chapter 7 forces in...

122 © John Bird & Carl Ross Published by Taylor and Francis

CHAPTER 7 FORCES IN STRUCTURES

EXERCISE 46, 03

1. Determine the internal forces in the following pin-jointed truss using a graphical method.

The spaces between the forces are shown by upper case letters in the diagram below.

In all such problems, joints with two or less unknowns should be tackled first.

The vector diagram, drawn to scale, for the whole framework is shown below.

By measurement, bc = - 2.0 kN (2-3), ca = - 3.5 kN (1-3), dc = 1.7 kN (1-2), da = 3.0 kN ( 1R ) and

bd = 1.0 kN ( 2R )

i.e. R 1 = 3.0 kN, R 2 = 1.0 kN, 1-2, 1.7 kN, 1-3, - 3.5 kN, 2-3, - 2.0 kN





By measurement, ac = + 3.0 kN (1-3), bc = - 5.2 kN (2-3), cd = - 1.5 kN (1-2), be = 6.0 kN ( 2H ),

ed = 2.6 kN ( 2R ) and da = - 2.6 kN ( 1R )

i.e. R 1 = - 2.6 kN, R 2 = 2.6 kN, H 2 = 6.0 kN, 1-2, - 1.5 kN, 1-3, 3.0 kN, 2-3, - 5.2 kN





By measurement, fa = - 7.1 kN (1-3), cf = - 1.3 kN (2-3), df = + 1.0 kN (1-2), cd = 1.0 kN ( 2R ),

de = 5.0 kN ( 1R ) and ea = 4.0 kN ( 1H )

i.e. R 1 = 5.0 kN, R 2 = 1.0 kN, H 1 = 4.0 kN, 1-2, 1.0 kN, 1-3, - 7.1 kN, 2-3, - 1.4 kN




As there is no joint with less than 3 unknowns, taking moments about joint 2 gives:

R1 × 12 = 2 × 9 + 4 × 6 + 6 × 3

i.e. 12 R1 = 18 + 24 + 18 = 60

and R1 = 6012

= 5 kN = de

and R 2 = 12 – 5 = 7 kN = cd

The vector diagram for the whole framework, drawn to scale, is shown below.

By measurement, bh = - 8.0 kN (4-5), hg = + 4.0 kN (4-6), ga = - 8.0 kN (3-4), df = + 8.7 kN (1-6),


jc = - 14 kN (5-2), fg = - 2.0 kN (3-6), dj = + 12.1 kN (6-2), ef = - 10.0 kN (1-3) and

jh = - 6.0 kN (5-6)

i.e. R 1 = 5.0 kN, R 2 = 7.0 kN, 1-3, - 10.0 kN, 1-6, 8.7 kN, 3-4, - 8.0 kN, 3-6, - 2.0 kN,

4-6, 4.0 kN, 4-5, - 8.0 kN, 5-6, - 6.0 kN, 5-2, - 14.0 kN, 6-2, 12.1 kN


EXERCISE 47, Page 109

1. Using the method of joints, determine the unknown forces for the following pin-jointed truss.

All unknown member forces are assumed to be in tension.

For joint 3 (see diagram):

Resolving vertically, 4 + 2 3F − cos 60º + 1 3F − cos 30º = 0 (1)

Resolving horizontally, 1 3F − sin 30º = 2 3F − sin 60º

i.e. 1 3F − = 2 3F − 2 3sin 60 0.866Fsin 30 0.5−

° = °

i.e. 1 3F − = 1.732 2 3F − (2)

Substituting equation (2) into equation (1) gives:

4 + 2 3F − (0.5) + 1.732 2 3F − (0.866) = 0

or 2 3F − (0.5 + 1.5) = - 4

and 2 3F − = - 2.0 kN (compression) (3)

From equation (2), 1 3F − = 1.732(- 2.0)


i.e. 1 3F − = - 3.46 kN (compression) (4)


Resolving horizontally, 1 2 1 3F F cos 60− −= − °

i.e. − = − − =1 2F ( 3.46)(0.5) 1.73kN

Resolving vertically, 1 1 3R F sin 60 0−+ ° =

or 1 1 3R F (0.866)−= − (5)

and from equation (4), 1R ( 3.46)(0.866)= − −

i.e. 1R 3.0kN= (6)

Resolving vertically, 1 2R R 4+ =

Hence, from equation (6), 2R 4 3 1.0kN= − =

Summarising, R 1 = 3.0 kN, R 2 = 1.0 kN, 1-2, 1.73 kN, 1-3, - 3.46 kN, 2-3, - 2.0 kN




Resolving vertically, 1 3 2 3F cos30 F cos 60 0− −° + ° =

i.e. 1 3 2 3(0.866)F (0.5)F 0− −+ = (1)

Resolving horizontally, 1 3 2 3F sin 30 F sin 60 6− −° = ° +

i.e. 1 3 2 30.5F 0.866F 6− −= +

i.e. 1 3 2 3F 1.732F 12− −= + (2)


2 3 2 3(0.866)(1.732F 12) (0.5)F 0− −+ + =

i.e. 2 3 2 31.5F 10.392 0.5F 0− −+ + =

and 2 32F 10.392− = −

i.e. 2 3F − = 10 3922.

− = - 5.2 kN (3)

From equation (2), 1 3F 1.732( 5.2) 12− = − +

i.e. 1 3F − = 3.0 kN (4)


Resolving vertically, 1 3 1F sin 60 R 0− ° + =


or 1 1 3R (0.866)F −= − (5)

From equation (4), 1R (0.866)(3.0)= −

i.e. 1R = - 2.6 kN

Resolving vertically (overall), 1 2R R 0+ =

i.e. = − = +2 1R R 2.6 kN

Resolving horizontally gives: 1 3 1 2F cos 60 F 0− −° + =

or 1 2 1 3F 0.5F− −= − (6)

From equation (4), 1 2F 0.5(3.0) 1.5kN− = − = −

Resolving horizontally (overall), 1H 6kN=

Summarising, R 1 = -2.6 kN, R 2 = 2.6 kN, H 2 = 6.0 kN, 1-2, - 1.5 kN, 1-3, 3.0 kN,

2-3, - 5.2 kN




Resolving horizontally, 1 3 2 3F sin 45 4 F sin 45− −° + = °

i.e. 1 3 2 3(0.707)F 4 (0.707)F− −+ =

and 1 3 2 34F F

0.707− −= −

or 1 3 2 3F F− −= - 5.657 (1)

Resolving vetically, 1 3 2 36 F cos 45 F cos 45 0− −+ ° + ° =

i.e. 1 3 2 36 F F 0

cos 45 − −+ + =°

and 1 3 2 38.485 F F 0− −+ + =

or 1 3 2 3F F 8.485− −+ = − (2)

Substituting equation (1) into equations (2) gives:

2 3 2 3F 5.657 F 8.485− −− + = −

i.e. 2 32F 2.828− = −

and 2 3F 1.414kN− = − (3)

Substituting equation (3) into equations (1) gives:

1 3F − -1.414 - 5.657 = - 7.07 kN



or 2 2 3R (0.707)F −= − (4)


From equation (3), 2R (0.707)(1.414) 1.0kN= − =

Resolving horizontally, 2 3 1 2F cos 45 F 0− −° + =

i.e. 1 2 2 3F (0.707)F− −= −

From equation (3), 1 2F (0.707)( 1.414)− = − − =1.0 kN

Resolving vertically (overall), 1 2R R+ = 6 or 1 2R 6 R 5.0kN= − =

Resolving horizontally (overall), 1H 4kN=

Summarising, R 1 = 5.0 kN, R 2 = 1.0 kN, H 1 = 4.0 kN, 1-2, 1.0 kN, 1-3, - 7.07 kN,

2-3, - 1.41 kN


As there are no joints which have two or less unknowns, it will be necessary to calculate 1R and

2R .

Taking moments about joint 2 gives: 1R × 12 = 2 × 9 + 4 × 6 + 6 × 3

= 18 + 24 + 18 = 60

i.e. 160R12

= = 5 kN (1)

Resolving vertically (overall), 1R + 2R = 2 + 4 + 6 = 12

Hence, 2R = - 1R + 12 = 12 – 5

i.e. 2R = 7 kN (2)

For joint 1 (as there are only 2 unknowns here) - see diagram:



i.e. 11 3

RF0.5− = −

From equation (1), 11 3

R 5F 10kN0.5 0.5− = − = − = − (3)

Resolving horizontally, 1 6 1 3F F cos30− −+ °= 0

i.e. 1 6 1 3F (0.866)F− −= − (4)

From equation (4), 1 6F (0.866)( 10) 8.66kN− = − − = (5)


Resolving horizontally, 1 3 3 6 3 4F cos30 F cos30 F cos30− − −° = ° + °

i.e. 3 6 1 3 3 4F F F− − −= − (6)

From equation (3), 3 6 3 4F 10 F− −= − − (7)

Resolving vertically, 1 3 3 6 3 4F sin 30 F sin 30 2 F sin 30− − −° + ° + = °

or 1 3 3 6 3 42F F F

0.5− − −+ + = (8)


1 3 3 4 3 4F 10 F 4 F− − −− − + =


or 1 3 3 4F 10 4 2F− −− + = (9)

From equation (3), 3 410 10 4 2F −− − + =

i.e. 3 42F 16− = −

from which, 3 4F 8kN− = − (10)

From equations (3) and (6), 3 6F 10 8 10 8 2kN− = − − − = − + = −


Resolving horizontally, 3 4 4 5F F− −= (11)

From equation (10), 4 5F 8kN− = − (12)

Resolving vertically, 4 6 3 4 4 54 F F cos 60 F cos 60 0− − −+ + °+ ° = (13)

From equations (10) and (12), 4 64 F 8cos 60 8cos 60 0−+ − °− ° =

i.e. 4 64 F 4 4 0−+ − − =

from which, 4 6F 4 4 4 4kN− = + − = (14)


Resolving horizontally, 4 5 5 6 5 2F cos30 F cos30 F cos30− − −° + ° = °

i.e. 5 6 5 2 4 5F F F− − −= − (15)


From equation (12), 5 6 5 2F F 8− −= + (16)

Resolving vertically, 4 5 5 6 5 2F sin 30 6 F sin 30 F sin 30− − −° = + °+ ° (17)

From equations (12) and (16), 5 2 5 28(0.5) 6 (F 8)(0.5) F (0.5)− −− = + + +

i.e. – 4 = 6 + 4 + 5 2F −

i.e. 5 2F 4 6 4 14kN− = − − − = − (18)

From equation (16), 5 6F 14 8 6kN− = − + = − (19)


Resolving horizontally, 2 6 5 2F F cos30− −+ °

i.e. 2 6 5 2F F cos30 ( 14)(0.866)− −= − ° = − −

i.e. 2 6F 12.1kN− =

Summarising, R 1 = 5.0 kN, R 2 = 7.0 kN, 1-3, - 10.0 kN, 1-6, 8.7 kN, 3-4, - 8.0 kN,

3-6, - 2.0 kN, 4-6, 4.0 kN, 4-5, - 8.0 kN, 5-6, - 6 kN, 5-2, - 14 kN, 2-6, 12.1 kN



1. Determine the internal member forces of the following truss, by the method of sections.

To calculate the reactions:

Let l = horizontal distance between joint (1) and joint (2), and let h = the perpendicular distance of

joint (3) from the base.

l = a + b where a = htan 60°

= 0.577 h and b = htan 30°

= 1.732 h

Taking moments about joint (2) gives:

1R l 4 b 4 1.732h× = × = ×

i.e. 1R (a b) 6.928h+ =

i.e. 1R (0.577h 1.732h) 6.928h+ =

from which, 1R = 6.928h 6.928h 6.9280.577h 1.732h 2.309h 2.309

= =+

= 3 kN (1)

Resolving vertically gives: 1 2R R+ = 4

Hence, since 1R 3kN= 2R = 1 kN (2)


Resolving vertically gives:

1 1 3R F sin 60 0−+ ° =

i.e. 11 3

RFsin 60− = −

°

From equation (1), −1 3F 30.866

− = - 3.46 kN (3)


i.e. 1 2 1 3F F cos 60− −= − °

From equation (3), −1 2F = - 3.46 × 0.5 = 1.73 kN

Consider section (2) (as shown):

Resolving horizontally gives: 1 2 2 3F F cos30 0− −+ ° =

i.e. −2 3F = 1 2F 1.73cos30 0.866

−− = −°

= - 2.0 kN (4)

Summarising, R 1 = 3.0 kN, R 2 = 1.0 kN, 1-2, 1.73 kN, 1-3, - 3.46 kN, 2-3, - 2.0 kN


Let l = horizontal distance between joint (1) and joint (2), and let h = the perpendicular distance of

joint (3) from the base, and l = a + b as in previous Problem.


l = a + b where a = htan 60°

= 0.577 h and b = htan 30°

= 1.732 h


1R l 6 h 0× + × =

i.e. 1R = ( )

6h 6h 6hl a b 0.577 1.732 h

− = − = −+ +

= - 2.6 kN (1)

Resolving vertically gives: 1 2R R+ = 0

Hence, since 1R 2.6kN= − 2R = 2.6 kN (2)

Consider section (1), as shown:


1 3 1F sin 60 R 0− ° + =

i.e. 11 3

RFsin 60− = −

°

From equation (1), −1 3F 2.60.866−

− = 3.0 kN (3)


i.e. 1 2 1 3F F cos 60− −= − °

From equation (3), −1 2F = - 3.0 × 0.5 = - 1.5 kN (4)



Resolving horizontally gives:

By inspection, H 2 = 6 kN

and 6 + 2 3 1 2F cos30 F 0− −° + =

and from equation (4), −2 3F = 1 2F 6 1.5 6cos30 0.866

−− − − − −=

° = - 5.2 kN

Summarising, R 1 = - 2.6 kN, R 2 = 2.6 kN, H 2 = 6 kN, 1-2, - 1.5 kN, 1-3, 3.0 kN,

2-3, - 5.2 kN


Let l = horizontal distance between joint (1) and joint (2).


1 1l lR l H 0 6 42 2

× + × = × + ×

i.e. 1R = 3 + 2 = 5.0 kN (1)

Hence, = − = − =2 1R 6.0 R 6.0 5.0 1kN (2)

By inspection, =1H 4kN




1 3 1F sin 45 R 0− ° + =

i.e. 11 3

RFsin 45− = −

°

From equation (1), −1 3F 5.00.707−

− = - 7.07 kN (3)

Resolving horizontally, 1 1 2 1 3H F F cos 45 0− −+ + ° =

i.e. 1 2 1 1 3F H 0.707F− −= − −

From equation (3), −1 2F = ( )4 0.707 7.07− − − = 1.0 kN (4)



2 3 1 2F cos 45 F 0− −° + =

and from equation (4), −2 3F = 1 2F 1.0cos 45 0.707

−− −=

° = - 1.41 kN

Summarising, R 1 = 5.0 kN, R 2 = 1.0 kN, H 1 = 4.0 kN, 1-2, 1.0 kN, 1-3, - 7.07 kN,

2-3, - 1.41 kN



Firstly, 1 2R and R have to be determined.


1R 12 2 9 4 6 6 3 60kN× = × + × + × =

from which, =160R12

= 5.0 kN (1)

Resolving vertically gives: 1 2R R 2 4 6+ = + =

2 1i.e. R 12 R 7.0 kN= − = (2) Consider section (1), as shown:

ha

= tan 30º from which, h = a tan 30º = 0.577 a (3)

Taking moments about joint (3) gives: 1 6 1F h R a− × = × (4)

From equations (3) and (4), 1 11 6

R a R 5.0F0.577a 0.577 0.577−

×= = =


i.e. 1 6F 8.7kN− = (5)


1 6 3 4 3 6F (F F )cos30 0− − −+ + ° =

Hence, 1 63 4 3 6

FF Fcos30

−− −= − −

° (6)

From equations (5) and (6), 3 4 3 68.7F F

0.866− −= − −

i.e. 3 4 3 6F 10.0 F− −= − − (7)


1 3 4 3 6R F sin 30 2 F sin 30− −+ ° = + ° (8)

and from equation (1), 3 4 3 65.0 0.5F 2 0.5F− −+ = + (9)

and from equation (7), ( )3 6 3 65.0 0.5 10 F 2 0.5F− −+ − − = +

i.e. 3 6 3 65.0 5.0 0.5F 2 0.5F− −− − = +

from which, 3 65.0 5.0 2 F −− − =

i.e. 3 6F 2.0kN− = − (10)

From equation (7), 3 4F 10.0 2.0 8.0 kN− = − − − = − (11)


Resolving vertically, 1 3 1F sin 30 R 0− ° + = (12)

from which, 11 3

R 5.0F 10.0 kNsin 30 0.5− = − = − = −

° (13)




2 2 6R a F h−× = ×

i.e. −2 6F = 7.0 a 7.0 a 7.0h 0.577a 0.577× ×

= = = 12.1 kN (14)

Resolving vertically gives: 2 4 5 5 6R F sin 30 6 F sin 30− −+ ° = + °

i.e. 4 5 5 67.0 0.5F 6 0.5F− −+ = +

i.e. 4 5 5 66 7F F0.5− −

−= +

or 4 5 5 6F 2 F− −= − + (15)

Resolving horizontally gives: 2 6 5 6 4 5F F cos30 F cos30 0− − −+ ° + ° = (16)

From equations (14) and (15), 5 6 4 512.1 0.866F 0.866F 0− −+ + =

i.e. 4 5 5 612.1F F0.866− −= − − (17)

Equating equations (15) and (17) gives:

5 62 F −− + = 5 614.0 F −− −

i.e. 5 62F − = - 14.0 + 2

i.e. −5 6F = - 6.0 kN (18)

From equation (15), 4 5F 2 6.0 8.0 kN− = − − = − (19)



Resolving horizontally gives: 2 6 2 5F F cos30 0− −+ ° =

i.e. 2 62 5

F 12.1F 14.0 kNcos 30 0.866

−− = − = − = −

° (20)


Resolving vertically gives: 4 6 3 4 4 54 F F cos 60 F cos 60 0− − −+ + °+ ° = (21) From equations (11) and (19), 4 64 F 8.0cos 60 8.0cos 60 0−+ − °− ° =

i.e. 4 6F 8.0cos 60 8.0cos 60 4− = ° + °−

i.e. 4 6F 4.0 kN− =

Summarising, R 1 = 5.0 kN, R 2 = 7.0 kN, 1-3, - 10.0 kN, 1-6, 8.7 kN, 3-4, - 8.0 kN,

3-6, - 2.0 kN, 4-6, 4.0 kN, 4-5, - 8.0 kN, 5-6, - 6.0 kN, 5-2, - 14.0 kN,

6-2, 12.1 kN



Answers found from within the text of the chapter, pages 98 to 110.


1. (b) 2. (a) 3. (c) 4. (c) 5. (b) 6. (a)

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