Topic 6.7: Rivets & Welds - Welded Joints
There are a number of types of welded joints. We will consider several common types in this discussion. One type of welded joint is the Butt Joint, where two plates are brought together with a small gap separating them, as shown in Diagram 1a.
The gap is then filled with weld material as from an arc welder, and the plates are welded together. Depending on the size and thickness of the plates, the end edges to be welded together may by sloped or vee-ed to generate a better weld. In theory, a butt weld can be made as strong or stronger than the plate, depending on the weld material used. In practice, the strength of a weld depends on how well the weld is made, whether or not there are voids or cracks in the weld. In some situations the weld may be x-rayed to determine its quality. For a welded butt joint loaded in tension as shown in Diagram 1b, the load, P, the weld can carry in tension is simply the product of the cross sectional area of the weld and the allowable tensile stress for the weld material, or:
Pweld = (w * t) * 
tension, where w and t are the width and thickness of the weld (plate).
A second type of weld which we will consider is the Lap Joint using a 45 degree fillet weld, shown in Diagrams 2, 3, and 4. In an example of this type of weld a top plate is welded to a bottom plate. The fillet weld runs from the top edge of the top plate to an equal distance horizontally outward on the top of the bottom plate, forming a triangular weld as shown in the cross sectional view in Diagram 2a. Since the two sides of the right triangle formed by the weld are equal forming a 45o right triangle, the weld is known as a 45o fillet weld. The perpendicular line bisecting the 90o angle and intersecting the hypotenuse of the 45o fillet weld is know as the throat of the weld, shown in Diagram 2a. When the weld fails, it is assumed to fail in shear across the throat region. To determine the load a weld can carry before failing we take the product of the area of the weld which fails in shear ( the throat distance times the length of weld, Diagram 2b) and the allowable shear stress for the weld material.
We can write:
Pweld = Area*
all = (throat * length) *all = (t sin 45o * L) * all, and finally
Pweld = (.707 t * L) *all, where
t = thickness of plate (and height & base of weld)
L = length of weld.all = allowable shear stress for the weld material
Welded Joint Example
Two steel plates are shown in Diagram 5. The top plate is 3/4 inch thick and 8 inches wide, and is to be welded to the bottom plate with a 45o fillet weld. The top plate is to be welded along sides AB and FG. We would first like to determine the minimum inches of weld need to carry the 80,000 lb. load and then to decide how the weld should be distributed along sides AB and FG.(There is only one way for the minimum inches of weld.) Notice also that the load is not applied symmetrically, that is, it is apply closer to one edge of the top plate than the other.
Following that, we will then determine the minimum 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: = 14,000 lb/in2 ; t = 24,000 lb/in2
Plate Material: = 15,000 lb/in2 ; t = 30,000 lb/in2
Dimensions: AB = GF = 20 inches; CD = 3 inches; DE = 5 inches
Solution:
Part 1. We first determine the minimum inches of weld needed to carry the 80,000 lb. load by using the weld formula and setting the load the weld can carry before failing to 80,000 lb..
Pweld = (.707 t * L) *all , or
80,000 lb = (.707 * 3/4" * L) * 14,000 lb/in2 = (7,424 lb./in.)L; then solving for L
L = 80,000 lb./(7,424 lb./in.) = 10.78 inches.
This is the minimum inches of weld needed to carry the load, however since the 80,000 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 6 we have shown the weld distributed with an amount LAB on side AB and an amount LGF on side GF. The lengths LAB and LGF must, of course sum to 10.78 inches ( the minimum inches of weld needed). The maximum force these weld can resist with is given by FAB = (7,424 lb./in.)* LAB , and FGF = (7,424 lb./in.)* LGF , as shown in Diagram 6.
We now apply static equilibrium conditions to the top plate:
Sum of Forces: 80,000 lb. - (7,424 lb./in.)* LAB - (7,424 lb./in.)* LGF = 0
Sum of Torque about G: -80,000 lb.* (5") + (7,424 lb./in.)* LAB * (8") = 0
(Notice that the force FGF does not produce a torque about point G, since its line of action passes through point G.)
Solving the torque equation for LAB = 6.73 inches, we can then solve for LGF since LAB and LGF sum to 10.78 inches. Therefore LGF = 10.78" - 6.73" = 4.05" This is how the weld must be distributed in order to carry the 80,000 lb. load (and satisfy static equilibrium conditions).
Part 3. The other question of interest in welded joints is exactly how much weld is needed to make the welded joint as strong as the plate itself. In this case we assume the plate is loaded in tension, and so the strength of the plate is Pplate = product of cross sectional area of plate and the allowable tensile stress for the plate material, or Pplate = (w * t ) t . We then set this equal to the strength of the riveted joint and solve for the length of weld needed.
Pweld = (.707 t * L) *all = Pplate =(w * t ) t , or
(.707 t * L) *all = (w * t ) t , and then
(.707 *3/4" * L) * 14,000 lb/in2 = (8" * 3/4" ) 30,000 lb/in2 , then solving for L
L = 24.25 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.
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