก for cross section (A) are as follows:
V(x = 6ft) = โ0.25๐ โ (2๐๐ก)๐ค = โ108.11๐๐๐
๐(๐ฅ = 6๐๐ก) = 2.5๐ + (6๐๐ก)๐ค = 1008.11๐๐๐๐ก
[4.9]
Shear stress values:
๐JFK = ๐\ =32๐๐ด = โ10.811๐๐ ๐
๐ยซ =๐๐๐ผ๐ก=
6๐ดโ&
โ&
4 โ ๐ฆ&ยก๐ =6
(2.5 โ 6)(6)& 6&
4 โ 1.5&ยก (โ108.11) = โ8.108๐๐ ๐
๐J๏ฟฝG = ๐ยฌ = 0
[4.10]
Flexural stress values:
๐J๏ฟฝG = ๐\ = 0๐๐ ๐
๐ยซ =โ๐๐ฆ๐ผ
=ยญ(โ1008.11 โ 12๐๐๐๐)(1.5๐๐)ยฎ
45๐๐8 = โ403.24๐๐ ๐
๐JFK = ๐ยฌ = โ806.49๐๐ ๐
[4.11]
The state of stress at three different points can be given as,