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Material Type: Assignment; Class: ELEMENTS OF MECH DESIGN; Subject: Mech, Aero, Nucl Engr; University: Rensselaer Polytechnic Institute; Term: Spring 2011;
Typology: Assignments
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4 /25/11: Welds
A steel side plate is to be welded to a vertical steel column according to the specifications in the figure (dimensions in mm). In this problem, you are to determine the required weld size if an E 60 welding electrode is to be used, and a safety factor of 2.5 guarding against yield is required.
(a) Think about what types of forces the welds see due to the point force and what stresses those forces will cause (direct shear, torsion, or bending?). Where do you think the critical location is? Is there just one location or more than one location that should be checked?
(b) Compute the stress due to the direct shear (force over total weld area) in terms of the weld size. This stress is assumed to be equal over the entire weld.
(c) There should also be a stress due to torsion. From the figure, think about where the welds are located. Imagine them as two long, thin rectangles, and find the centroid of the combined weld area (you can use the weld table handout for this). Also compute the unit second polar moment of area (see handout).
(d) Compute the moment (torque) due to the force about that centroid. Using that torque, the unit second polar moment, and the weld centroid location, compute the shear stress due to torsion at the critical location(s) in terms of the weld size. Also determine the direction(s) by assuming that the direction is perpendicular to the line from the centroid to the critical location(s).
(e) Using vector addition, add the shear stresses at the critical location(s) and determine the maximum shearing stress in terms of the weld size.
(f) Using the safety factor and the value of the weld yield strength from Table 13.13, find the required weld size.
April 25, 2011
(a) Direct shear and torsion leading to shear stress. The welds are along A-B and B-C. The furthest points from the weld centroid are points A and C, so these are where the torsion induced shear stress will be highest. At both points, the direct shear stress acts straight down. Based on the direction of the torsion, at A, the torsion induced shear stress will act down and to the right, and at C, it will act up and to the left. While C is further from the weld centroid and will thus have the higher torsion induced shear stress, the angle between the torsion induced shear stress and direct shear stress is smaller at A than at C (both have downward components that combine). So it is not possible to determine in advance which location will have the higher overall shear stress. (b) The weld length is Lw = 250mm.
ฯw = (^) tLP w
= (^) t^20 (250,^000 mm^ N) =^80 N/mmt
(c) Using the weld table for properties of fillet welds (Table 13.14), we can find that the centroid of the weld is located 20 mm to the right and 45 mm down from point B. The unit second polar moment of area can also be computed (noting that in this case b = 100 mm and d = 150 mm)
Ju = (b^ +^ d)
(^4) โ 6 b (^2) d 2 12(b + d) =
(250 mm)^4 โ 6(100 mm)^2 (150 mm)^2 12(250 mm) = 8.^521 ร^10
(^5) mm 3
(d) T = (20, 000 N )(280 mm) = 5. 6 ร 106 N-mm
The torsional shear stresses at A and C are:
ฯ (^) tA = T rA Jut (
rA^ i^ โ^
rA^ j) =
(5. 6 ร 106 N-mm)(91. 79 mm) (8. 521 ร 105 mm^3 )t (0.^4903 i^ โ^0.^8716 j) =
t (295.^8 i^ โ^525.^8 j^ N/mm) ฯ (^) tC = T rC Jut
rC i +^20 rC j) = (5.^6 ร^10
(^6) N-mm)(106. 9 mm) (8. 521 ร 105 mm^3 )t (โ 0. 9823 i + 0. 1871 j)
= (^1) t (โ 690. 1 i + 131. 4 j N/mm)