Sample of Final Exam asfdsfg asdg adsfg adsg ergdfgbsdfgberg er ger gsdfgbfgnbfxb dgsdfg dsz gdafg dsf gsdf PDF

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Question 1 (30%) Multiple Choice Questions (Answer will be only one among A to D for each question.) (1) The shaft in the figure below has a uniform cross section. It is fixed at end A and is subjected to four axial forces. For this shaft, which of the following statements is incorrect?

6 kN

(A) (B) (C) (D)

Maximum internal force occurs in the shaft between A and B. Normal stress in the shaft between A and B is less than that between C and D. Internal force in the shaft between B and C is zero. Shaft between C and D has larger deformation than that between B and C.

(2) For a cantilever beam subjected to a point load at the free end, as shown in the figure below, which of the following statements is correct? The cross-section of the beam is a square. P

(A) (B) (C) (D)

Shear force varies linearly along the beam. Maximum bending moment occurs at the free end of the beam. Bending moment is zero at the fixed end. Shear force is constant along the beam.

(3) For shear force and bending moment diagrams, which of the following statements is correct? (A) With a distributed load applied on a beam, shear force varies linearly along the beam. (B) At the section with a concentrated load applied, the difference of the shear forces through the section equals half of the concentrated load. (C) For a simply supported beam subjected to a moment load at the middle of the beam, there must be a change of bending moment in both magnitude and sign at the middle section. (D) Between two concentrated loads and no distributed load, bending moment is constant along the beam.

1

(4) The connection shown in the figure below consists of five steel plates, jointed by a single bolt. What is the maximum shear force of the bolt?

10 kN 20 kN

70 kN

80 kN

20 kN

(A) (B) (C) (D)

70 kN 60 kN 80 kN 20 kN

(5) For the pin-jointed structure in the figure below, if friction of the joints is negligible, which of the following statements is incorrect?

(A) (B) (C) (D)

Horizontal reaction at joint E is zero. Horizontal reaction at joint D is not zero. Resultant force at joint B is equal to the load P in magnitude. Resultant force at joint A is zero.

2

(6) In the pulley system of the figure below, the friction between wheel and cable is negligible. What is the magnitude of force F required to hold the 50 lb weight in equilibrium?

(A) (B) (C) (D)

100 lb 200 lb 150 lb 50 lb

(7) For three-dimensional stress state, which of the following statements is incorrect? (A) (B) (C) (D)

Volume strain is the ratio of volume change to original volume. Volumetric change is only caused by normal strains. Volume strain is the sum of normal and shear strains. Shear strains do not cause volumetric change of the body.

(8) For stress ~ strain behavior of materials, which of the following statements is incorrect? (A) Strain hardening means that after yielding stress increases with further deformation. (B) Stress increases when necking occurs until rupture or fracture of the materials. (C) When yielding occurs, the deformation of the material cannot be totally recovered when the load is removed. (D) Tensile strength is the maximum stress a material can sustain. 3

(9)

Bracket ABO is welded at B and pin-jointed at O, as shown below. The weights of members are neglected. The structure is in force equilibrium under the loads P and w. To determine the internal components at C, which of the FBDs below is correct?

B

A C P

w

O

Cy

Cy

M Oy O

C

Cx Oy

Cx

C

Ox

Ox

O (A)

(B)

Cy

Cy M

Oy

C

Cx

Oy

Ox O Mo

O

(C)

C

Cx

Ox Mo (D)

(10) In buckling analysis of a strut/column, which of the following statements is incorrect? (A) Critical buckling load increases with the Young’s modulus of the strut material. (B) Critical buckling load increases with the length of the strut. (C) For safe application of the strut, the stress due to critical buckling load must be lower than the yielding stress of the strut material. (D) Critical buckling load increases with the dimensions of the cross section of the strut.

4

(11) For a cantilever beam subjected to a moment load at the free end, as shown in the figure below, which of the following statements is incorrect? The cross section is a square.

M0

(A) (B) (C) (D)

x

Shear force varies linearly along the beam. Bending moment is constant along the beam. Shear stress is zero throughout the beam. Maximum bending stress (normal stress) is constant along the beam.

(12) Which of the following statements is incorrect for beam bending? (A) (B) (C) (D)

At neutral axis normal stress is zero. At the top and bottom surfaces of the beam shear stress is zero. At neutral axis shear stress is zero. At neutral axis normal strain is zero.

(13) In the figure below, a rigid beam rests in the horizontal position on two aluminum cylinders before the load 80 kN is applied on the beam. What is the value of x which results in the internal force of cylinder A is three times as high as that of cylinder B?

(A) (B) (C) (D)

1/4 m 1m 2/3 m 3/4 m

5

(14) The shaft in the figure below is fixed at end C and is subjected to two torques in opposite directions. For this shaft, which of the following statements is incorrect?

(A) Twisting angle at B is in the same direction with external torque of 3 kNm. (B) Internal torque of part AB equals 2 kNm. (C) Twisting angle at A is the difference between twisting angle at B and the twisting angle of part AB due to external torque of 2 kNm. (D) Internal torque of part BC equals 3 kNm.

(15) For a simply supported beam under a uniformly distributed load, as shown below, which of the following statements is incorrect? w A

B L

(A) (B) (C) (D)

Shear force varies linearly along the beam. Bending moment varies non-linearly along the beam. The slope at the center of the beam is the maximum. Shear force at the center of the beam is zero.

6

Question 2 (15%) The structure in Figure 2 is pin-jointed at A, B, C, and D. A non-uniformly distributed load is applied over member AD. A cable is fixed at joint D and carries a weight W = 50 kg through a wheel. The friction at all the joints is neglected. The friction coefficient between the cable and wheel is 0.5. All the members have a circular cross section with a diameter of 6 mm and the pins have a diameter of 3 mm. The members and pins are all made of steel with the Young’s modulus E = 200 GPa. (1) (2) (3) (4)

Find the reactions at the supports B and C. Determine the stress and deformation of member BD. Compute the maximum shear stress of pin B. Check the buckling behavior of member CD.

w0 = 10 N/m

A

D 30

1.5 m

C

B

2m

Figure 2

7

W = 50 kg

Solution: (1) T

 T2 = e  ,  = T1 3 T1 = T and T2 = W = 50 kg = 490 N 0.5 490 = e  = e 3 = 1.6876 T T = 290.35 N

T

RCy

RBy

RBx RBx + T cos 30 = 0

1 RBy + RCy − T sin 30 −  2  10 = 0 2 2 1 1.5 T cos 30  + 2 T sin 30  +   2 10 − 2R Cy = 0 3 2 RBx = −251.45 N RBy + RCy − 155.18 = 0 RCy = 337.11N and RBy = −181.93 N

8

(2) RCy fCD fBC

f BC = 0 f CD + RCy = 0 , f CD = − RCy = −337.11N (Compressive)

RBy fAB

fBD  RBx

1.5 = 0.75 ,  = 36.87 2 RBx + f BD cos 36.87 = 0 , − 251.45 + fBD cos 36.87 = 0 , f BD = 314.31 N

tan =

 =

=

f BD 314.31 = = 11.12 MPa A   32

 E

=

11.12 − = 5.56 10 5 200 1000

l =  l = 5.56 10 −5 1.52 + 22 =13.9 10 −5 m = 0.139 mm

(3)

R  = B = AB

(4)

I=

  34 4

RBx2 + RBy2 AB

=

251.45 2 + 181.93 2 = 43.93 MPa   1.5 2

= 63.59 mm4

For pin-ended strut of AB, Pc =

 2 EI L2

=

 2  200  10 3  63.59

(1500) 2

= 55.73 N

9

For fixed-fixed strut, Pc =

 2EI

(0.5 L) 2

=

 2  200  103  63.59

( 0.5 1500) 2

= 222.92 N

f CD = 337.11N > Pc, member CD will buckle.

Question 3 (15%) A rigid bar ABCD is pinned at end A and supported by two cables at points B and C, as shown in Figure 3. The cable at B has nominal diameter dB = 12 mm, the Young’s modulus EB = 140 GPa, the allowable stress B = 210 MPa, and the coefficient of thermal expansion B = 12  10-6/C. The cable at C has nominal diameter dC = 20 mm, the Young’s modulus EC = 120 GPa, the allowable stress  C = 150 MPa, and the coefficient of thermal expansion C = 14  10-6/C. A load P acts at end D of the bar. Determine the allowable load P if the temperature rises by 60C.

L

Figure 3

10

Solution: fB

fC

Ry

Rx

2b  f B + 4b  f C = 5b  P , 2 f B + 4 f C = 5P

B

C

 B 2b , C = 2 B =  C 4b      B = L B +  B  T  , C = L  C + C  T    EB   EC      2 B L  C + C  T  = 2L B +  B  T  , C + C   T = + 2 B  T EB   EC  EC  EB C 2 B + 2  12  10 −6  60 14  10 −6  60 = 3 + 140  103 120 10 8.33 C − 14.29 B = 600

For cable B,  B = 210 MPa 8.33 C −14.29  210 = 600,  C = 432.28 MPa >  C = 150 MPa, not safe!

For cable C, C = 150 MPa 8.33 150 −14.29 P=

2 f B + 4 fC 5

=

B

=600 ,  B = 45.45 MPa < B = 210 MPa, safe!

2  45.45   6 2 + 4  150    102 = 39735 N = 39.735 kN 5

11

Question 4 (15%) In the gear system of Figure 4, a torque of 5 Nm is applied on gear E. Shaft (1) and shaft (2) are both solid shafts with the length of 300 mm and diameter of 10 mm. They are made of the same material with the shear modulus G = 50 GPa. The bearings allow free rotation of the shafts. (1) Determine the reaction torque at gear A. (2) Compute the shear stress in shaft (1) and shaft (2). (3) Determine the twist angle between gear A and gear E. 70 teeth B

30 teeth C (1)

TE = 5 Nm (2) A

E D

20 teeth 60 teeth Figure 4 Solution: (1) TD = TE = 5 Nm TD TC rC n T n , = = C , then D = D rD rC rD nD TC nC TD  nC 5 30 = 2.5 Nm = nD 60 TA TB rA n A T n = = , , then A = A r A r B rB n B T B nB TB = TC =

TA =

TB  n A 2.5  20 = 0.714 Nm = nB 70

12

(2)

𝜏1 = 𝜏2 =

𝐽 𝑇2 ∙𝑟 𝐽

= =

2500×5 𝜋×54 2

5000×5 𝜋×54 2

= 12.74 MPa = 25.48 MPa

5000  300 = 0.03 rad 54 3 50  10  2  n n 0.03 60 = 0.06 rad = D , C = D D = nC nC 30

(3)  D =

C D

𝑇1 ∙𝑟

Te L = GJ 2

2500  300 = 0.075 rad 4 3  5 50  10  2  A nB  B  n B 0.075 70 , = = 0.2625 rad = = 20 B n A A nA

B = C +

TC L = 0.06 + GJ1

 A−E = 0. 2625 rad

Question 5 (15%) The beam in Figure 5 is subjected to a uniformly distributed load and two moment loads. (1) Draw the shear force and bending moment diagrams for this beam. (2) Compute the maximum normal stress of the beam. (3) Is the normal stress at point D of the middle section of the beam tensile or compressive?

40 kN/m 18 kNm

27 kNm

1.2 m

2.4 m

13

Beam cross-section

z

y

Figure 5 Solution: (1)

40 kN/m 18 kNm

R1

27 kNm

1.2 m

2.4 m

R2

R1 + R 2 = 40  2.4 = 96 kN 18 − 27 − 40  2.4  2.4 + 3.6R 2 = 0 , R2 = 66.5 kN, then R1 = 29.5 kN

14

M

18 kNm V R1 x

V = R1 = 29.5 kN M + 18 − R1  x = 0 , M = 29.5x − 18

Let M = 0 , x = 0.61 m M

40 kN/m 27 kNm

V x

R2 V + R 2 = 40 (3.6 − x ) , V = 77.5 − 40 x

Let V = 0 , x = 1.9375 m 40 2 2 M + 27 − R 2 (3.6 − x) + (3.6 − x) = 0 , M = 212.4 − 66.5x − 20(3.6 − x ) 2 2 M max = 212.4− 66.5 1.9375− 20( 3.6− 1.9375) = 28.278 Nm Let M = 0, x = 3.1265 m V 29.5 kN

1.937.5 m 3.6 m

0 1.2 m

66.5 kN

15

x

M 28.278 kNm

17.4 kNm 3.6 m

0 18 kNm

x

1.2 m 1.9375 m 3.1265 m 0.61 m

27 kNm

(2)

Iz =

240  240 3 200  200 3 20  200 3 = 1.56  108 mm4 − + 12 12 12

=

My 28.278 106 120 = = 21.75 MPa 1.56  10 8 I

(3) The normal stress at point D of the middle section of the beam is compressive.

Question 6 (10%) The cantilever beam AB in Figure 6 is made of steel with cross-sectional dimensions 6  6 in and the Young’s modulus E = 30  106 psi. The simply supported beam DE is a wood beam with cross-sectional dimensions 8  8 in and the Young’s modulus E = 1.5  106 psi. A steel rod AC with a diameter 0.2 in, a length 10 ft, and the Young’s modulus E = 30  106 psi serves as a hanger joining the two beams. The hanger fits snugly between the beams before the uniform load is applied to beam DE. Determine the deflection of beam DE at point C and the defection of beam AB at point A.

16

Figure 6 Solution: For simply supported beam under a distributed load 4 wL DE 5  384 E DE I DE

v C1 =

For simply supported beam under a concentrated load at the center of the beam vC 2 =

fL3DE 48 EDE I DE

For cantilever beam subjected to a point load at the free end fL3AB vA = 3E AB I AB

For rod AC, f = A = EA =

v C = v C 1 − vC 2 =

4 DE

 EA l

=

(vC − vA )EA l fL3DE

wL 5 −  384 E DE I DE 48EDE I DE

400 3 (20  12 )4 f (20  12 ) 12 − = 84 84 6 6       48 1.5 10 384 1.5 10 12 12 5

vC = 2.81− 5.6 10 −4 f

vA =

f (6  12)

3

64 3  30  10 6  12

= 3.84 10 −5 f

17

2

(

 0.2  30  106    2.81 − 5.6  10 −4 f − 3.84 10 −5 f EA(vC − v A )  2  f = = l 10 12 f = 3877 lb

vC = 2.81 − 5.6  10− 4 f = 2.81 − 5.6  10−4  3877 = 0.64 in

v A = 3.84  10− 5 f = 3.84  10− 5  3877 = 0.15 in

18

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