A double-angle Shape, 2 L 6 × 6 × 5 8 , is connected to a 5 8 -inch gusset plate as shown in Figure P7.6-l. Determine the maximum total service lead that can be applied if the ratio of dead load to live load is 8.5. The bolts are 7 8 -inch-diameter, Group A bearing-type bolts. A572 Grade 50 steel is used for the angle, and A36 steel is used for the gusset plate. a . Use LRFD. b . Use ASD.

(a)

The maximum total load by using Load and Resistance Factor Design (LRFD) method.

Given:

Double angle section is 2L6×6×58.

Thickness of the gusset plate is 58 inch.

Ratio of the dead load to live load is 8.5.

Diameter of the bolts is 78 inch.

Steel used for angle is A572, Grade 50.

Steel Used for gusset plate is A36.

Calculation:

The properties for 2L6×6×58 section from the AISC steel table are as follows:

The gross area is 7.13 inch2.

The distance from the plane of connection to the centroid is 1.72 inch.

The properties for A36 steel from the AISC steel table are as follows:

The ultimate tensile stress is 58 ksi.

Write the expression for cross-sectional area of the bolt.

Ab=π4d2     ...... (I)

Here, cross-sectional area of the bolt is Ab and diameter of the bolt is d.

Substitute 78 inch for d in Equation (I).

Ab=π4(78 inch)2=0.601 inch2

Write the expression for nominal shear capacity one bolt.

Rnv=FnvAb     ...... (II)

Here, nominal shear stress of one bolt is Rnv and nominal shear stress of the bolt material is Fnv.

Substitute 0.601 inch2 for Ab and 54 ksi for Fnv in Equation (II).

Rnv=(54 ksi)(0.601 inch2)×2=64.94 kips

Write the expression to calculate the strength.

S=n×Rnv     ...... (III)

Here, number of bolts is n and strength is S.

Substitute 8 for n and 64.94 kips for Rnv in Equation (III).

S=8×64.94 kips=519.52 kips

Write the expression to calculate the nominal bearing strength of a bolt Rn.

Rn=1.2lctFu     ...... (IV)

Here, nominal bearing strength is Rn, thickness of the gusset plate is t, ultimate tensile stress is Fu, and the distance between the edge of the adjacent hole to the edge of the bolt hole is lc.

With the upper limit of nominal bearing strength of the bolt,

Rn=2.4dtFu

Write the expression to calculate the value of lc for the edge bolts.

lc=le−h2     ...... (V)

Here, height of the bolt is h.

Write the expression to calculate the value of lc for inner bolts.

lc=s−h     ...... (VI)

Here, spacing between the bolts is s.

Write the expression to calculate the height of the bolt.

h=d+116 inch     ...... (VII)

Calculate the value of h.

Substitute 78 inch for d Equation (VII).

h=78 inch+116 inch=1516 inch

Calculate the value of lc for edge bolts.

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

lc=2 inch−(1516 inch)2=1.531 inch

Calculate the nominal bearing strength for edge bolts.

Substitute 1.531 inch for lc, 58 inch for t, and 58 ksi for Fu in Equation (IV).

Rn=1.2(1.531 inch)(58 inch)(58 ksi)=66.6 kips

Calculate the upper limit of nominal bearing strength for edge bolts.

Substitute 78 inch for d, 58 ksi for Fu, and 58 inch for t.

Rn=2.4(78 inch)(58 inch)(58 ksi)=76.13 kips

Thus, the minimum value of Rn is considered and is equal to 66.6 kips.

Calculate the value of lc for inner bolts.

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

lc=3 inch−1516 inch=2.063 inch

Calculate the nominal bearing strength for inner bolts.

Substitute 2.063 inch for lc, 58 inch for t, and 58 ksi for Fu in Equation (IV).

Rn=1.2(2.063 inch)(58 inch)(58 ksi)=89.74 kips

Calculate the upper limit of nominal bearing strength for inner bolts.

Substitute 78 inch for d, 58 ksi for Fu and 58 inch for t.

Rn=2.4(78 inch)(58 inch)(58 ksi)=76.13 kips

Thus, the minimum value of Rn is considered and is equal to 76.13 kips.

Write the expression to calculate the nominal strength in yielding.

Pn=FyAg    

(b)

The maximum total allowable load by using Allowable Stress Design (ASD) method.