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 3, Problem 3.5.4P
To determine

(a)

The maximum factored load using Load and Resistance Factor Design (LRFD).

Expert Solution
Check Mark

Answer to Problem 3.5.4P

The maximum factored load using LRFD is 125.789kips.

Explanation of Solution

Given:

The following figure shows the A36 steel connection with 34in bolts.

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 3, Problem 3.5.4P

Figure-(1)

Concept Used:

Write the expression for the factored strength in yielding.

Py=Φy×Pny …… (I)

Write the expression for the factored strength in rupture.

Pr=Φf×Pnf …… (II)

Here, the factored yielding strength of the material is Py, the factored strength in rupture is Pr, the resistance factor for yielding is Φy, the resistance factor for rupture is Φf, the nominal strength in yielding is Pny and the nominal strength in rupture is Pnr.

Write the expression for block shear.

Ru=0.6FuAnv+UbsFuAnt …… (III)

Write the expression for the upper limit of block shear.

UL=0.6FyAgv+UbsFuAnt …… (IV)

Here, the upper limit is UL, the block shear is Ru, the ultimate stress is Fu, the yield stress is 0.6Fy, the factor for tension stress is Ubs, the gross area along the shear surface is Agv, the net area along the shear surface is Anv, and the net area along the tension surface is Ant .The design strength of LRFD is the minimum of Equation (I) and (II).

Write the expression for the design block shear strength.

P=RΦ …… (V)

Here the design block shear strength, the minimum of Equation (III) and (IV), is R and the resistance factor is Φ.

The maximum factored load is the minimum of Equation (I), (II), and (V).

Write the expression for the nominal strength in yielding for the tension member.

Pny=Fy×Ag …… (VI)

Here, the yield strength in yielding is Fy, the gross area of the gross member is Ag.

Write the expression for the nominal strength in rupture for the tension member.

Pnf=Fu×Ae …… (VII)

Here, the yield strength in rupture is Fu, the effective area is Ae.

Write the expression for the effective area.

Ae=AnU …… (VIII)

Here, the area reduction factor is U, the effective area is Ae, the net area is An

Write the expression for the area reduction factor.

U=1x¯l …… (IX)

Here the distance from the centroid of the connected area x¯ and the length of the connection is l.

Write the expression for the net area of the tension member.

An=Agt×dholes …… (X)

Here, the thickness of the tension member is t, the diameter of holes is dholes.

Write the expression for the diameter of the holes.

dholes=d+18in …… (XI)

Here the diameter of the bolts is d and 316in is the allowance for side clearance for bolts greater than 1in.

Calculation:

Calculate the nominal shear strength of the tension member in yielding.

Substitute 50ksi for Fy and 3.60in2 for Ag in the Equation (VI).

Pny=50ksi×3.60in2=180kips

Calculate the diameter of the holes.

Substitute 34in for d in the Equation (XI).

dholes=34in+18in=0.875in

Calculate the net area of the tension member.

Substitute 3.60in2 for Ag and 0.875in for dholes in the Equation (X).

An=3.60in22×(0.314×0.875)in2=3.602×0.278=3.602×0.550=3.051in2

Calculate the length of the connection.

l=3in+3in=6in

Calculate the area reduction factor.

Substitute 6in for l and 0.525in for x¯ in the Equation (IX).

U=10.525in6in=10.088=0.913

Calculate the effective area of the member.

Substitute 3.051in2 for An and 0.913 for U in Equation (VIII).

Ae=(3.051in2)×0.913=2.784in2

Calculate the nominal shear strength of the tension member in rupture.

Substitute 65ksi for Fu and 2.784in2 for Ae in the Equation (VII).

Pnf=(65ksi)×(2.784in2)=180.963kips

Calculate the net area along the shear surface of the tension member.

An=0.314in2[2×(112+3+3)in2×(2.5×0.875)in]=0.314[154.375]in2=0.314[10.625]in2=3.336in2

Calculate the net area along the tension surface of the tension member.

Ant=0.314×3in22×0.314×0.5(34+18)in2=0.942in20.314×0.875in2=0.942in20.278in2=0.667in2

Calculate the gross area along the shear surface in the gusset plate.

Agv=2×(38in×(112in+3in+3in))=2×(38in×7.5in)=2×2.813in2=5.625in2

Calculate the net area along the shear surface of the gusset plate.

Anv=2×[38in×(112in+3in+3in)38in(2.5×(34+18)in)]=2×[38in×(7.5in)38in(2.5×0.875)]=2×[2.813in2-38×2.188in2]=2×[2.813in20.820in2]

Substitute further.

Anv=2×1.993in2=3.985in2

Calculate the net area along the tension surface of the gusset plate.

Ant=38in×(32×0.5×(34+18))in=38in×[30.875]in=38×2.125in2=0.797in2

Calculate the shear strength for the tension member.

Substitute 3.336in2 for Anv, 0.667in2 for Ant, 65ksi for Fu and 1 for Ubs in Equation (III).

Ru=(0.6×65ksi×3.336in2)+(1×65×0.667in2)=130.104kips+43.355kips=173.459kips

Calculate the upper limit.

Substitute 0.667in2 for Ant

4.71in2 for Agv, 50ksi for Fy, 65ksi for Fu and 1 for Ubs in the Equation(IV).

UL=0.6×(50ksi)×(4.71in2)+1×(65ksi)×(0.667in2)=141.3kips+43.355kips=184.655kips

The value of the shear is larger than the upper limit. Hence, it is not feasible.

Adopt the shear strength of the tension member to be 173.459kips.

Calculate the shear strength for the gusset plate.

Substitute 3.985in2 for Anv, 0.797in2 for Ant, 58ksi for Fu, and 1 for Ubs in the Equation (III)

Ru=(0.6×58ksi×3.985in2)+(1×58×0.797in2)=136.678kips+46.219kips=184.897kips

Calculate the upper limit

Substitute 0.797in2 for Ant

5.625in2 for Agv, 36ksi for Fy, 58ksi for Fu and 1 for Ubs in the Equation (IV).

UL=0.6×(36ksi)×(5.625in2)+1×(58ksi)×(0.797in2)=121.5kips+46.219kips=167.719kips

The value of the shear is larger than the upper limit. Hence, it is not feasible.

Adopt the shear strength of the gusset plate to be 167.719kips.

Compare the shear strength of the tension member and that of the gusset plate.

Thus the block shear strength is 167.719kips.

Calculate the design block shear strength of the connection.

Calculate the factored yielding strength.

Substitute 0.90 for Φy and 180kips for Pny in the Equation (I).

Py=0.90×180kips=162kips

Calculate the factored rupture strength.

Substitute 0.75 for Φy and 180.963kips for Pny in the Equation (I).

Pr=0.75×180.963kips=135.722kips

Calculate the factored strength of block shear.

Substitute 167.719kips for Ru and 0.75 for Φ in the Equation (V).

P=(167.719kips)×0.75=125.789kips

Conclusion:

Thus, the maximum factored load is 125.789kips.

To determine

(b)

The allowable block shear strength of the connection.

Expert Solution
Check Mark

Answer to Problem 3.5.4P

The allowable block shear strength of the connection is 83.860kips.

Explanation of Solution

Concept used:

Write the expression for the factored design block shear strength.

Pa=RuΩ …… (XII)

Here the safety factor is Ω, the allowable block shear strength value is Pa.

Calculation:

Calculate the allowable block shear strength of the connection.

Calculate the allowable yielding strength.

Substitute 2 for Ω and 180kips for Ru in the Equation (XII).

Pa=1802=90kips

Calculate the allowable rupture strength.

Substitute 2 for Ω and 180.963kips for Ru in the Equation (XII).

Pa=180.9632=90.482kips

Calculate the factored strength of block shear.

Substitute 167.719kips for Ru and 2 for Ω in the Equation (XII).

Pa=167.719kips2=83.860kips

Conclusion:

Thus, the maximum allowable block shear strength of the connection is 83.860kips.

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Students have asked these similar questions
In the connection shown in the figure, ASTM A572 Grade 50 steel is used for the tension member, A36 steel is used for the gusset plate, and the holes are ¾-inch-bolts. a. What is the maximum factored load that can be applied if LRFD is used? Consider all limit states. b. What is the maximum service load that can be applied if ASD is used? Consider all limit states.
Can u solve me step by step please . thank u
Topic:Welded Connection - Civil Engineering -Steel Design   *Use latest NSCP/NSCP 2015 formula to solve this problem *Use attached formula picture for block shear  *Please use hand written to solve this problem     A channel is used as a tension member with the web of the channel welded to a 9.5 mm thick gusset plate as shown in the figure. The tension member is subjected to the following axial loads. Use LRFD   Service dead load = 200 kN Wind load = 276 kN Service live load = 260 kN   For channel: Fy = 345 MPa   For gusset plate: Fy = 248 MPa Fu = 400 MPa   Size of E 70 electrodes = 4 mm Ultimate tensile strength of E 70 electrodes = 480; Fw = 0.6(480)   Questions : a) Determine the design factored tensile force. b) Determine the length of longitudinal welds "L". c) Determine the block shear strength of the gusset plate.
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