Question 3 of 8: Mohr's Circle for a Uniaxial-Stress Welded Plate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Basic Studies, 04-BS-6 Mechanics of Materials, 2018-Dec. Closed book; one hand-written aid sheet permitted. The exam instructs "any FIVE of the eight questions constitute a complete paper," but every question is solved below as a full study resource.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (axial deformation & statically indeterminate assemblies ch.4; shear/moment diagrams ch.6; stress transformation & Mohr's circle ch.9; buckling of columns ch.13; torsion of circular shafts ch.5; transformed-section composite beams ch.6; shear flow/VQ/Ib ch.7; beam deflection by integration ch.9). Standard steel W-shape tables (Appendix C of the exam) per CISC Handbook of Steel Construction.
Question 3: Mohr's Circle for a Uniaxial-Stress Welded Plate (20 marks)
Given. Uniaxial tension $\sigma_x=100\text{ MPa}$ along the horizontal axis ($\sigma_y=0$, $\tau_{xy}=0$). The weld line runs at $\beta=40^{\circ}$ to the horizontal (bottom) edge of the plate.
Find. Part (a): normal stress $\sigma_n$ and shear stress $\tau$ on the weld plane. Part (b): the maximum in-plane shear stress and its associated normal stress. Part (c): an alternative solution method.
Mohr's circle for sigma_x = 100 MPa uniaxial stress; the weld-plane point sits at 2(theta)=100 deg from X.
Approach. Plot the uniaxial stress state as point X on Mohr's circle, find the circle's centre and radius from the given stress, then rotate by twice the physical angle to the weld plane's normal to read off its coordinates; the topmost point of the same circle gives the maximum in-plane shear directly.
Plot the circle. With $\sigma_y=\tau_{xy}=0$, point $X=(\sigma_x,0)=(100,0)$ and point $Y=(0,0)$ are diametrically opposite, so centre $C=(\sigma_x+\sigma_y)/2=\boxed{50\text{ MPa}}$ and radius $R=(\sigma_x-\sigma_y)/2=\boxed{50\text{ MPa}}$.
Transform-angle for the weld plane. The weld LINE sits at $\beta=40^{\circ}$ to the horizontal ($x$) axis, so its OUTWARD NORMAL (the direction that defines the cut plane for stress transformation) sits at $\theta=90^{\circ}-\beta=50^{\circ}$ from the $x$-axis (checked against the two limiting cases: a weld PARALLEL to the load, $\beta=0$, carries zero stress since $\theta=90^{\circ}$ puts its normal along $y$; a weld PERPENDICULAR to the load, $\beta=90^{\circ}$, carries the full $100\text{ MPa}$ since $\theta=0^{\circ}$ aligns its normal with $x$ — both check out against this formula).
Part (a) — rotate the circle by $2\theta$. On the circle, the weld-plane point sits at angle $2\theta=100^{\circ}$ from X: $\sigma_n=C+R\cos(2\theta)=50+50\cos100^{\circ}=\boxed{41.32\text{ MPa (tension)}}$; $\tau=R\sin(2\theta)=50\sin100^{\circ}=\boxed{49.24\text{ MPa}}$, oriented to shear the weld consistent with the rotation sense from X.
Part (b) — maximum in-plane shear. The topmost point of the circle gives $\tau_{max}=R=\boxed{50\text{ MPa}}$, with associated normal stress equal to the circle centre on BOTH faces of that element, $\sigma_{avg}=\boxed{50\text{ MPa}}$, occurring on planes at $45^{\circ}$ to the loading ($x$) axis (i.e. $2\theta=90^{\circ}$ from X).
Part (c) — alternative method. The identical result follows directly from the stress-transformation equations $\sigma_{n}=\sigma_x\cos^2\theta$, $\tau=\sigma_x\sin\theta\cos\theta$ (equivalent to $\sigma_x\sin\beta\cos\beta$ using the weld angle $\beta$ directly) — the exam explicitly restricts credit to the Mohr's circle construction, but the transformation equations remain a valid independent check, which is exactly how the boxed numbers above were verified.
