This set of Machine Design quiz focuses on “Unsymmetrically Loaded Welded Joints”.

1. The two welds shown in the figure experience equal resisting force?

a) True
b) False

Explanation: As the welded joints are unsymmetrically loaded hence resisting forces are different in the two welds.

2. The moment of forces about the CG is?

a) Zero
b) Infinite
c) P₁y₁+P₂y₂
d) None of the listed

Explanation: External force passes through CG hence moment is zero.

3. Which equations can be used to find the resisting force in the two welds?

a) P=P₁+P₂
b) P₁+P₂=0
c) Both P=P₁+P₂ and P₁y₁=P₂y₂
d) P₁y₁=P₂y₂

Explanation: One equation is of equilibrium and the other conservation of moment about CG.

4. For the following welded joint l₁y₁=l₂y₂.

a) True
b) False

Explanation: P₁y₁=P₂y₂ and P₁=0.707hl₁τ and P₂=0.707hl₂τ.

5. Find the total length of weld required to with stand the load of 100kN. Permissible shear stress in weld is 70kN/mm².

a) None of the listed
b) 101.03mm
c) 202.06mm
d) 30309mm

Explanation: 100000=0.707x10xlx70.

6. Find the length of weld 2 required to with stand the load of 100kN. Permissible shear stress in weld is 70kN/mm².

a) None of the listed
b) 102.22mm
c) 132.4mm
d) 70.16mm

Explanation: By finding CG y₂=103.9mm, therefore l₁y₁=l₂y₂ and l₁+l₂=202.06mm.
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7. Find the length of weld 1 required to with stand the load of 100kN. Permissible shear stress in weld is 70kN/mm².

a) None of the listed
b) 102.22mm
c) 132.4mm
d) 70.16mm

Explanation: By finding CG y₂=103.9mm, therefore l₁y₁=l₂y₂ and l₁+l₂=202.06mm.

8. The following welded joint is an example of

a) Eccentric Load in plane of welds
b) Axial load in plane of welds
c) Axial load in plane perpendicular to plane of welds
d) None of the mentioned

Explanation: The figure clearly depicts.

9. What kind of stresses does the welded joint undergo?

a) Torsional shear stress
b) Direct shear stress
c) Direct and torsional shear stress
d) None of the listed

Explanation: After shifting force to CG, we have a force and a moment about CG.

10. While considering moment of inertia for calculating torsional shear stress, J=I(xx) + I(yy), which of the following can be neglected in context with the following figure?

a) I(xx)
b) I(yy)
c) Both I(xx) and I(yy)
d) None of the listed

Explanation: I(xx) is negligible as compared to I(yy) as I(xx)=ltᵌ/12 and t is very less as compared to l so I(xx) is neglected in comparison to I(yy).

11. Calculate the direct shear stress in the welds by ignoring torsional shear stress if P=8kN and thickness of weld is 5mm.

a) None of the listed
b) 12.33N/mm
c) 13.33N/mm
d) 14.33N/mm

Explanation: Direct Shear=P/2xtx60.

12. Determine the torsional shear stress in the welds if P=8kN and thickness of the welds is assumed t.

a) None of the listed
b) 456.34/t
c) 543.13/t
d) 589.31/t

Explanation: M=8000×150,r(farthest point)=√25²+25²,J=2x[tlᵌ/12 + Ar₁²] ; τ=Mr/J.

13. Find the thickness of the weld if P=8kN and permissible shear stress in the welds is 100N/mm².

a) 4mm
b) 5mm
c) 6mm
d) 7mm

Explanation: Direct shear stress and torsional shear stress act at an angle of 45’,135’,225’ and 315’. Maximum stress will be at the point where they act at 45’. Direct=8000/120t or 66.67/t and torsional =543.13/t. Hence net shear=592.1/t =100.

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