### Problem Statement:Given:- Allowable stress, \(\sigma_{allow} = 160 \, \text{MPa}\)Tasks:a) Select the most economical metric wide-flange shape that should be used to support the loads shown.b) Calculate the Von Mises equivalent stress at point \(Q\) (top of beam’s cross-section).c) Calculate the Von Mises equivalent stress at point \(P\) (junction between flange and web).d) Determine if points \(P\) and \(Q\) are safe based on \(\sigma_{allow}\).### Diagram Explanation:#### Beam Setup:- The beam is horizontally positioned.- Supports are indicated at points \(A\) and \(C\).- There is a concentrated load of \(40 \, \text{kN}\) acting vertically downward at the midpoint between \(A\) and \(C\).- A distributed load of \(2.2 \, \text{kN/m}\) extends from \(A\) to \(C\).#### Dimensional Details:- Distance from \(A\) to \(B\) is \(4.5 \, \text{m}\).- Distance from \(B\) to \(C\) is \(2.7 \, \text{m}\).#### Cross-Section Details:- The cross-section is depicted as a wide-flange shape (typical metric I-beam).- Point \(Q\) is at the top of the beam’s cross-section.- Point \(P\) is at the junction between the flange and the web.### Analysis Requirement:Acquire the necessary stress calculations and ensure design safety by comparing equivalent stresses at key structural points against the given allowable stress.

Data given: σallow = 160 MPa Asked: Select most economical metric-wide flange shape Calculate Von Mises equivalent stress at point Q Calculate Von Mises equivalent stress at point P Are points P and Q safe based pn the allowable stressSupport reactions at A and C can be calculated as, RA = wL2 + PbL = 2.2 × 7.2 2 + 40 × 2.77.2 = 22.92 kNSimilarly,RC = wL2 + PaL = 2.2 × 7.2 2 + 40 × 4.57.2 = 32.92 kN Shear force diagram,As we know that, maximum bending moment occur when the value of shear force is zero, As shown above the value of shear force = 0 at point B. Ergo, the maximum bending moment will occur at point B, ∴ MB = 32.92 × 2.7 - 2.2 × 2.722∴ MB = 80.865 kN-mConsider sagging moment as positive and hogging moment as negativeAs we know that, maximum bending stress, σmax = MZ Here, σmax = 160 MPaM = 80.865 kN-m ∴ Z = Mσmax ∴ Z = 80.865 × 106160∴ Z =505406.25 mm3 = 505.406 cm3 Let's consider a wide flange section with Z > 505.406 cm3. Therefore, let's consider, W 200 x 200 x 52 with Z = 512 cm3 As per Von Mises stress formula, σV = 0.5 σX - σY2 + σY - σZ2 + σZ - σX2 + 6τXY2 + τYZ2 +τZX2 0.5 Now, at Q σX = MBZ = 80.865 × 106512000 = 157.94 MPa However, σY = σZ = τXY = τYZ = τZX =0 ∴ σV = 0.5 (157.94)2 + (-157.94)2 0.5∴ σV = 157.94 MPa < 160 MPa.... Hence, OkayStress at P can be obtained by, σX = MmaxINA × y INA = Moment of inetia = 5268 × 104 mm4 Y = h2 - tf = 2062 - 7.9 = 95.1 σX = 80.865 × 1065268 × 104 × 95.1 = 145.98 MPa < 160 MPa..... Hence, Okay Thus, with the section W200 x 200 x 52 both the points P and Q are safe.