Megson_StructlStressAnalysis_2nd_solman.pdf

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Structural and Stress

Analysis

Second Edition

by

Dr. T.H.G. Megson

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Solutions Manual

Solutions to Chapter 2 Problems

S.2.1

(a) Vectors representing the 10 and 15 kN forces are drawn to a suitable scale as shown

in Fig. S.2.1. Parallel vectors AC and BC are then drawn to intersect at C. The

resultant is the vector OC which is 21.8 kN at an angle of 23.4 ◦ to the 15 kN force.

B

C

10 kN

R

60

°

u

O

15 kN

A

F IGURE S.2.1

(b) From Eq. (2.1) and Fig. S.2.1

R 2 = 15 2 + 10 2 + 2 × 15 × 10 cos 60 ◦

which gives

R = 21 . 8kN

Also, from Eq. (2.2)

10 sin 60 ◦

15 + 10 cos 60 ◦

tan θ =

so that

θ = 23 . 4 ◦ .

3

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4

•

Solutions Manual

S.2.2

(a) The vectors do not have to be drawn in any particular order. Fig. S.2.2 shows the

vector diagram with the vector representing the 10 kN force drawn ﬁrst.

12 kN

8kN

u

10 kN

R

20 kN

F IGURE S.2.2

The resultant R is then equal to 8.6 kN and makes an angle of 23.9 ◦ to the negative

direction of the 10 kN force.

(b) Resolving forces in the positive x direction

F x = 10 + 8 cos 60 ◦ − 12 cos 30 ◦ − 20 cos 55 ◦ =−7 . 9kN

Then, resolving forces in the positive y direction

F y = 8 cos 30 ◦ + 12 cos 60 ◦ − 20 cos 35 ◦ =−3 . 5kN

The resultant R is given by

R 2 = (−7 . 9) 2 + (−3 . 5) 2

so that

R = 8 . 6kN

Also

tan θ = 3 . 5

7 . 9

which gives

θ = 23 . 9 ◦ .

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Solutions to Chapter 2 Problems

•

5

S.2.3 Initially the forces are resolved into vertical and horizontal components as shown

in Fig. S.2.3.

y

(1.0, 1.6)

x

69.3 kN

30

°

80 kN

40 kN

20.0 kN

(1, 1.25)

R x

50 kN

30

°

35.4 kN

34.6 kN

40 kN

45

°

y

(0, 0.5)

35.4 kN

R y

(1.25, 0.25)

O

x

F IGURE S.2.3

60 kN

Then

R x = 69 . 3 + 35 . 4 − 20 . 0 = 84 . 7kN

Now taking moments about the x axis

R x y = 35 . 4 × 0 . 5 − 20 . 0 × 1 . 25 + 69 . 3 × 1 . 6

which gives

y = 1 . 22 m

Also, from Fig. S.2.3

R y = 60 + 40 + 34 . 6 − 35 . 4 = 99 . 2kN

Now taking moments about the y axis

R y x = 40 . 0 × 1 . 0 + 60 . 0 × 1 . 25 − 34 . 6 × 1 . 0

so that

x = 0 . 81 m

The resultant R is then given by

R 2 = 99 . 2 2 + 84 . 7 2

from which

R = 130 . 4kN

Finally

θ = tan −1 99 . 2

84 . 7 = 49 . 5 ◦ .

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