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Bolted Connections

In the previous sections, we have already discussed the failure modes based on the tension of the plate (gross area, net area, and staggered connections), as well as block shear. Here, we will discuss the bearing of the plate, which is the shear tear out at the end of a connected element, and the effects of torsion on bolts due to eccentric loadings.

From the idealized failure surface, the total nominal strength is:

Where:

• $L_c$ = clear distance, measured in the direction parallel to the applied load, from the edge of the bolt hole to the edge of the adjacent hole or to the material edge
• $t$ = thickness of the connected part
• $F_u$ = ultimate tensile stress of the connected part (not the bolt)
• $d$ = bolt diameter
• $2.4dtF_u$ = upper limit of $R_n$

For LRFD:

For ASD:

Spacing and Edge - Distance Requirements

Minimum spacing, S, and edge distance, L_e, in any direction, both in the line of force and transverse to the line of force.

Bolt shear strength

For LRFD:

For ASD:

Other bearing-strength cases from NSCP/AISC notes

Combined bolt shear and tension

Eccentric Bolted Connections (Elastic Analysis)

Consider the bolted connection acted by a force P as shown.

1. Resolve P into components P_x and P_yand construct the equivalent loading
2. Locate the center of gravity of the bolt group. For symmetrical arrangements of bolts, the center of gravity is located at the geometric center. For non=symmetrical arrangements, it can be located by using Varignon's Theorem
3. The equivalent effect is that the system of bolts is acted by direct shears P_x and P_y and a torsional moment T at the center of gravity.

Equivalent Loading

Bolt Group: Direct Shear + Torsion (Procedure)

5. For direct shear loads, distribute equally among the bolts:
   $$ P_{Dx} = \frac{P_x}{n}, \qquad P_{Dy} = \frac{P_y}{n} $$
   where $n$ = number of bolts.
6. Compute the torsional moment about the bolt group centroid:
   $$ T = P_x e_y + P_y e_x $$
7. The most stressed bolt is where the forces due to direct shear and torsion are of similar directions. Forces due to direct shear have the same directions as the components of the applied force. For the forces due to torsion, draw lines (indicated as green lines) from the c.g. going to the center of each bolt. Next, draw forces perpendicular to such line (indicated as purple lines) that cause a rotation opposite to the torsional moment.
   
   
   Since the torsional moment is clockwise, the forces, P_T1, P_T2, and P_T3 must cause a counterclockwise rotation.

8. Consider bolt 1. The torsion-induced load on bolt 1 is $P_T$, resolved into components:
   $$ P_{Tx1} = \frac{T\,y_1}{\sum (x^2 + y^2)}, \qquad P_{Ty1} = \frac{T\,x_1}{\sum (x^2 + y^2)} $$
   Define the polar moment of the bolt group:
   $$ J = \sum (x^2 + y^2) $$
9. Select the bolt with the maximum combined shear (direct + torsion). The total resultant on that bolt is:
   $$ P = \sqrt{\left(P_{Tx} + P_{Dx}\right)^2 + \left(P_{Ty} + P_{Dy}\right)^2} $$

Problem: Eccentrically Loaded Bolted Connection

For the bolted connection shown below, use the following data:

• a = 80 mm
• b = 220 mm
• c = 80 mm
• P = 240 kN
• Bolt diameter = 16 mm

1. Which of the following most nearly gives the direct shear stress of each bolt in megapascals?
   1. 136.30
   2. 149.20
   3. 178.80
   4. 189.10
2. Which of the following most nearly gives the reaction of bolt #8 in kN?
   1. 133
   2. 110
   3. 122
   4. 105
3. When the reaction of the most stressed bolt is 180 kN, compute the required diameter of the most stressed bolt if its allowable shear stress is limited to 300 MPa.
   1. 36
   2. 32
   3. 25
   4. 28

Use direct shear plus torsional shear about the centroid of the bolt group.

$$ J=\sum x^2+\sum y^2 $$ $$ J=8(80)^2+4(40)^2+4(120)^2=115200\text{ mm}^2 $$

Direct shear per bolt:

$$ P_{Dy}=\frac{240}{8}=30\text{ kN} $$ $$ \tau_D=\frac{240(1000)}{8\left(\frac{\pi}{4}\right)(16)^2}=149.208\text{ MPa} $$ $$ \tau_D\approx 149.20\text{ MPa} $$

The eccentric load produces a clockwise torsional moment about the bolt-group centroid. The torsional force at each bolt is taken perpendicular to the straight line from the centroid to that bolt, and its direction is chosen so the bolt group develops a resisting rotation opposite to the rotation caused by the load.

For bolt #8, the eccentricity is 300 mm, so the torsional moment is:

$$ T=240(1000)(300)=72000000\text{ N-mm} $$ $$ P_{Tx}=\frac{T(120)}{J}=75000\text{ N} $$ $$ P_{Ty}=\frac{T(80)}{J}=50000\text{ N} $$ $$ R_8=\frac{\sqrt{P_{Tx}^2+(P_{Dy}+P_{Ty})^2}}{1000} $$ $$ R_8=\frac{\sqrt{75000^2+(30000+50000)^2}}{1000}=109.659\text{ kN} $$ $$ R_8\approx 110\text{ kN} $$

Required diameter when the most stressed bolt carries 180 kN:

$$ 300=\frac{180(1000)}{\frac{\pi}{4}d^2} $$ $$ d=\sqrt{\frac{4(180)(1000)}{300\pi}}=27.64\text{ mm} $$ $$ d\approx 28\text{ mm} $$

Answers: a. B. 149.20; b. B. 110; c. D. 28.

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