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 𝑥𝑦 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, 𝐼$$% and 𝐼$$& . 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 𝑏 × 0.4𝑏) from the area moment of inertia of the larger rectangle (of size 1.4𝑏 × 𝑏) about the 𝑧 axis. Note that the holes also have their centroids at the 𝑧 axis.

𝐼$$% =112 𝑏

(1.2𝑏)2 − 2𝐼4567 = 0.144𝑏8 − 2 9112 × 0.45𝑏 × 𝑏

2;

=> 𝐼$$% = 0.144𝑏8 − 0.075𝑏8 = 0.069𝑏8

[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 𝑧 axis.

∴ 𝐼$$& = 𝐼B7C + 2𝐼E6FGH7 =112(𝑏)(0.1𝑏)2 +

212(0.1𝑏)(𝑏)2 = 0.0168𝑏8 [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.

𝑦JFK,(%) = 0.6𝑏; 𝑦JFK,(&) = 0.5𝑏 [1.3]

The loading conditions being same for both the configurations, 𝑀OPQ can be considered as the maximum bending moment

𝜎OPQ% =𝑀JFK𝑦JFK,(%)

𝐼$$%= 8.695𝑏S2𝑀JFK

[1.4]

𝜎JFK& =𝑀JFK𝑦JFK,(&)

𝐼$$&= 29.76𝑏S2𝑀JFK

[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 𝐿 is subjected to a uniformly distributed downward load 𝑝V(Force/length) between A and B, a concentrated moment 𝑀W at C and a point force 𝑃acting 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 𝑞V =Z[\]^_

,𝑀W = 𝑘&𝑀V, 𝑃 =Za\]^

a) Determine the numerical values of 𝑘%, 𝑘&, and 𝑘2. b) Determine the second area moment of inertia of the cross section about its centroid 𝐼b in

terms of 𝑏8. c) Determine the location (𝑥) where the cross-section experiences zero flexural stress. d) Determine the (𝑥, 𝑦) coordinates where maximum tensile stress is experienced, and the

(𝑥, 𝑦) 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:

Σ𝐹e :𝑞V𝐿2 + 𝑃 − 𝐸e = 0

=>𝐸e = 𝑃 −𝑞V𝐿2 =

𝑀V

𝐿 9−𝑘%2 + 𝑘2;

[2.1]

Σ𝑀h :𝑞V𝐿2 9

3𝐿4 ; −𝑀W − 𝑃 9

𝐿8; + 𝑀h = 0

=> 𝑀h = 𝑘&𝑀V +𝑘2𝑀V

8 −38𝑘%𝑀V = 𝑀V 9−

38𝑘% + 𝑘& +

𝑘28 ;

[2.2]

Part a)

The shear force and bending moment equations for each section can be written using the following two formulae:

𝑉(𝑥&) = 𝑉(𝑥%) + k 𝑝(𝜁)𝑑𝜁

K_

K[

[2.3]

Fig. 6.2 (c)

𝑀(𝑥&) = 𝑀(𝑥%) + k 𝑉(𝜁)𝑑𝜁

K_

K[

[2.4]

Section AB:

𝑉(𝑥) = 𝑉(0) + k−𝑞V𝑑𝑥K

V

= 0 − 𝑞V𝑥; 𝑉 9𝐿2; = −

𝑘%𝑀V

2𝐿 [2.5]

𝑀(𝑥) = 𝑀(0) + k−𝑞V𝑥𝑑𝑥K

V

= −𝑞V𝑥&

2 ;𝑀 9𝐿2; = −

𝑘%𝑀V

8 [2.6]

Section BC:

𝑉(𝑥) = 𝑉 9𝐿2; +

k0𝑑𝑥K

^&

= −𝑘%𝑀V

2𝐿 ; 𝑉 93𝐿4 ; = −

𝑘%𝑀V

2𝐿 [2.7]

𝑀(𝑥) = 𝑀 9𝐿2; +

k𝑉(𝑥)𝑑𝑥K

^&

= −𝑘%𝑀V

8 −𝑘%𝑀V

2𝐿 9𝑥 −𝐿2; =

𝑘%𝑀V

8 −𝑘%𝑀V𝑥2𝐿 ;

𝑀93𝐿4 ; = −

𝑘%𝑀V

4

[2.8]

𝑉 93𝐿4 ;

n

= 𝑉 93𝐿4 ;

S

= −𝑘%𝑀V

2𝐿 ; 𝑀 93𝐿4 ;

n

= 𝑀 93𝐿4 ;

S

+𝑀W = 𝑀V 9−𝑘%4 + 𝑘&;

[2.9]

Section CD:

𝑉(𝑥) = 𝑉 93𝐿4 ;

n

+ k0𝑑𝑥K

2^8

= −𝑘%𝑀V

2𝐿 ; 𝑉 97𝐿8 ; = −

𝑘%𝑀V

2𝐿 [2.10]

𝑀(𝑥) = 𝑀 93𝐿4 ;

n

+ k𝑉(𝑥)𝑑𝑥K

2^8

= 𝑀V 9−𝑘%4 + 𝑘&; −

𝑘%𝑀V

2𝐿 9𝑥 −3𝐿4 ;

= 𝑀V 9𝑘%8 + 𝑘&; −

𝑘%𝑀V𝑥2𝐿 ;

𝑀97𝐿8 ; = 𝑀V 9

−5𝑘%16 + 𝑘&;

[2.11]

𝑉 97𝐿8 ;

n

= 𝑉 97𝐿8 ;

S

+ 𝑃 =𝑀V

𝐿 9−𝑘%2 + 𝑘2; ;

𝑀 97𝐿8 ;

n

= 𝑀 97𝐿8 ;

S

= 𝑀V 9−5𝑘%16 + 𝑘&;

[2.12]

Section DE:

𝑉(𝑥) = 𝑉 97𝐿8 ;

n

+ k0𝑑𝑥K

o^p

=𝑀V

𝐿 9−𝑘%2 + 𝑘2;; 𝑉(𝐿) = 𝐸e =

𝑀V

𝐿 9−𝑘%2 + 𝑘2;

[2.13]

𝑀(𝑥) = 𝑀 97𝐿8 ;

n

+ k𝑉(𝑥)𝑑𝑥K

o^p

= 𝑀V 9−5𝑘%16 + 𝑘&; +

𝑀V

𝐿 9−𝑘%2 + 𝑘2; q𝑥 −

7𝐿8r

= 𝑀V 9𝑘%8 + 𝑘& −

7𝑘28 ; +

𝑀V𝑥𝐿 9−

𝑘%2 + 𝑘2; ;

𝑀(𝐿) = 𝑀h = 𝑀V 9−3𝑘%8 + 𝑘& +

𝑘28 ;

[2.14]

Using the moment values at certain lengths from the equations above,

𝑀s^&t = −Z[\]

p= −0.3𝑀V (from Fig. 6.2 (a))

⟹ 𝑘% = 2.4

[2.15]

𝑀so^ptn= 𝑀V s−

vZ[%w+ 𝑘&t = −0.7𝑀V (from Fig. 6.2 (a))

⟹ 𝑘& = 0.05

[2.16]

𝑀(𝐿) = 𝑀V s−2Z[p+ 𝑘& +

Zapt = 0.2𝑀V (from Fig. 6.2 (a)) [2.17]

⟹ 𝑘2 = 8.4

Part b)

To find the centroid of the area, by considering the top surface to be 𝑦 = 0 just for this purpose,

𝐴% = 0.3𝑏&;𝐴& = 0.35𝑏& [2.18]

𝑦% = −0.15𝑏;𝑦& = −0.53𝑏; [2.19]

The y-coordinate of the centroid, 𝑦y can be found out by using,

𝑦y =𝐴%𝑦% + 𝐴&𝑦&𝐴% + 𝐴&

= −0.355𝑏 [2.20]

From the fig below, second area moment of inertia 𝐼$

𝐼% =𝑏ℎ2

12 + 𝐴%(𝑦y − 𝑦%)& =112𝑏

(0.3𝑏)2 + 0.3𝑏&(0.082𝑏)& = 0.015𝑏8 [2.21]

𝐼& =𝑏ℎ2

36 + 𝐴&(𝑦y − 𝑦&)& =136𝑏

(0.7𝑏)2 + 0.35𝑏&(0.301𝑏)& = 0.0202𝑏8

𝐼b = 𝐼% + 𝐼& = 0.0355𝑏8

[2.22]

Part c)

The flexural stress can be denoted as 𝜎K = −\e{

, hence the flexural stress is zero across the whole

cross-section when 𝑀 is 0. Here, 𝑀 = 0 at 𝑥 = 0 and o^p< 𝑥 < 𝐿. To find another 𝑥, the moment

in section DE is:

Fig. 6.2 (d)

𝑀(𝑥) = 𝑀V 9𝑘%8 + 𝑘& −

7𝑘28 ; +

𝑀V𝑥𝐿 9−

𝑘%2 + 𝑘2; = 0

⟹ 𝑥 = 0.972𝐿

[2.23]

Hence, 𝜎K = 0 at 𝑥 = 0 and 𝑥 = 0.927𝐿.

Part d)

Maximum bending moment from the Fig. 6.2 (a) is 𝑀JFK = −0.7𝑀V

𝜎}7G~�67 =−𝑀𝑦𝑡𝑜𝑝

𝐼=0.7𝑀V(0.355𝑏)0.0355𝑏8 = 7𝑀V𝑏S2

𝜎�5J��7~~��7 =−𝑀𝑦𝑏𝑜𝑡𝑡𝑜𝑚

𝐼=−0.7𝑀V(𝑏 − 0.355𝑏)

0.0355𝑏8 = −12.72𝑀V𝑏S2

[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:

Σ𝐹e:𝐴e − 1200 − 1600 + 𝐸e − 600 = 0

=> 𝐴e + 𝐸e = 3400𝑁

[2.1]

Σ𝑀� :− 1200(1.5) − 1200 − 160097 +43; + 𝐸e

(8) − 600(11) = 0

=>𝐸e = 2867𝑁

⟹ 𝐴e = 533𝑁

[2.1]

Fig. 6.3 (c)

Parts a), b)

Part c)

The maximum magnitude of bending moment from the fig. above is at E, 𝑀JFK = 2703𝑁𝑚. 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 (𝑦y) of the cross-section must be obtained.

Fig. 6.3 (d)

𝑦y =𝐴%𝑦% − 𝐴&𝑦&𝐴% − 𝐴&

=15.75𝑎&(−1.75𝑎) − 6.25𝑎&(−2.25𝑎)

9.5𝑎& = −1.421𝑎 [3.1]

Therefore, the centroid is at 1.421𝑎from the top surface of the cross-section. Similarly, to find the second area moments of inertia,

𝐼% =𝑏ℎ2

12 + 𝐴%(𝑦y − 𝑦%)& =1124.5𝑎

(3.5𝑎)2 + 15.75𝑎&(0.329𝑎)& = 17.785𝑎8 [3.2]

𝐼& =𝑏ℎ2

12 + 𝐴&(𝑦y − 𝑦&)& =112(2.5𝑎)8 + 6.25𝑎&(0.829𝑎)& = 7.55𝑎^4

𝐼b = 𝐼% − 𝐼& = 1.64𝑋10Sw𝑚8

[3.3]

𝑦}5� = 1.421𝑎; 𝑦C5}}5J = 2.079𝑎

𝜎}7G~�67 = 46.84𝑀𝑃𝑎; 𝜎�5J��7~~��7 = 68.53𝑀𝑃𝑎

[3.4]

Part d)

The maximum shear stress can be calculated using the following formula,

𝜏JFK =|𝑉|JFK𝑄

𝐼𝑡=|𝑉|JFK𝐴∗𝑦∗

𝐼b𝑡=1500 ∗ {4.5𝑎(1.421𝑎) − 2.5𝑎(0.421𝑎)} ∗ {0.809𝑎}

10.235𝑎8 ∗ 2𝑎

∴ 𝜏JFK = 791.21𝑘𝑃𝑎

[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 𝑤 of the beam is to be considered, using 𝛾 = 50𝑙𝑏 𝑓𝑡2⁄ for the specific weight of the wood. The maximum shear stress in any cross-section (along the y-direction) must not exceed 𝜏F665B =90𝑝𝑠𝑖.

a) Calculate the maximum allowable value of the loading 𝑃 for the solid rectangular cross-section (A), with dimensions 𝑏 = 2.5𝑖𝑛 and ℎ = 6𝑖𝑛.

b) Calculate the maximum allowable value of the loading 𝑃 for the solid circular cross-section (B), with diameter 𝑑 = 6𝑖𝑛.

c) Calculate the maximum allowable value of the loading 𝑃 for the hollow circular cross-section (C), with outer diameter diameter 𝑑5 = 6𝑖𝑛 and inner diameter 𝑑� = 3𝑖𝑛

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 𝑤 for each of the cross-sections A, B & C:

For (A): 𝑤 = �s&.v%&𝑖𝑛t s w

%&𝑖𝑛t�s50 6CE

E}at ≈ 5.21 6CE

E} [4.1]

For (B): 𝑤 = �(𝜋)s w�G&∗%&

t&� s50 6CE

E}at ≈ 9.82 6CE

E} [4.2]

Fig. 6.4 (b)

Fig. 6.4 (c)

For (C): 𝑤 =  (𝜋)s w�G&∗%&

t&− (𝜋)s 2�G

&∗%&t&¡s50 6CE

E}at ≈ 7.36 6CE

E}

[4.3]

Equilibrium Equations:

Σ𝐹e:𝐴e − (8𝑓𝑡)𝑤 − 𝑃 − 2𝑃 + 𝐷e = 0

=> 𝐴e + 𝐷e = 3𝑃 + (8𝑓𝑡)𝑤

[4.4]

Σ𝑀� :− (8𝑓𝑡)𝑤(4𝑓𝑡) − 𝑃(4𝑓𝑡) − 2𝑃(7𝑓𝑡) + 𝐷e(8𝑓𝑡) = 0

=>𝐷e = 2.25𝑃 + (4𝑓𝑡)𝑤

𝐴e = 0.75𝑃 + (4𝑓𝑡)𝑤

[4.5]

Shear Force Diagram:

The maximum magnitude of shear stress is seen to |𝑉|JFK = 𝐷e = 2.25𝑃 + (4𝑓𝑡)𝑤 and occurs at 𝑥 = 8𝑓𝑡.

Part a)

Fig. 6.4 (d)

The maximum shear stress of a rectangular cross-section can by calculated using the reduced formula,

𝜏JFK =|𝑉|JFK𝑄

𝐼𝑡 =32|𝑉|JFK𝐴 = 90𝑝𝑠𝑖

⟹3

2

2.25𝑃𝑎𝑙𝑙𝑜𝑤 + (4𝑓𝑡) s5.21𝑙𝑏𝑓𝑓𝑡 t

(2.5𝑖𝑛)(6𝑖𝑛)= 90𝑝𝑠𝑖

∴ 𝑃F665B ≈ 390.74𝑙𝑏𝑓

[4.6]

Part b)

𝜏JFK =|𝑉|JFK𝑄

𝐼𝑡 =34|𝑉|JFK𝐴 = 90𝑝𝑠𝑖

⟹43

2.25𝑃F665B + (4𝑓𝑡) 99.82𝑙𝑏𝑓𝑓𝑡 ;

(𝜋) s6𝑖𝑛2 t& = 90𝑝𝑠𝑖

∴ 𝑃F665B ≈ 830.77𝑙𝑏𝑓

[4.7]

Part c)

Calculating the 𝜏JFK at the neutral axis for this cross-section,

For (C): 𝜏JFK is expected to occur at the neutral axis. Calculating the maximum shear stress:

𝜏JFK =|𝑉|JFK𝑄

𝐼𝑡 = 90𝑝𝑠𝑖

𝑄 = 𝐴∗𝑦∗ = [𝐴% − 𝐴&] �𝐴%𝑦% − 𝐴&𝑦&𝐴% − 𝐴&

� = 𝐴%𝑦% − 𝐴&𝑦&

∴ 𝑄 = ¥𝜋296𝑖𝑛2;&¦ q9

43𝜋;96𝑖𝑛2;r − ¥

𝜋293𝑖𝑛2;&¦ q9

43𝜋;93𝑖𝑛2;r = 15.75𝑖𝑛2

[4.8]

𝐼 =𝜋4  9

6𝑖𝑛2 ;

8

− 93𝑖𝑛2 ;

8

¡ = 59.64𝑖𝑛8; 𝑡 = 𝑑5 − 𝑑� = 3𝑖𝑛

∴ 𝑃F665B ≈ 439.94𝑙𝑏𝑓

Part d)

The shear force 𝑉 and the bending moment 𝑀 at 𝑥 = 6𝑓𝑡 for cross section (A) are as follows:

V(x = 6ft) = −0.25𝑃 − (2𝑓𝑡)𝑤 = −108.11𝑙𝑏𝑓

𝑀(𝑥 = 6𝑓𝑡) = 2.5𝑃 + (6𝑓𝑡)𝑤 = 1008.11𝑙𝑏𝑓𝑡

[4.9]

Shear stress values:

𝜏JFK = 𝜏\ =32𝑉𝐴 = −10.811𝑝𝑠𝑖

𝜏« =𝑉𝑄𝐼𝑡=

6𝐴ℎ&  

ℎ&

4 − 𝑦&¡𝑉 =6

(2.5 ∗ 6)(6)&  6&

4 − 1.5&¡ (−108.11) = −8.108𝑝𝑠𝑖

𝜏J�G = 𝜏¬ = 0

[4.10]

Flexural stress values:

𝜎J�G = 𝜎\ = 0𝑝𝑠𝑖

𝜎« =−𝑀𝑦𝐼

=­(−1008.11 ∗ 12𝑙𝑏𝑖𝑛)(1.5𝑖𝑛)®

45𝑖𝑛8 = −403.24𝑝𝑠𝑖

𝜎JFK = 𝜎¬ = −806.49𝑝𝑠𝑖

[4.11]

The state of stress at three different points can be given as,

Fig. 6.4 (e)