Download - Chapter04 04
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 1/29
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 2/29
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$$ %Pa, and a #actor o# sa#et& o# 4.$ is
re'uired with respect to #racture. (etermine the minimumallowable bolt diameter D re'uired to support an applied load o# P
) 35$ *+.
i!ure E4.4–1
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 3/29
Solution
"o support a load o# V ) 5,333.333 +, the area sub-ected to
shear stress #or each bolt must e'ual or exceed
"he allowable shear stress #or the bolts is
"o support a load o# P ) 35$ *+, each o# the six bolts must support a load o
τ allow
= τ
ult
S
=3$$ %Pa
4
= /5 %Pa
V ≥ P
6 bolts=
35$,$$$ +
6 bolts= 5,333.333 +0bolt
AV ≥ V
τ allow
=5,333.333 +0bolt
/5 +0mm2 = ///.// mm2 0bolt
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 4/29
Solution
Each bolt acts in double shear there#ore, the area sub-ected to shear stress #
"he minimum bolt diameter can be #ound be e'uatin! the two expressions #o
AV = 2 sur#aces per bolt
π
4d
bolt
2
2 sur#acesπ
4d
bolt
2≥ ///.// mm2
∴d bolt
≥ 22.252 mm = 22.3 mm
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 5/29
Example Problem 4.4–2 steel plate is to be attached to a support with three bolts as shown in i!ur
crosssectional area o# the plate is $$ 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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 6/29
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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 7/29
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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 8/29
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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 9/29
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°
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 10/29
Solution
"he allowable normal stress in member 2 is
"he allowable normal stress in member 1 is
and the allowable #orce in member 1 is
"he allowable normal stress in member 2 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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 11/29
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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 12/29
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$$ *si
= 1.//$
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 13/29
Example Problem 3.3–4
:i!id bar ABD in i!ure E3.3–4 is
supported b& a pin connection at A and a
tension lin* BC . "he mmdiameter pin at
A is supported in a double shear
connection, and the 12mmdiameter pins
at B and C are both used in sin!le shear
connections. =in* BC is 3$mm wide and
6mm thic*. "he ultimate shear stren!th o#
the pins is 33$ %Pa and the &ield stren!th
o# lin* BC is 25$ %Pa.a (etermine the #actor o# sa#et& in pins A
and B with respect to the ultimate shear
stren!th.
b (etermine the #actor o# sa#et& in lin*
BC with respect to the &ield stren!th.
i!ure E3.3–4
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 14/29
Solution
;onsider a #reebod& dia!ram o# the ri!id bar. =in*
BC is a two#orce member that is oriented at an
an!le o#θ
with respect to the hori>ontal axis
"he e'uilibrium e'uations #or the ri!id bar can be
written as
tanθ = $.4 m
$.35 m∴θ = 4.14°
ΣF x = −F
BCcos4.14°+ 6.2 *+cos/$°+ A
x = $
ΣF y = F
BCsin4.14°− 6.2 *+sin/$°+ A
y = $
ΣM A = $.6 mF
BCcos4.14°+ $.35 mF
BCsin4.14°
−1.35 m6.2 *+sin/$°− $.6 m6.2 *+cos/$° = $
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 15/29
Solution
Solin! these three e'uations simultaneousl& !ies
"he resultant #orce at pin A is
<e#ore the #actors o# sa#et& can be determined, we must compute the shearstresses in pins A and B as well as the normal stress in lin* BC .
F BC =13./6 *+ A
x = /.$1/ *+ A
y = −4.61/ *+
A = A x
2 + A y
2 = /.$1/ *+2 + −4.61/ *+2 =.4$$ *+
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 16/29
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$$$ +0*+
5$.265 mm2
= 3.552 +0mm2= 3.552 %Pa
A pin B
= π 4
12 mm2 =113.$8/ mm2
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 17/29
Solution
and thus, the shear stress in pin B is
Normal stress in link:
"he normal stress in lin* BC is
τ B =
V B
AV =
13./6 *+1$$$ +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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 18/29
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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 19/29
Example Problem 3.3–5
"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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 20/29
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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 21/29
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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 22/29
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$3 *ips
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 23/29
Solutionrom E'. e, the maximum load that ma& be
applied to the ri!id bar based on the limitations
on the pin shear stress is
Maximum load P:
;omparin! the results in E's. # and !, the
maximum load P that ma& be applied to the ri!id
bar is
Pmax
≤C allow
2.125$=
1$.6$3 *ips
2.125$= 4.88$ *ips
Pmax
= 4.88$ *ips = 4.88 *ips
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 24/29
Example Problem 3.3–6
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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 25/29
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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 26/29
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
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 27/29
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.
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 28/29
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
≥ $.8/6 in.
7/21/2019 Chapter04 04
http://slidepdf.com/reader/full/chapter04-04 29/29
Solution
c "he pin at A is a sin!le shear connection there#ore, AV ) A pin. "he shear ar
AV re'uired #or this pin is
"he pin diameter can be computed #rom
AV ≥ Aτ
allow
= /6.53$ *ips4$.$ *si
=1.813 in.2
1 sur#aceπ
4 d pin
2
≥1.813 in.2
∴d pin ≥ 1.561 in.