WORLD FOR CIVIL

Topic 6.7a: Welded Joints - Example 1

Two steel plates are shown in Diagram 1. The top plate is .5 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 welded completely across end AG and partially along sides AB and FG.

A.) Determine the minimum inches of weld need to carry the 90,000 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: = 15,000 lb/in² ; t = 24, 000 lb/in²
Plate Material: = 14,000 lb/in² ; t = 28,000 lb/in²
Dimensions: AB = GF = 20 inches; CD = 4 inches; DE = 6 inches

Solution:
Part 1. We first determine the minimum inches of weld needed to carry the 90,000 lb. load by using the weld formula and setting the load the weld can carry before failing to 90,000 lb..
P_weld = (.707 t * L) *_all , or
90,000 lb = (.707 * .5" * L) * 15,000 lb/in² = (5,300 lb./in.)*L; then solving for L
L = 90,000 lb./(5,300 lb./in.) = 16.98 inches.
This is the minimum inches of weld needed to carry the load, however since the 90,000 pound load is not applied symmetrically to the plate, we can not apply the weld symmetrically. That is, if we try to put equal amounts of weld on sides AB and GF, the weld will fail in this case. To determine how the weld should be distributed, we once again apply static equilibrium conditions.

Part 2. In Diagram 2 we have shown the weld distributed with an amount L_AB on side AB, L_GF on side GF, and 10 inches of weld completely across the end AG. The lengths L_AB , L_GF , and the 10 inches of weld must sum to 16.98 inches ( the minimum inches of weld needed). The maximum force these weld can resist with is given by F_AB = (5,300 lb./in.)* L_AB , F_GF = (5,300 lb./in.)* L_GF , and F_AG = 5,300 lb/in * 10" = 53,000 lb. as shown in Diagram 2.

We now apply static equilibrium conditions to the top plate:
Sum of Forces: 90,000 lb. - (5,300 lb./in.)* L_AB - (5,300 lb./in.)* L_GF - 53,000 lb. = 0
Sum of Torque _{about G}: -90,000 lb.* (6") + (5,300 lb./in.)* L_AB * (10") + 53,000 lb. * 5" = 0
(Notice that the force F_GF does not produce a torque about point G, since its line of action passes through point G.)
Solving the torque equation for L_AB = 5.19 inches, we can then solve for L_GF since L_AB + L_GF + 10" must sum to 16.98 inches. Therefore L_GF = 16.98" - 5.19" - 10" = 1.79" This is how the weld must be distributed in order to carry the 80,000 lb. load (and satisfy static equilibrium conditions).

Part 3.. We assume the plate is loaded in tension, and so the strength of the plate is P_plate = product of cross sectional area of plate and the allowable tensile stress for the plate material, or P_plate = (w * t ) t . We then set this equal to the strength of the riveted joint and solve for the length of weld needed.
P_weld = (.707 t * L) *_all = P_plate =(w * t ) t , or
(.707 t * L) *_all = (w * t ) t , and then
(.707 *.5" * L) * 15,000 lb/in² = (10" * .5" ) 28,000 lb/in² , then solving for L
L = 26.4 inches
This is the minimum inches of weld needed to make the weld as strong as the plate (in tension). We can not state how the weld should be distributed, since we are not dealing with a specific loading.