WORLD FOR CIVIL

Statics & Strength of Materials

Topic 6.8a - Problem Assignment 1 - Rivets/Welds

1. A riveted lap joint is shown in Diagram 1. The diameter of the rivets is 3/4 inch. The width of the plates is 8 inches, and the thickness of the plates is 1/2 inch. The allowable stresses are as follows:

Rivets: = 15, 000 lb/in², _t = 22, 000 lb/in², _c = 25, 000 lb/in²
Plate: = 14,000 lb/in², _t = 20, 000 lb/in², _c = 24, 000 lb/in²
Determine the Strength of the Joint, and the Efficiency of the Joint.
(P_s = 79,520 lb., P_b = 108,000 lb, P_r1 = 72,500 lb, P_r2 = 70,910 lb, P_r3 = 76,700 lb)
[Joint Strength = 70,910 (lowest of failure loads above); eff. = .886]

2. A riveted butt joint is shown in Diagram 2. The diameter of the rivets is 1/2 inch. The width of the plates is 6 inches, and the thickness of the plates is 3/4 inch. The allowable stresses are as follows:

Rivets: = 18, 000 lb/in², _t = 22, 000 lb/in², _c = 24, 000 lb/in²
Plate: = 17,000 lb/in², _t = 20, 000 lb/in², _c = 21, 000 lb/in²
Determine the Strength of the Joint, and the Efficiency of the Joint.
(P_s = 63,620 lb., P_b = 70,880 lb, P_r1 = 82,500 lb, P_r2 = 84,375 lb, P_r3 = 101,250 lb)
[Joint Strength = 63,620 (lowest of failure loads above); eff. = .707]

3. A lap joint (Diagram 3) is to connect two steel plates both with a width of 7 inches and a thickness of 5/8 inch. The rivets to be used have a diameter of 3/4 inch. The maximum allowable stresses for the rivet and plate materials are as follows:

Rivets: = 20, 000 lb/in², _t = 21, 000 lb/in², _c = 23, 000 lb/in²
Plate: = 17,000 lb/in², _t = 22, 000 lb/in², _c = 25, 000 lb/in²
A. Determine the number of rivets for the most efficient joint. (ans. 9.7 = 10 rivets)
B. Select the best pattern from those shown in Diagram 4. (ans. 10,11,12 pattern)
C. Calculate the strength and efficiency of the joint. (ans. P_r2 = 82,500, eff = .857)

4. A double cover plate butt joint (Diagram 5) is to connect two steel plates both with a width of 10 inches and a thickness of 5/8 inch.

The rivets to be used have a diameter of 3/4 inch. The maximum allowable stresses for the rivet and plate materials are as follows:
Rivets: = 16, 000 lb/in², _t = 22, 000 lb/in², _c = 23, 000 lb/in²
Plate: = 14,000 lb/in²,_t = 20, 000 lb/in², _c = 21, 000 lb/in²
A. Determine the number of rivets for the most efficient joint. (ans. 11.75 = 12 rivets)
B. Select the best pattern from the patterns in Diagram 6. (11-12 pattern)
C. Calculate the strength and efficiency of the joint. (ans. P_r1 = 115, 625 lb; eff = .925)

5. A lap joint is to connect two steel plates each with thickness of 3/4 inch (Diagram 7). The rivets to be used have a diameter of 1/2 inch. The maximum allowable stresses for the rivet and plate materials are as follows:

Rivets: = 15, 000 lb/in², _t = 22, 000 lb/in², _c = 27, 000 lb/in²
Plate: = 14,000 lb/in², _t = 24, 000 lb/in², _c = 26, 000 lb/in²
A. Calculate the width needed for the steel plate such that the joint will fail in bearing and plate tearing at row 1at the same load. (ans. w = 4.83")
B. Using the width found in part A, determine the strength and efficiency of the joint.
(P_s = 23,562 lb., P_b = 78,000 lb, P_r1 = 77,940 lb, P_r2 = 78,790 lb)
[Joint Strength = 23,562 (lowest of failure loads above); eff. = .27, not a very good joint]

6. A a double cover plate butt joint is to connect two steel plates each with thickness of 3/4 inch ( See Diagram 8). The rivets to be used have a diameter of 1/2 inch. The maximum allowable stresses for the rivet and plate materials are as follows:

Rivets: = 16, 000 lb/in², _t = 24, 000 lb/in², _c = 28, 000 lb/in²
Plate: = 15,000 lb/in², _t = 25, 000 lb/in², _c = 26, 000 lb/in²
A. Calculate the width needed for the steel plate such that the joint will fail in rivet shear and plate tearing at row 1 at the same load. (ans. 2.5", very small width, not very realistic.)
B. Using the width found in part A, determine the strength and efficiency of the joint.
(P_s = 37,700 lb., P_b = 58,500 lb, P_r1 = 37,500 lb, P_r2 = 33,750 lb, P_r3 = 37,500 lb)
[Joint Strength = 33,750 (lowest of failure loads above); eff. = .72]

7. Two steel plates are shown below. The top plate is 1/2 inch thick and 10 inches wide, and is to be welded to the bottom plate with a 45^o fillet weld. The top plate is to be weld completely across end AG and partially along sides AB and FG.

A.) Determine the minimum inches of weld need to carry the 120,000 lb. load and specify how many inches of weld should be placed along each side AB and FG.
B.) Determine the number of inches of weld needed to make the weld strength as great as the plate strength.

The allowable stresses are as follows;
Weld Material: = 16,000 lb/in² ; t = 25, 000 lb/in²
Plate Material: = 15,000 lb/in² ; t = 26,000 lb/in²
Dimensions: AB = GF = 20 inches; CD = 3 inches; DE = 7 inches

(ans.Part A. L_total =21.2"; L_AB = 9.85"; L_AG = 10" (given); L_GF = 1.35")
(ans.Part B. L_total = 22.98")