Show all work 6.18 For the beam and loading shown, determine the minimum requiredwidth b, knowing that for the grade of timber used, oall = 12 MPaand Tall = 825 kPa.2.4 kN4.8 kN7.2 kNBDA150 mm1 m1m1 m0.5 mFig. P6.18

To obtain the length of side b, we must first calculate the maximum bending moment and maximum shear force in the beam. And then we will apply the pure bending equation to obtain the value of side b. According to pure bending equation, σy=MI where σ is the normal stress acting on the beam fiber y is the vertical distance of the fiber in observation from the neutral axis M is the maximum bending moment acting on the beam I is the area moment of inertia of the cross section of the beamSo to obtain the desired value of side b of the cross-section, we must calculate the maximum bending moment and maximum shear force in the beam. First of all calculating the support reactions at point A and Point D. Since the point A is knife edge supported, thus there must be 2 support reactions present at that point. And the point D is roller supported, thus there must be only vertical support reaction. Let us denote the vertical support reactions at point A and D to be VA and VD. And denoting the horizontal support reaction at point A to be HA. Now, taking the summation of all vertical forces to be zero: ∑Fy=0VA+VD=2.4+4.8+7.2=14.4kN.........................(1) Now, taking the summation of all horizontal forces to be zero: ∑Fx=0HA=0 kN Now, taking the summation of moments produced by all the forces about point A to be zero: ∑MA=0VD×3=2.4×1+4.8×2+7.8×3.5VD=13.1kN Thus, VA=14.4-VDVA=14.4-13.1=1.3kN It can be seen clearly that the maximum shear force in the beam is 7.2 kN. Thus, the shear stress becomes, ζallowable=32×ζavg=32×Fbd825×103=32×7.2×103b×0.150b=11.45m The Maximum bending moment occurs at at point D because the single largest load is acting very close to it. Thus, maximum bending moment be: BMmax=7.2×0.5=3.6kNm The area moment of inertia of the cross-section of the beam is: I=b×0.150312=2.8125×10-4b We will consider the topmost fiber because it will experience the maximum normal stress due to bending. Thus, the distance of topmost fiber of the beam from the neutral axis be: y=0.1502=0.075m Now, applying the pure bending equation as: σy=BMmaxI12×1060.075=3.6×1032.8125×10-4bb=80mmThus, the minimum required width of the beam is 11.45 meters.