Chapter04 04

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Example Problem 4.4–1Six bolts are used in the connection between the axial member and

the support, as shown in i!ure E4.4–1. "he ultimate shear

stren!th o# the bolts is 3$$ %Pa, and a #actor o# sa#et& o# 4.$ is

re'uired with respect to #racture. (etermine the minimumallowable bolt diameter D re'uired to support an applied load o# P

) 35$ *+.

i!ure E4.4–1

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Solution

"o support a load o# V ) 5,333.333 +, the area sub-ected to

shear stress #or each bolt must e'ual or exceed

"he allowable shear stress #or the bolts is

"o support a load o# P ) 35$ *+, each o# the six bolts must support a load o

τ allow

= τ

ult

S

=3$$ %Pa

4

= /5 %Pa

V ≥ P

6 bolts=

35$,$$$ +

6 bolts= 5,333.333 +0bolt

AV ≥ V

τ allow

=5,333.333 +0bolt

/5 +0mm2 = ///.// mm2 0bolt

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Solution

Each bolt acts in double shear there#ore, the area sub-ected to shear stress #

"he minimum bolt diameter can be #ound be e'uatin! the two expressions #o

AV = 2 sur#aces per bolt

π

4d

bolt

2

2 sur#acesπ

4d

bolt

2≥ ///.// mm2

∴d bolt

≥ 22.252 mm = 22.3 mm

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Example Problem 4.4–2 steel plate is to be attached to a support with three bolts as shown in i!ur

crosssectional area o# the plate is $$ mm2 and the &ield stren!th o# the stee

"he ultimate shear stren!th o# the bolts is 425 %Pa. #actor o# sa#et& o# 1.6

&ield is re'uired #or the plate. #actor o# sa#et& o# 4.$ with respect to the ultstren!th is re'uired #or the bolts. (etermine the minimum bolt diameter re'u

the #ull stren!th o# the plate. Note: consider onl& the !ross crosssectional a

7not the net area.

i!ure E4.4–2

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Solution

"he allowable shear stress o# the bolts is

"he allowable normal stress is

"he #ull stren!th o# the plate based on the !ross crosssectional area is ther

σ allow

= σ

Y

S=

25$ %Pa

1.6/=148./$1 %Pa

Pmax

=σ allow

A= 148./$1 %Pa$$ mm2 =118,/6$. + = 118. *+

τ allow

= τ

ult

S=

425 %Pa

4.$=1$6.25 %Pa

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Solution

E'uatin! the two expressions #or the shear area, the minimum bolt diameter

computed

Each o# the bolts supports an e'ual share o# the load thus, each bolt must sup

#orce o#

or each bolt, the minimum shear area re'uired to support a load o# V ) 38,8

Each bolt acts in sin!le shear, thus

V ≥ P

max

3 bolts

=118,/6$. +

3 bolts

= 38,82$.26/ +0bolt

AV ≥ 38,82$.26/ +

1$6.25 +0mm2 = 3/5./2$ mm

20bolt

AV = 1 sur#ace per boltπ

4d

bolt

2

1 sur#aceπ

4d

bolt

2≥ 3/5./2$ mm2

∴d bolt

≥ 21./2 mm = 21.8 mm

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Example Problem 4.4–3

9n i!ure i!ure E4.4–3, member 1 is a steel

bar with a crosssectional area o# 1./5 in.2 and

a &ield stren!th o# 5$ *si. %ember 2 is a pair

o# 6$61"6 aluminum bars hain! a combinedcrosssectional area o# 4.5$ in.2 and a &ield

stren!th o# 4$ *si. #actor o# sa#et& o# 1.5

with respect to &ield is re'uired #or both

members. (etermine the maximum allowable

load P that ma& be applied to the structure.

:eport the #actors o# sa#et& #or both membersat the allowable load.

i!ure E4

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Solution

and rearran!in! E'. c !ies an expression #or P in terms o# F 2

;onsider a <( o# -oint B and write the #ollowin! e'uilibrium e'uations

Substitutin! E'. c into E'. a !ies an expression #or P in terms o# F 1

ΣF x = F

2cos55°− F

1 = $........................................a

ΣF y = F

2sin55°− P = $.........................................b

rom E'. b

F 2 =

P

sin55°...........................................................c

F 2

cos55°− F 1

= $

P

sin55°

÷cos55°− F

1 = $

P = F 1tan55°

P = F 2 sin55°

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Solution

"he allowable normal stress in member 2 is


and the allowable #orce in member 1 is


Substitute the allowable #orce in member 1 #rom E'. # into E'. d to ob

maximum load P based on the capacit& o# member 1

σ allow,1

=σ Y ,1

S1

=5$ *si

1.5= 33.333 *si

F allow,1

=σ allow,1

A1 = 33.333 *si1./5 in.2 = 5.333 *ips

σ allow,2

=σ Y ,2

S2

=4$ *si

1.5

= 26.66/ *si

F allow,2

=σ allow,2

A2 = 26.66/ *si4.5$ in.2 =12$.$ *ips

P ≤ F allow,1 tan55° = 5.333 *ipstan55° =3.3$ *ips

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Solution

"he #actor o# sa#et& in member 1 is thus

:epeat with the allowable #orce in member 2 #rom E'. ! substituted into E

obtain the maximum load P based on the capacit& o# member 2

;ompare the results in E's. h and i to #ind that the maximum load P that

controlled b& the capacit& o# member 1

or an applied load o# P ) 3.3 *ips, the #orce in member 2 can be compute

#rom E'. c

P ≤ F allow,2

sin55° = 12$.$ *ipssin55° = 8.28 *ips

Pmax

= 3.3 *ips

S1 = 1.5

F 2 = P

sin55° = 3.3$ *ips

sin55° =1$1./$$ *ips

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Solution

which results in a normal stress o#

and thus, its #actor o# sa#et& is

σ 2 = F

2

A2

=1$1./$$ *ips

4.5$ in.2 = 22.6$$ *si

S2 =

σ Y ,2

σ 2

=4$ *si

22.6$$ *si

= 1.//$

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:i!id bar ABD in i!ure E3.3–4 is

supported b& a pin connection at A and a

tension lin* BC . "he mmdiameter pin at

A is supported in a double shear

connection, and the 12mmdiameter pins

at B and C are both used in sin!le shear

connections. =in* BC is 3$mm wide and

6mm thic*. "he ultimate shear stren!th o#

the pins is 33$ %Pa and the &ield stren!th

o# lin* BC is 25$ %Pa.a (etermine the #actor o# sa#et& in pins A

and B with respect to the ultimate shear

stren!th.

b (etermine the #actor o# sa#et& in lin*

BC with respect to the &ield stren!th.

i!ure E3.3–4

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Solution

;onsider a #reebod& dia!ram o# the ri!id bar. =in*

BC is a two#orce member that is oriented at an

an!le o#θ

with respect to the hori>ontal axis

"he e'uilibrium e'uations #or the ri!id bar can be

written as

tanθ = $.4 m

$.35 m∴θ = 4.14°

ΣF x = −F

BCcos4.14°+ 6.2 *+cos/$°+ A

x = $

ΣF y = F

BCsin4.14°− 6.2 *+sin/$°+ A

y = $

ΣM A = $.6 mF

BCcos4.14°+ $.35 mF

BCsin4.14°

−1.35 m6.2 *+sin/$°− $.6 m6.2 *+cos/$° = $

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Solution

Solin! these three e'uations simultaneousl& !ies

"he resultant #orce at pin A is

<e#ore the #actors o# sa#et& can be determined, we must compute the shearstresses in pins A and B as well as the normal stress in lin* BC .

F BC =13./6 *+ A

x = /.$1/ *+ A

y = −4.61/ *+

A = A x

2 + A y

2 = /.$1/ *+2 + −4.61/ *+2 =.4$$ *+

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Solution Pin shear stresses:

"he mmdiameter pin at A is supported in a double shear connection ther

#orce actin! on one shear plane which is simpl& e'ual to the crosssectional

is hal# o# the resultant #orce at pin A V A ) 4.2$$ *+. "he crosssectional aris

and there#ore, the shear stress in pin A is

"he 12mmdiameter pin at B and C is supported in a sin!le shear connectio

and so the shear #orce actin! on one shear plane is the entire #orce in lin* B

V B ) 13./6 *+. "he crosssectional area o# the pin at B is

A pin A

= π

4 mm2

= 5$.265 mm2

τ A =V A

AV

=4.2$$ *+1$$$ +0*+

5$.265 mm2

= 3.552 +0mm2= 3.552 %Pa

A pin B

= π 4

12 mm2 =113.$8/ mm2

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Solution

and thus, the shear stress in pin B is

Normal stress in link:

"he normal stress in lin* BC is

τ B =

V B

AV =

13./6 *+1$$$ +0*+

113.$8/ mm2 =122.683 +0mm2

=122.683 %Pa

σ BC = F

BC

ABC=

13./6 *+1$$$ +0*+

3$ mm6 mm

= //.$8$ +0mm2= //.$8$ %Pa

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Solution

Factors of safety:

or pin A, the #actor o# sa#et& is

or pin B, the #actor o# sa#et& is

"he #actor o# sa#et& in lin* BC is

S A = τ u

τ A

= 33$ %Pa3.552 %Pa

= 3.85

SB = τ

u

τ B

=33$ %Pa

122.683 %Pa= 2.68

SBC =

σ y

σ BC

=25$ %Pa

//.$8$ %Pa= 3.24

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"he simple pinconnected structure carries

a concentrated load P as shown in i!ure

E3.3–5. "he ri!id bar is supported b& strut

AB and b& a pin support at C . "he steel

strut AB has a crosssectional area o# $./5

in.2 and a &ield stren!th o# 6$ *si. "he

diameter o# the steel pin at C is $.5 in., and

the ultimate shear stren!th is 54 *si. 9# a

#actor o# sa#et& o# 2.$ is re'uired in both

the strut and the pin at C , determine themaximum load P that can be supported b&

the structure.

i!ure E3.3

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Solution

Substitute this result into E'. a to express C x in terms o# P

rom a <( o# the ri!id bar, the #ollowin!

e'uilibrium e'uations can be written

rom E'. c, express F 1 in terms o# the un*nown load P

"he resultant reaction #orce at pin C can now be expressed as a #unction o# P

ΣF x = −F

1+C

x = $..................................a

ΣF y =C

y − P = $....................................b

ΣMC = in.F

1− 15 in.P = $...............c

F 1 = 15 in.

in.

P =1./5P

C x = F

1 = 1./5P

C = C x

2+C

y

2= 1./5P2

+ P2= 2.125$P

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Solution

rom E'. d, the maximum load that ma& be applied to the ri!id bar based

limitations on the strut normal stress is

Strut AB:

"he allowable normal stress #or strut AB is

"here#ore, the allowable axial #orce #or strut AB is

σ allow

= σ

Y

S=

6$ *si

2.$= 3$ *si

F allow,1

=σ allow A

1 = 3$ *si$./5 in.2 = 22.5$ *ips

Pmax

≤F

allow,1

1./5=

22.5$ *ips

1./5=12.$ *ips

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Solution

thus, the allowable reaction #orce at C is

Pin C:

"he allowable shear stress #or the pin at C is

"he $.5in.diameter double shear pin at C has a

shear area o#

τ allow

= τ

ult

S=

54 *si

2.$= 2/ *si

AV = 2 sur#aces

π

4D

pin

2= 2 sur#aces

π

4$.5 in.

2= $.382/ in.

2

Callow

= τ allow

AV = 2/ *si$.382/ in.

2 =1$.6$3 *ips

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Solutionrom E'. e, the maximum load that ma& be

applied to the ri!id bar based on the limitations

on the pin shear stress is

Maximum load P:

;omparin! the results in E's. # and !, the

maximum load P that ma& be applied to the ri!id

bar is

Pmax

≤C allow

2.125$=

1$.6$3 *ips

2.125$= 4.88$ *ips

Pmax

= 4.88$ *ips = 4.88 *ips

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9n i!ure E3.3–6, the ri!id member ABDE is

supported at A b& a sin!le shear pin connection

and at B b& a tie rod 1. "he tie rod is attached

at B and C with double shear pin connections.

"he pins at A, B, and C each hae an ultimate

shear stren!th o# $ *si, and tie rod 1 has a

&ield stren!th o# 6$ *si. concentrated load o#

P ) 24 *ips is applied perpendicular to DE , as

shown. #actor o# sa#et& o# 2.$ is re'uired #or

all components. (etermine

a the minimum re'uired diameter #or the tie

rod.

b the minimum re'uired diameter #or the pin at

B.

c the minimum re'uired diameter #or the pin at

A.

i!ure E3.3–6

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Solution

rom E'. c

%ember 1 is a two#orce member that is oriented at θ with respect to

the hori>ontal axis

rom a <( o# ri!id structure ABDE , the #ollowin!e'uilibrium e'uations can be written

tanθ =8 #t

6 #t=1.5 ∴θ = 56.31$°

ΣF x = P cos/$°− F

1cos56.31$°+ A

x = $..........a

ΣF y = −P sin/$°− F

1sin56.31$°+ A

y = $.........b

ΣM A = F

1cos56.31$°8 #t

−Pcos/$°?13 #t + #tsin2$°@

−P sin/$°? #tcos2$°@= $..........................c

F 1 = P cos/$°?13 #t + #tsin2$°@+ sin/$°? #tcos2$°@

8 #tcos56.31$°

= 48.34 *ip

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Solution

"he resultant pin #orce at A is #ound #rom A x and A y

Substitute F 1 into E'. a to obtain A x

and substitute F 1 into E'. b to obtain A y

A x

= F 1

cos56.31$°− P cos/$°

= 58.34 *ipscos56.31$°− 24 *ipscos/$°= 24.82 *ips

A y = F

1sin56.31$°+ Psin/$°

= 58.34 *ipssin56.31$°+ 24 *ipssin/$°

= /2.33 *ips

A = A x

2+ A

y

2= 24.82 *ips2

+ /2.33 *ips2= /6.53$ *ips

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Solution

"here#ore, the minimum tie rod diameter is

a "he allowable normal stress #or tie rod 1 is

"he minimum crosssectional area re'uired #or the tie rod is

σ allow

= σ

Y

S

=6$ *si

2.$

= 3$.$ *si

Amin ≥ F

1

σ allow

=58.34 *ips

3$.$ *si=1.884 in.

2

π

4d

1

2 ≥1.884 in.2 ∴d1 ≥1.584 in.= 1.584 in.

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Solution

"he pin diameter can be computed #rom

b "he allowable shear stress #or the pins at A, B, and C is

"he doubleshear pin connection at B and C must support a load o# F 1 ) 5

*ips. "he shear area AV re'uired #or these pins is

τ allow

= τ

ult

S

=$ *si

2.$

= 4$.$ *si

AV ≥ F

1

τ allow

=58.34 *ips

4$.$ *si=1.486 in.2

2 sur#acesπ

4d

pin

2 ≥1.486 in.2 ∴d pin

≥ $.8/6 in.

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Solution

c "he pin at A is a sin!le shear connection there#ore, AV ) A pin. "he shear ar

AV re'uired #or this pin is

"he pin diameter can be computed #rom

AV ≥ Aτ

allow

= /6.53$ *ips4$.$ *si

=1.813 in.2

1 sur#aceπ

4 d pin

2

≥1.813 in.2

∴d pin ≥ 1.561 in.

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Chapter04 04