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 8.6.3P
To determine

(a)

The available shear strength of the connection based on Allowed Stress Design.

Expert Solution
Check Mark

Answer to Problem 8.6.3P

The available shear strength of the connection based on Allowed Stress Design.

is 51.75kips.

Explanation of Solution

Given:

A W16×45 beam is connected to a W10×45 column.

The 20 number of Group A bolts of diameter equal to 78in are used.

Calculation:

The available shear strength of the connections is least of the following,

  1. Strength of the bolt.
  2. Plates shear yielding strength.
  3. Plate shear rupture strength.
  4. Plates block shear strength.
  5. Strength of the weld.

Write the expression to obtain the bolt shear strength.

RnΩ=Fnv×AbΩ=Fnv×πd24Ω ..... (I)

Here, the cross section area of the unthreaded part of the bolt is Ab, the nominal shear strength of the bolt is Fnv, safety factor is Ω and diameter of the bolt is d.

Substitute 48ksi for Fnv from manual for A325N type of bolts, 78in for d and 2 for Ω in Equation (I).

RnΩ=48ksi×π(7/8in)242=14.4312kips/bolt

The shear strength of the 4 bolts is given as,

4×(RnΩ)=14.4312kips/bolt×4bolts=57.7248kips ..... (II)

Write the expression to obtain the bearing strength of the bolts on the plate.

RnΩ=(1.2LctFu)Ω(2.4dtFu)Ω ..... (III)

Here, the safety factor is Ω, the ultimate tensile strength of the connected part is Fu, thickness of the connected part is t, diameter of the bolt is d and the clear distance between the edge of the bolt hole to the edge of the material.

Substitute 2 for Ω, 5/16in for t, 58ksi for Fu, 1.7812in for Lc and 7/8in for d in Equation (III).

RnΩ=(1.2×1.7812in×516in×58ksi)2(2.4×78in×516in×58ksi)2=19.371kips/bolt19.031kips/bolt

Thus,

(RnΩ)1=19.031kips/bolt

The bearing strength for the two bolts adjacent to the plate.

2×(RnΩ)1=2×19.031kips/bolt=38.062kips

Write the expression to obtain the bearing strength of the bolts other than on the plate.

RnΩ=(1.2LctFu)Ω(2.4dtFu)Ω ..... (IV)

Here, the safety factor is Ω, the ultimate tensile strength of the connected part is Fu, thickness of the connected part is t, diameter of the bolt is d and the clear distance between the edge of the bolt hole to the edge of the material.

Substitute 2 for Ω, 5/16in for t, 58ksi for Fu, 2.0625in for Lc and 7/8in for d in Equation (IV).

RnΩ=(1.2×2.0625in×516in×58ksi)2(2.4×78in×516in×58ksi)2=22.4297kips/bolt19.031kips/bolt

Thus,

(RnΩ)2=19.031kips/bolt

The bearing strength for the two such bolts is,

2×(RnΩ)2=2×19.031kips/bolt=38.062kips

Thus, the total bearing strength of the bolts is:

RnΩ=38.062kips+38.062kips=76.124kips ..... (V)

Thus, the bolt strength is least of the shear strength and bearing strength of the bolts.

From Equations (II) and (V),

ϕRn=57.7248kips ..... (VI)

Write the expression to obtain the plate shear yielding strength.

RnΩ=(0.60FyAg)Ω=(0.60Fy(wg×t))Ω ..... (VII)

Here, the gross cross section area of the plate perpendicular to the applied load is Ag safety factor is Ω, web plate yield strength is Fy, thickness of the plate is t and gross width of the plate is wg.

Substitute 1.50 for Ω, 36ksi for Fy, 13.5in for wg and 5/16in for t in Equation (VII).

RnΩ=(0.60×36ksi(13.5in×516in))1.50=60.751kips ..... (VIII)

Write the expression to obtain the shear rupture strength of the plate.

RnΩ=(0.60FuAnv)Ω ..... (IX)

Here, the net area along the shear surface is Anv safety factor is Ω and plate ultimate strength is Fu.

Substitute, 2.0 for Ω, 58ksi for Fu and 2.9688in2 for Anv in Equation (IX).

RnΩ=(0.60×58ksi×2.9688in2)2.0=51.6571kips ..... (X)

Write the expression to obtain the block shear strength of the plate.

RnΩ=(0.6FuAnv+UbsFuAnt)Ω(0.6FyAgv+UbsFuAnt)Ω ..... (XI)

Here, the net area along the shear surface is Agv, ultimate strength of the plate is Fu, the net area of the tension surface is Ant, the reduction factor for the tensile strength is Ubs and the gross area along the shear surface is Agv and yield strength of plate is Fy.

Substitute 2.0 for Ω, 58ksi for Fu, 2.422in2 for Anv, 1.0 for Ubs, 0.547in2 for Ant, 3.5156in2 for Agv and 36ksi for Fy in Equation (XI).

RnΩ=(0.6×58ksi×2.422in2+1.0×58ksi×0.547in2)2.0(0.6×36ksi×3.5156in2×1.0×58ksi×0.547in2)Ω=58kips53.8315kips

Thus,

RnΩ=53.8315kips ..... (XII)

Write the expression to obtain the design shear strength of weld.

RnΩ=(0.9279kips/in)w ..... (XIII)

Here, the size of the fillet weld is w in terms of sixteenths.

Substitute 0.25in for w in Equation (XIII).

RnΩ=(0.9279kips/in)0.25in=(0.9279kips/in)×14in=(0.9279kips/in)×4×(116)in=(0.9279kips/in)×4sixteenths

Further solve the above equation.

RnΩ=3.7116kips/in

For 2 weld on both sides of the plate is given as,

2×RnΩ=2×3.7116kips/in=7.4232kips/in ..... (XV)

Write the expression to obtain the shear yielding strength for the base metal for 1in length.

RnΩ=0.40×Fy×t ..... (XVI)

Here, the thickness of the base metal is t and the yield strength of the base metal is Fy.

Substitute 36ksi for Fy and 5/16in for t in Equation (XVI).

RnΩ=0.40×36ksi×516in=4.5kips/in ..... (XVII)

Write the expression to obtain the shear rupture strength of the base metal.

RnΩ=0.3×Fu×t ..... (XVIII)

Here, the thickness of the base metal is t and the ultimate strength of the base metal is Fy

Substitute 58ksi for Fu and 5/16in for t in Equation (XVIII).

RnΩ=0.3×58ksi×516in=5.4375kips/in ..... (XIX)

Write the expression to obtain the yielding strength for the base metal.

RnΩ=0.40×Fy×t×L ..... (XX)

Here, the thickness of the base metal is t, the yield strength of the base metal is Fy and the length of the plate is L.

Substitute 36ksi for Fy, 5/16in for t and 13.5in for L in Equation (XX).

RnΩ=0.40×36ksi×516in×13.5in=60.75kips ..... (XXI)

From Equation (VI), (VIII), (X) and (XXI), the least value will be the available shear strength of the connections.

Thus,

RnΩ=51.75kips

Conclusion:

Thus, the available shear strength of the connection is 51.75kips.

To determine

(b)

The flexural strength of the member based on Allowed Stress Design.

Expert Solution
Check Mark

Answer to Problem 8.6.3P

The available flexural strength of the member based on Allowed Stress Design.

is 121.59ft-kips.

Explanation of Solution

Calculation:

The available flexural strength of the connections is least of the following,

  1. Strength of the bolt.
  2. Tension on gross section of the plate.
  3. Tension on net section of the plate.
  4. The block shear of the flange plate.
  5. The block shear of the beam flange.
  6. The compression of the bottom flange plate.

Write the expression to obtain the tension force due to yielding of the gross section area of the plate.

PnΩt=Fy×AgΩt=Fy×(b×t)Ωt ..... (XXII)

Here, the gross sectional area of the plate is Ag, the yield strength of the flange plate is Fy, safety factor in tension is Ωt, breadth of the plate is b and the thickness of the plate is t.

Substitute, 36ksi for Fy for A36 steel, 5/8in for t, 7in for b and 1.67 for Ωt in Equation (XXII).

PnΩt=36ksi×(7in×58in)1.67=94.3114kips ..... (XXIII)

Write the expression to obtain the tension rupture strength of the plate.

PnΩ=(Fu×An)Ω ..... (XXIV)

Here, the net area of the plate is An safety factor in tension is Ω and plate ultimate strength is Fu.

Substitute, 2.0 for Ω, 58ksi for Fu and 3.125in2 for An in Equation (XXIV).

PnΩ=(58ksi×3.125in2)Ω=90.625kips ..... (XXV)

Write the expression to obtain the bolt shear strength.

RnΩ=Fnv×AbΩ=Fnv×πd24Ω ..... (XXVI)

Here, the cross section area of the unthreaded part of the bolt is Ab, the nominal shear strength of the bolt is Fnv, safety factor is Ω and diameter of the bolt is d.

Substitute 48ksi for Fnv from manual for A325N type of bolts, 7/8in for d and 2.0 for Ω in Equation (I).

RnΩ=48ksi×π(7/8in)24Ω=14.4312kips/bolt

The shear strength of the 8 bolts is given as,

8×RnΩ=14.4312kips/bolt×8bolts=115.4496kips ..... (XXVII)

Write the expression to obtain the bearing strength of the bolts on the plate.

RnΩ=(1.2LctFu)Ω(2.4dtFu)Ω ..... (XXVIII)

Here, the safety factor is Ω, the ultimate tensile strength of the connected part is Fu, thickness of the connected part is t, diameter of the bolt is d and the clear distance between the edge of the bolt hole to the edge of the material.

Substitute 2.0 for Ω, 5/8in for t, 58ksi for Fu, 1.7812in for Lc and 7/8in for d in Equation (XXVIII).

RnΩ=(1.2×1.7812in×58in×58ksi)2.0(2.4×78in×58in×58ksi)2.0=38.7411kips38.0625kips

Thus,

(RnΩ)1=38.0625kips/bolt

The bearing strength for the 4 bolts adjacent to the plate is,

4×(RnΩ)1=4×38.0625kips/bolt=152.25kips

Write the expression to obtain the bearing strength of the bolts other than on the plate.

RnΩ=(1.2LctFu)Ω(2.4dtFu)Ω ..... (XXIX)

Here, the safety factor is Ω, the ultimate tensile strength of the connected part is Fu, thickness of the connected part is t, diameter of the bolt is d and the clear distance between the edge of the bolt hole to the edge of the material.

Substitute 2.0 for Ω, 5/8in for t, 58ksi for Fu, 2.0625in for Lc and 7/8in for d in Equation (XXIX).

RnΩ=(1.2×2.0625in×58in×58ksi)2.0(2.4×78in×52in×58ksi)2.0=44.8594kips/bolt38.0625kips/bolt

Thus,

(RnΩ)2=38.0625kips/bolt

The bearing strength for 4 such bolts is,

4×(RnΩ)2=4×38.025kips/bolt=152.25kips

Thus, the total bearing strength of the bolts is:

RnΩ=152.25kips+152.25kips=304.5kips ..... (XXX)

Thus, the bolt strength is least of the shear strength and bearing strength of the bolts.

From Equation (XXVII) and (XXX),

RnΩ=115.4496kips ..... (XXXI)

Write the expression to obtain the block shear strength of the plate.

RnΩ=(0.6FuAnv+UbsFuAnt)Ω(0.6FyAgv+UbsFuAnt)Ω ..... (XXXII)

Here, the net area along the shear surface is Agv, ultimate strength of the plate is Fu, the net area of the tension surface is Ant, the reduction factor for the tensile strength is Ubs and the gross area along the shear surface is Agv and yield strength of plate is Fy.

Substitute 2.0 for Ω, 58ksi for Fu, 9.6875in2 for Anv, 1.0 for Ubs, 0.9375in2 for Ant, 14.0625in2 for Agv and 36ksi for Fy in Equation (XXXII).

RnΩ=(0.6×58ksi×9.6875in2+1.0×58ksi×0.9375in2)2.0(0.6×36ksi×14.0625in2×1.0×58ksi×0.9375in2)2.0=195.75kips179.0625kips

Thus,

RnΩ=179.0625kips ..... (XXXIII)

Write the expression to obtain the block shear strength of the beam flange.

RnΩ=(0.6FuAnv+UbsFuAnt)Ω(0.6FyAgv+UbsFuAnt)Ω ..... (XXXIV)

Here, the net area along the shear surface is Agv, ultimate strength of the plate is Fu, the net area of the tension surface is Ant, the reduction factor for the tensile strength is Ubs and the gross area along the shear surface is Agv and yield strength of plate is Fy.

Substitute 2.0 for Ω, 65ksi for Fu, 8.7575in2 for Anv, 1.0 for Ubs, 1.9775in2 for Ant, 12.7125in2 for Agv and 50ksi for Fy in Equation (XXXIV).

RnΩ=(0.6×65ksi×8.7575in2+1.0×65ksi×1.9775in2)2.0(0.6×50ksi×12.7125in2×1.0×65ksi×1.9775in2)2.0=235.04kips254.956kips

Thus,

RnΩ=235.04kips ..... (XXXV)

Write the expression to obtain the compressive strength of the member.

PnΩc=(Fy×Ag)Ωc ..... (XXXVI)

Here, the safety factor for compression is Ωc, yield strength of the member is Fy and the gross sectional area of the plate is Ag.

Substitute, 1.67 for Ωc, 36ksi for Fy and 4.375in2 for Ag in Equation (XXXVI).

PnΩ0=(36ksi×4.375in2)1.67=94.3114kips ..... (XXXVII)

From Equations (XXIII), (XXV), (XXXI), (XXXV) and (XXXVII), the least value will be the tension at the net section as H.

Thus,

H=90.625kips

Write the expression to obtain the moment transferred by flange plate.

Mn=H×Z ..... (XXXVIII)

Here, lever arm distance is Z.

Substitute 90.625kips for H and 16.1in for Z in Equation (XXXVIII).

Mn=90.625kips×16.1in=1459.0625in-kips=1459.0625in-kips×(1ft12in)=121.59ft-kips

Conclusion:

Thus, the available flexural strength of the member is 121.59ft-kips.

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