   # 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. ### Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740

#### Solutions

Chapter
Section ### Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
Chapter 7, Problem 7.6.6P
Textbook Problem
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## 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. To determine

(a)

If the members and its connections are satisfactory using LRFD.

### Explanation of Solution

Given:

Tension member is L6×312×12.

Steel Used is A36.

Thickness of gusset plate is 38inch.

Diameter of the bolt is 118inch.

Length of the member is 9ft.

Calculation:

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

The gross area is 4.5inch2.

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

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

The ultimate tensile stress is 58ksi.

The yield strength is 36ksi.

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

Lrmin=(9ft)(12inch1ft)0.756inch=142.857143

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 1.125inch for d in Equation (II).

Ab=π4(1.125inch)2=0.994inch2

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

Rnv=(54ksi)(0.994inch2)=53.68kips

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 56kips for Tb in Equation (IV).

Rn=0.3×1.0×1×1.13×56kips=18.98kips

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=leh2     ...... (VI)

Here, height of the bolt is h.

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

lc=sh     ...... (VII)

Here, spacing between the bolts is s.

Write the expression to calculate the height of the bolt.

h=d+116inch     ...... (VIII)

Calculate the value of h.

Substitute 98inch for d Equation (VIII).

h=98inch+116inch=1.188inch

Calculate the value of lc for edge bolts.

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

lc=2inch1.188inch2=1.406inch

Calculate the nominal bearing strength for edge bolts.

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

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

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

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

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

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

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

lc=3.5inch1.188inch=2.312inch

Calculate the nominal bearing strength for inner bolts.

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

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

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

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

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

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

Write the expression for total strength of connection.

Rnt=6×Rn     ...... (X)

Substitute 18.98kips for Rn in Equation (X).

Rnt=6×18.98kips=113.9kips

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

Pn=36ksi×4.5inch2=162kips

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

dh=(98inch+18inch)=1

To determine

(b)

If the members and its connections are satisfactory using ASD.

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