The tension member shown in Figure P7.6-6 is an L 6 × 3 1 2 × 1 2 . It is connected with 1 1 8 -inch-diameter Group A slip-critical bolts to a 3 8 -inch-thick gusset plate. It must resist a service dead load of 20 kips, a service live load of 60 kips, and a service wind load of 20 kips. The length is 9 feet and all structural steel is A36. Are the member and its connection satisfactory? a . Use LRFD. b . Use ASD.

(a)

If the members and its connections are satisfactory using LRFD.

Given:

Tension member is L6×312×12.

Steel Used is A36.

Thickness of gusset plate is 38 inch.

Diameter of the bolt is 118 inch.

Dead load is 20 kips.

Live load is 60 kips.

Wind load is 20 kips.

Length of the member is 9 ft.

Calculation:

The properties for L6×312×12 section from the AISC steel table are as follows:

The gross area is 4.5 inch2.

The distance from the plane of connection to the centroid is 0.829 inch.

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

The ultimate tensile stress is 58 ksi.

The yield strength is 36 ksi.

Write the expression for slenderness ratio.

Lrmin=Lrz     ...... (I)

Here, radius of gyration is rmin, length of the member is L, and radius of gyration about z-axis is rz.

Substitute 9 ft for L and 0.756 inch for rz in Equation (I).

Lrmin=(9 ft)(12 inch1 ft)0.756 inch=142.857≈143

The slenderness ratio is less than 300, so it is acceptable.

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

Ab=π4d2     ...... (II)

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

Substitute 1.125 inch for d in Equation (II).

Ab=π4(1.125 inch)2=0.994 inch2

Write the expression for nominal shear capacity of one bolt.

Rnv=(FnvAb)     ...... (III)

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

Substitute 0.994 inch2 for Ab and 54 ksi for Fnv in Equation (III).

Rnv=(54 ksi)(0.994 inch2)=53.68 kips

Write the expression for slip critical shear strength per bolt for class A surfaces.

Rn=μDuhfnsTb     ...... (IV)

Here, the ratio of mean actual bolt pre-tensioned to specified bolt pre-tensioned is Du, slip shear strength per bolt is Rn, filler factor is hf, number of slip planes is ns, mean slip coefficient is μ, and minimum tension force per bolt is Tb.

Substitute 1.13 for Du, 1.0 for hf, 0.3 for μ, 1 for ns and 56 kips for Tb in Equation (IV).

Rn=0.3×1.0×1×1.13×56 kips=18.98 kips

Write the expression to calculate the nominal bearing strength of an edge bolt Rn1.

Rn1=1.2lctFu     ...... (V)

Here, nominal bearing strength is Rn1, 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 edge bolt,

Rn1=2.4dtFu

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

lc=le−h2     ...... (VI)

Here, height of the bolt is h.

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

lc=s−h     ...... (VII)

Here, spacing between the bolts is s.

Write the expression to calculate the height of the bolt.

h=d+116 inch     ...... (VIII)

Calculate the value of h.

Substitute 98 inch for d Equation (VIII).

h=98 inch+116 inch=1.188 inch

Calculate the value of lc for edge bolts.

Substitute 2 inch for le and 1.188 inch for h in Equation (VI).

lc=2 inch−1.188 inch2=1.406 inch

Calculate the nominal bearing strength for edge bolts.

Substitute 1.406 inch for lc, 38 inch for t and 58 ksi for Fu in Equation (V).

Rn1=1.2(1.406 inch)(38 inch)(58 ksi)=36.70 kips

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

Substitute 1.125 inch for d, 58 ksi for Fu and 38 inch for t.

Rn1=2.4(1.125 inch)(38 inch)(58 ksi)=58.73 kips

Thus, the minimum value of Rn1 is considered and is equal to 36.70 kips.

Write the expression to calculate the nominal bearing strength of inner bolt Rn2.

Rn2=1.2lctFu     ...... (IX)

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

Rn1=2.4dtFu

Calculate the value of lc for inner bolts.

Substitute 3.5 inch for s and 1.188 inch for h in Equation (VII).

lc=3.5 inch−1.188 inch=2.312 inch

Calculate the nominal bearing strength for inner bolts.

Substitute 2.312 inch for lc, 38 inch for t, 1.125 inch for d, and 58 ksi for Fu in Equation (IX).

Rn2=1.2(2.312 inch)(38 inch)(58 ksi)=60.34 kips

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

Substitute 1.125 inch for d, 58 ksi for Fu and 38 inch for t.

Rn2=2.4(1.125 inch)(38 inch)(58 ksi)=58.73 kips

Thus, the minimum value of Rn2 is considered and is equal to 58.73 kips.

Write the expression for total strength of connection.

Rnt=6×Rn     ...... (X)

Substitute 18.98 kips for Rn in Equation (X).

Rnt=6×18.98 kips=113.9 kips

Write the expression to calculate the nominal strength in yielding.

Pn=FyAg     ...... (XI)

Here, yield strength is Fy, nominal strength is Pn and gross area of the section is Ag.

Substitute 4.5 inch2 for Ag and 36 ksi for Fy in Equation (XI).

Pn=36 ksi×4.5 inch2=162 kips

Write the expression to calculate the diameter of the hole.

dh=db+Sc     ...... (XI)

Here, diameter of the hole is dh, side clearance is Sc, and diameter of the bolt is db.

Substitute 98 inch for db and 18 inch for Sc in Equation (XI).

dh=(98 inch+18 inch)=1

(b)

If the members and its connections are satisfactory using ASD.