Unit 3 Structures Solutions

5/12/2018 Unit 3 Structures Solutions - slidepdf.com

http://slidepdf.com/reader/full/unit-3-structures-solutions 1/5

Unit

3■■■■■■■■■

Detailed study 3.2:

Materials and their use

in structures

Jacaranda Physics 2, 3rd Edition TSK

Part A — WORKED

SOLUTIONS

Chapter 8

Investigating

materials1. (a) Reading from the graph,

Δl = 0.60 mm

(b) Young’s modulus is determined from the slope of

the σ vs ε graph. However, in this case use the

relationship σ = Y × ε :

6 3

6

σ

ε

150 0.100

50 10 0.6 10

500 10 Pa

500 MPa

F LY

A l

− −

×= =

× ∆×=

× × ×

= ×=

2. Estimate the area of your feet. Taking a mass of 60 kg

supported on two feet, each approximately 25 cm × 6cm, the stress would be

460 102 10 Pa.

2 0.25 0.06

×= ×

× ×

The stress in the column is5

2

4000 1.3 10 Paπ(0.100)

= ×

The column causes the larger stress.

3. (a)

Strainε

100

10 000

0.01

l

L

=∆

=

=

=

(b)3

2

8

Stressσ

50 10

(0.004)

9.9 10 Pa

F

A= =

×=

π ×

= ×

4.

4

σ

ε

80 MPa

0.002

4 10 MPa

40 GPa

Y =

=

= ×

=

5. (a) Young’s modulus is given by the slope of the σ vs

ε graph. A has the larger Young’s modulus as it is

the steepest.

(b) B(c) Toughness is given by the area under the σ vs εgraph. B is therefore the tougher material.

(d) B exhibits more plastic behaviour and is therefore

more ductile.

(e) A

6. (a)

3

2

6

Stressσ

100 10

π 0.100

3.2 10 Pa

3.2 MPa

F

A= =

×=

×

= ×=

(b)3

50.05 10ε 1 10

5

l

L

−

−∆ ×= = = ×

7. σ = Y × ε= 110 × 109 × 3 × 10

−4

= 3.3 × 107 Pa

= 33 MPa

8. From:

σ

ε

F

A

Y

Y l

L

=

= ×× ∆

=

F Ll

A Y

×⇒ ∆ =

×3

2 9

15 10 5

π (0.001) 200 10

0.119 m

120 mm

× ×=

× × ×=≅

9. (a) Using stress = σ , F

A=

where A = 12.7 × 12.7 × 10−6 m2 ≅ 161 × 10

−6

m2,

and strain = εl

L

∆= = , where L = 5.08 × 10

−2

m

Stress (MPa) Strain

0 0

449 1.97 × 10−3

673 2.95 × 10−3

898 3.94 × 10−3

999 4.92 × 10−3

Jacaranda Physics 2 TSK 31 © John Wiley & Sons Australia, Ltd 2009



1170 5.91 × 10−3

Jacaranda Physics 2 TSK 32 © John Wiley & Sons Australia, Ltd 2009



(b)

3

3

680 MPa

3 10

227 10 MPa

227 GPa

Y −

=

×

= ×

=

(c) A strain of 0.5% = 0.005 is beyond the elastic limit.

Therefore use the graph rather than the relationship

σ = Y × ε . From the graph σ ≅ 1030 MPa.

(a)

Area under curve A = area under curve B.

(b) As A and B are equally tough, the areas under thegraphs must be equal. This shows that A is stiffer by

a factor of 2.

11. σ = Y × ε= 110 × 103 MPa × 5 × 10

−4

= 55 MPa

(a)

(b) It is ductile because of its noticeable plastic behaviour.

(c) From the slope of the linear portion of the graph

before yielding, 2

24 MPa200 MPa.

12 10

Y −

= =

×(d) When stretched to twice its original

length, Δ L = L; that is, ε 1 100%. L

L= = =

The stored energy is found from the area under

the σ vs ε graph.

Stored energy ≅ 20 × 106 × 1 = 2 × 107 J m−3

13. The stiffness of fishing line (a) is constant but thestiffness of (b) and (c) changes. When the stress–strain

graph is steeper, the fishing line is stiffer. In each case the

fisherman will feel the same increasing force, assumingthat the stress in each case is applied at a constant rate.

14.

1

2

6 3

1

2

4 3

Energyσ ε

54 10 2 10

5

1.1 10 J m

−

−

= ×

× × ×= ×

= ×

15. (a) Using stress = σ , F

A=

where 2

3

2 6 212.5 10π m 122 10 m ,

2 A

−

− ×

= ≅ ×

and strainε ,

l

L

∆

= = where L

= 50.00 × 10

−3

m

Stress (MPa) Strain

0.0 0.0000

36.7 0.0004

109 0.0014

182 0.0026

254 0.0036

272 0.0150

287 0.0400

291 0.0600

291 0.0800

275 0.1030



(b) and (c)

(d) (i)

3

6

31.2 10σ 254 MPa

122 10Y −

×= =

×(ii) The ultimate tensile strength is given by the

largest stress = 291 MPa.

(iii) Breaking strength = stress at fracture = 275

MPa

16. (a) The minimum force to cause plastic deformation will be just larger than the force at yield.

That is,

F = σ Y × A

= 55 × 106 × π × (1.5 × 10−3)2

≅ 3.9 × 102 N

(b)6 3

9

5

2

ε

σ

55 10 40 10

76 10

2.9 10 m

2.9 10 mm

Y

l L

LY

−

−

−

∆ = ×

×=

× × ×=

×

= ×

= ×(c)

F = σ × A = 125 × 106 × π × 0.00152 = 884 N

17.

6

3

9σ 90 10ε 1.2 10

75 10Y

−×= × = ××

Δl = ε × L = 1.2 × 10−3 × 1 = 1.2 × 10

−3

m = 1.2 mm

18. (a)

3

3 2

6 10σ 850 MPa

π (1.5 10 )

F

A−

×= = =

× ×

(b)40.4

ε 8 10500

l

L

−∆= = = ×

(c)

6

12

4

σ 850 101.1 10 Pa

ε 8 10Y

−

×= = = ×

×

(d) From the area under the graph of force vs extension,3 31

2strain energy 6.0 10 0.4 10

1.2 J

−= × × × ×

=

(e)

3 31

2Total energy 6.0 10 0.4 10

1.2 J

−= × × × ×

=19. (a) (i) Tension

(ii) Shear

(iii) Tension

(iv) Compression(v) Tension

(vi) Tension

(vii) Tension(viii) Compression

(ix) Tension

(x) Compression(b) (i) Nylon — tough, elastic, stiff

(ii) Steel — strong, hard, stiff

(iii) Skin — elastic (decreases with age), soft, tough

(iv) Plastic — tough, stiff (v) Rubber — elastic, tough

(vi) Plastic — tough(vii) Nylon — elastic

(viii) Rubber — tough, elastic

(ix) Plastic — plastic, tough

(x) Putty or silicon — depends on material and

purpose(a)

0 F Σ =

Σ =Στ ≠

0

0

F

(b) No. Rotational equlibrium is not satisfied.

21. Because of the lever arm, Vince will get maximum effect

when the pedals are horizontal and he pushes verticallydown. The force he applies will have no effect when the

pedals are vertical because the lever arm is zero and

consequently the torque is zero.

22. To pick the rod up, a force equal to 500 + 300 + 800 N

must be applied. If the rod is not going to rotate it must be

picked up at the CM where rotational equilibrium would be satisfied i.e. Σ τ = 0. Assuming the CM is x cmfrom the bucket on the left and clockwise torques are

positive:

Σ τ = 0:300 × 1.5 − 800 × x = 0

1.5300 0.56 m

800 x = × =

23. As Sam moves beyond the left of the fulcrum, the seesaw

will rotate anticlockwise. This will happen when the

torque from Sam about the fulcrum is larger than the

torque of the bag.24. (a) The reaction from the left abutment decreases and the

reaction from the right abutment increases.



(b)

Σ F = 0: RL + RR = 12 tonne (1)

Σ τ = 0: taking torques about the left hand support

12 × 16 − 20 RR = 0 (2)

R

1612 9.6 tonne

20 R = × =

Substituting in (1), RL = 2.4 tonne.

25. (a) The wall resists the loads from the balcony and the

person with an upwards force and an anticlockwise

torque.(b) As the person moves towards the wall, the total

reaction remains constant. However, the torque on

the wall decreases.

26.

Equilibrium must be satisfied at each joint if the tent is to

stay up. The forces can be simply resolved into their vertical and horizontal components. Let the force in the

guy = T and the force in the pole = P :

(a)

horizontal: 500 cos60 0

500

cos60

1000 N tension

F T

T

T

Σ − ° =

=°

=

(b)vertical

: sin 60 0

1000 sin 60 0

Σ ° + =

° + =

F T P

P

P = −866 N

i.e. compression

The guys create a downward force on the poles.

(c) The poles would sink into the ground until the groundcan resist the downward force from the poles.

27. Taking torques about the left support:1.8 20(5 1.8) 0

72

1.8

35.6 tonnes 36 tonnes

W

W

W

− − =

=

= ≅

28. This question once again relies on applying the equations

of equilibrium.

Σ F = 0: 2000 + 800 + 600 + 200 − RL − RR = 0

that is, RL + RR = 3600 N (1)

Σ τ = 0: taking torques about the left hand support

800 × 2.0 + 2000 × 3.0 + 600 × 4.0 + 200 × 5.0 − RR × 6.0 = 0 (2)

29. Pirate Bill will tip into the water when the overturning

torque caused by his weight is greater than the stabilisingtorque of the plank’s own weight. This will occur when

the pirate is x m past the edge of the boat. The CM of the

plank is 1 m inside the edge of the boat.

Σ τ = 0: −500 × + 800 × 1 = 0

8001.6 m

500 x = =

30. If the tension in the cable isT

, the force can be found byconsidering vertical equilibrium.

2T sin 8° = 4 kN

T = l4.4 kN

31. No, the wire did not return to its original length. Theforce–extension graph shows that the wire was loaded

beyond its elastic limit. When it was unloaded, some of

the energy had been used to permanently deform the wire.

Documents

Unit 3 Structures Solutions