Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
Question
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Chapter 8, Problem 8.2.7P
To determine

(a)

The required diameter of group A bearing-type bolt by elastic analysis using LRFD.

Expert Solution
Check Mark

Answer to Problem 8.2.7P

The required diameter of the bolt by elastic analysis using LRFD is 78in.

Explanation of Solution

Given data:

The live load is 18kips.

The dead load is 40kips.

Concept Used:

Write the equation of nominal bolt area.

Ab=π4d2 ...... (I)

Here, the diameter of bolt is d, the nominal bolt area is Ab.

Calculation:

Write the equation to calculate the factored load for LRFD.

Pu=1.2(DL)+1.6(LL) ...... (II)

Here, the dead load is DL and the live load is LL.

Substitute 18kips for DL and 40kips for LL in Equation (II).

Pu=1.2(18kips)+1.6(40kips)=21.6kips+64kips=85.6kips

Determine the centroid with respect to the lower left bolt.

Write the equation to calculate the distance of centroid in x direction.

x¯=(n×d)T ..... (III)

Here, the number of bolts is n, the distance of the bolts is d and the total number of bolts is T.

Substitute the values in Equation (III).

x¯=(2×4in)+(3×6in)6=8in+18in6=26in6=4.33in

Write the equation to calculate the distance of the centroid in y direction.

y¯=(n×d)T ..... (IV)

Substitute the values in Equation (IV).

y¯=((2×4in)+(1×8in))6=16in6=2.67in

Write the expression to calculate the load applied in the x direction.

Px=Pcosθ ....... (V)

Here, the total load is P and load component in x direction is Px.

Substitute 45 for cosθ and 85.6kips for P in Equation (V).

Px=85.6kips×(45)=85.6kips×0.8=68.48kips

Write the expression to calculate the load applied in the y direction.

Py=Psinθ     ...... (VI)

Here, the and load component in y direction is Py.

Substitute 35 for sinθ and 85.6kips for P in Equation (VI).

Py=85.6kips×(35)=85.6kips×0.6=51.36kips

Write the equation to calculate the horizontal direct shear component.

Pcx=Pxn ...... (VII)

Here, the load in x direction is Px and the total number of bolts is n.

Substitute 68.48kips for Px and 6 for n in Equation (VII).

Pcx=68.48kips6=11.41kips

Write the equation to calculate the vertical direct shear component.

Pcy=Pyn ...... (VIII)

Here, the load in y direction is Py and the total number of bolts is n.

Substitute 51.36kips for Py and 6 for n in Equation (VIII).

Pcy=51.36kips6=8.56kips

Calculate the horizontal and vertical components of loads due to eccentricity.

(x2+y2)=(3×(1.67in)2+2×(0.33in)2+1×(4.33in)2+3×(2.67in)2+2×(1.33in)2+1×(5.33in)2)=80.67in2

Calculate the distance of the load from centroid of connection.

ex=8in+2in+2in+4in4.33in=11.67in

ey=2in+4in+4in2.67in=7.33in

Here, the eccentricity in x direction is ex and the eccentricity in y direction is ey.

Write the expression for the moment for x axis about midpoint of the connection between the bolts.

Mx=Py×ex ....... (IX)

Here, moment generated is x direction is Mx.

Substitute 51.36kips for Py and 11.67in for ex in Equation (IX).

Mx=51.36kips×11.67in=599kips-in

Write the expression for the moment for y axis about midpoint of the connection between the bolts.

My=Px×ey ....... (X)

Here, moment generated is y direction is My.

Substitute 68.48kips for Px and 7.33in for ey in Equation (X).

My=68.48kips×7.33in=502kips-in

Write the expression for the moment for y axis and x axis about midpoint of the connection between the bolts.

M=(MxMy) ....... (XI)

Here, the total moment generated is M.

Substitute 599kips-in for Mx and 502kips-in for My in Equation (XI).

M=599kips-in502kips-in=97kips-in

Write the equation to calculate the total fastener force.

P=(Px)2+(Py)2 ...... (XII)

Here, the total fastener force is P and the x and y direction of the total connection load P are Px and Py.

Write the equation to calculate the x component of total connection load.

Px=Pcx+Pmx     ...... (XIII)

Here, the horizontal direct shear component is Pcx, the load coming on critical bolt in x direction due to eccentricity is Pmx.

Write the equation to calculate the y component of total connection load.

Py=Pcy+Pmy     ...... (XIV)

Here, the vertical direct shear component is Pcy, the load coming on critical bolt in y direction due to eccentricity is Pmy.

The top right bolt is critical, thus check for its safety.

Calculate the distance of the bolt from the centroid.

x=4in+2in4.33in=1.67in

y=4in+4in2.67in=5.33in

Here, the distance of the top right bolt from the centroid in the x and y directions are x and y respectively.

Write the equation to calculate the load coming on the critical bolt in x direction due to eccentricity.

Pmx=M×y(x2+y2)     ...... (XV)

Substitute 97kips-in for M, 5.33in for y and 80.67in2 for (x2+y2) in Equation (XV).

Pmx=97kips-in×5.33in80.67in2=517.01kipsin280.67in2=6.41kips

Write the equation to calculate the load coming on critical bolt in y direction due to eccentricity.

Pmy=M×x(x2+y2)     ...... (XVI)

Substitute 97kips-in for M, 1.67in for x and 80.67in2 for (x2+y2) in Equation (XVI).

Pmy=97kips-in×1.67in80.67in2=161.99kipsin280.67in2=2.01kips

Calculate the x component of the total connection load.

Substitute 11.41kips for Pcx and 6.41kips for Pmx in Equation (XIII).

Px=11.41kips+6.41kips=5kips

Calculate the y component of total connection load.

Substitute 8.56kips for Pcy and 2.01kips for Pmy in Equation (XIV).

Py=8.56kips+2.01kips=10.57kips

Calculate the total fastener force.

Substitute 5kips for Px and 10.57kips for Py in Equation (XII).

P=(5kips)2+(10.57kips)2=(25kips2)+(111.73kips2)=11.693kips

Check the safety for the bottom right bolt.

Calculate the distance of the bolt from the centroid.

x=4in+2in4.33in=1.67in

y=2.67in

Calculate the load coming on the critical bolt in x direction due to eccentricity.

Substitute 97kips-in for M, 2.67in for y and 80.67in2 for (x2+y2) in Equation (XV).

Pmx=97kips-in×2.67in80.67in2=258.99kipsin280.67in2=3.21kips

Calculate the load coming on critical bolt in y direction due to eccentricity.

Substitute 97kips-in for M, 1.67in for x and 80.67in2 for (x2+y2) in Equation (XVI).

Pmy=97kips-in×1.67in80.67in2=161.99kipsin280.67in2=2.01kips

Calculate the x component of the total connection load.

Substitute 11.41kips for Pcx and 3.21kips for Pmx in Equation (XIII).

Px=11.41kips+3.21kips=14.62kips

Calculate the y component of total connection load.

Substitute 8.56kips for Pcy and 2.01kips for Pmy in Equation (XIV).

Py=8.56kips+2.01kips=10.57kips

Calculate the total fastener force.

Substitute 14.62kips for Px and 10.57kips for Py in Equation (XII).

P=(14.62kips)2+(10.57kips)2=(213.74kips2)+(111.73kips2)=18.041kips

Therefore, the total shear force in the fastener is 18.041kips.

Write the equation to calculate the nominal shear capacity of bolt.

ϕRnv=ϕFnvAb     ...... (XVII)

Here, the nominal shear capacity is Rnv, the nominal shear strength is Fnv, the load resistance factor is ϕ and the nominal bolt area is Ab.

Substitute 18.041kips for ϕRnv, 54ksi for Fnv and 0.75 for ϕ in Equation (XVII).

18.041kips=0.75×54ksi×AbAb=18.041kips0.75×54ksiAb=0.445in2

Calculate the diameter of bolt.

Substitute 1.28in2 for Ab in Equation (I).

0.445in2=π4d2d2=0.445in2×4πd=0.567in2d=0.753in

Therefore, adopt the bolt diameter of 78in.

Check for safety:

Write the equation to calculate the bearing deformation strength of the bolt.

Rbd=2.4dtFu ...... (XVIII)

Here, the bearing deformation strength is Rbd, the thickness of the connected part is t and the ultimate tensile stress of gusset plate is Fu.

Assume A36 steel and the value of Fu is 58ksi.

Assume the thickness of the plate is t and its value is 38inch.

Calculate the bearing deformation strength of bolt.

Substitute 0.875in for d, 58ksi for Fu and 38inch for t in Equation (XVIII).

Rbd=2.4×(0.875in)×(38in)×58ksi=0.7875in2×58ksi=45.675kips

Write the equation to calculate the tear-out strength of inner bolt.

Rti=1.2lctFu ...... (XIX)

Here, the tear-out strength of inner bolt is Rti and the clear distance, in the direction parallel to the applied load from the edge of bolt hole to the edge of the adjacent hole or to the edge of the material is lc and the thickness of the plate is t.

Write the equation to calculate the clear distance for edge bolt.

lc=leh2 ...... (XX)

Here, the bolt edge distance is le and the bolt hole diameter is h.

Write the equation to calculate the hole diameter.

h=1516in ...... (XXI)

Calculate the clear distance for edge bolt.

Substitute 1516in for h and 2in for le in Equation (X).

lc=2in(1516in)2=4in0.9375in2=1.53in

Calculate the tear-out strength of inner bolt.

Substitute 1.53in for lc, 58ksi for Fu and 38inch for t in Equation (XIX).

Rti=1.2×1.53in×38in×58ksi=0.6885in2×58ksi=39.933kips

Write the equation to calculate the tear-out strength of edge bolt.

Rte=1.2lctFu ...... (XXII)

Here, the tear-out strength of edge bolt is Rte.

Write the equation to calculate the clear distance for edge bolt.

lc=sh ...... (XXIII)

Here, the spacing between bolts is s.

Substitute 3in for s and 1516in for h in Equation (XXIII).

lc=3in1516in=2.0625in

Calculate the tear-out strength of edge bolt.

Substitute 2.0625in for lc, 58ksi for Fu and 38inch for t in Equation (XXII).

Rte=1.2×2.0625in×38in×58ksi=0.928in2×58ksi=53.83kips

The minimum of Equations (XXII), (XIX) and (XVIII) is the value of Rn.

Calculate the design of strength of bolt by substituting 0.75 for the resistance factor ϕ.

ϕRn=0.75×39.933kips=29.95kips

The strength is greater than the design strength.

29.95kips>18.041kips

Therefore, adopt 78in diameter bolt.

Conclusion:

Thus, the size of bolt is 78in.

To determine

(b)

The size of bolt by elastic analysis using ASD.

Expert Solution
Check Mark

Answer to Problem 8.2.7P

The size of bolt by elastic analysis using ASD is 78in.

Explanation of Solution

Calculation:

Write the equation to calculate the factored load.

Pu=DL+LL ...... (XXIV)

Here, the dead load is DL and the live load is LL.

Substitute 18kips for DL and 40kips for LL in Equation (XXIV).

Pu=(18kips)+(40kips)=58kips

Write the expression to calculate the load applied in x direction.

Px=Pcosθ ....... (XXV)

Here, the total load is P and load component in x direction is Px.

Substitute 45 for cosθ and 58kips for P in Equation (XXV).

Px=58kips×(45)=58kips×0.8=46.4kips

Write the expression to calculate the load applied in y direction.

Py=Psinθ     ...... (XXVI)

Here, the and load component in y direction is Py.

Substitute 35 for sinθ and 58kips for P in Equation (XXVI).

Py=58kips×(35)=58kips×0.6=34.8kips

Calculate the horizontal direct shear component.

Substitute 46.4kips for Px and 6 for n in Equation (VII).

Pcx=46.4kips6=7.73kips

Calculate the vertical direct shear component.

Substitute 0.890kips for Py and 6 for n in Equation (VIII).

Pcy=34.8kips6=5.8kips

Write the expression for the moment for x axis about midpoint of the connection between the bolts.

Mx=Py×ex ....... (XXVII)

Here, moment generated is x direction is Mx.

Substitute 34.8kips for Py and 11.67in for ex in Equation (XXVII).

Mx=34.8kips×11.67in=406.116k-in

Write the expression for the moment for y axis about midpoint of the connection between the bolts.

My=Px×ey ....... (XXVIII)

Here, moment generated is y direction is My.

Substitute 46.4kips for Px and 7.33in for ey in Equation (XXVIII).

My=46.4kips×7.33in=340.112k-in

Write the expression for the moment for y-axis and x-axis about midpoint of the connection between the bolts.

M=(MxMy) ....... (XXIX)

Here, the total moment generated is M.

Substitute 406.116k-in for Mx and 340.112k-in for My in Equation (XXIX).

M=(406.116k-in340.112k-in)=66.004k-in

Check the safety for the bottom right bolt.

Calculate the distance of the bolt from the centroid.

x=1.67in

y=2.67in

Calculate the load coming on the critical bolt in x direction due to eccentricity.

Substitute 66.004kips-in for M, 2.67in for y and 80.67in2 for (x2+y2) in Equation (XV).

Pmx=66.004kips-in×2.67in80.67in2=176.23kipsin280.67in2=2.185kips

Calculate the load coming on critical bolt in y direction due to eccentricity.

Substitute 66.004kips-in for M, 1.67in for x and 80.67in2 for (x2+y2) in Equation (XVI).

Pmy=66.004kips-in×1.67in80.67in2=110.23kipsin280.67in2=1.366kips

Calculate the x component of the total connection load.

Substitute 7.73kips for Pcx and 2.185kips for Pmx in Equation (XIII).

Px=7.73kips+2.185kips=9.915kips

Calculate the y component of total connection load.

Substitute 5.8kips for Pcy and 1.366kips for Pmy in Equation (XIV).

Py=5.8kips+1.366kips=7.166kips

Calculate the total fastener force.

Substitute 9.915kips for Px and 7.166kips for Py in Equation (XII).

P=(9.915kips)2+(7.166kips)2=(98.307kips2)+(51.35kips2)=12.233kips

Write the equation to calculate the nominal shear capacity of bolt.

RnvΩ=FnvAbΩ     ...... (XXX)

Here, the nominal shear capacity is Rnv, the nominal shear strength is Fnv, the safety factor is Ω and the nominal bolt area is Ab.

Substitute 12.233kips for RnvΩ, 54ksi for Fnv and 2 for Ω in Equation (XXX).

12.233kips=54ksi×Ab2Ab=12.233kips×254ksiAb=0.453in2

Calculate the diameter of bolt.

Substitute 0.453in2 for Ab in Equation (I).

0.453in2=π4d2d2=0.453in2×4πd=0.577in2d=0.759in

Therefore, adopt the bolt diameter of 78in.

Check for safety:

Calculate the bearing deformation strength of bolt.

Substitute 0.875in for d, 58ksi for Fu and 38inch for t in Equation (XVIII).

Rbd=2.4×(0.875in)×(38in)×58ksi=0.7875in2×58ksi=45.675kips

Calculate the hole diameter.

h=1516in ...... (XXI)

Calculate the clear distance for edge bolt.

Substitute 1516in for h and 2in for le in Equation (X).

lc=2in(1516in)2=4in0.9375in2=1.53in

Calculate the tear-out strength of inner bolt.

Substitute 1.53in for lc, 58ksi for Fu and 38inch for t in Equation (XIX).

Rti=1.2×1.53in×38in×58ksi=0.6885in2×58ksi=39.933kips

Calculate the clear distance for edge bolt.

Substitute 3in for s and 1516in for h in Equation (XXIII).

lc=3in1516in=2.0625in

Calculate the tear-out strength of edge bolt.

Substitute 2.0625in for lc, 58ksi for Fu and 38inch for t in Equation (XXII).

Rte=1.2×2.0625in×38in×58ksi=0.928in2×58ksi=53.83kips

The minimum of Equations (XXII), (XIX) and (XVIII) is the value of Rn.

Calculate the allowable strength of bolt by substituting 2 for the safety factor Ω.

RnΩ=39.933kips2=19.967kips

The strength is greater than the design strength.

19.967kips>12.233kips

Therefore, adopt 78in diameter bolt.

Conclusion:

Thus, the size of bolt is 78in.

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