Bending Stress p375. Bending-Stress Page 1
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1 Bending Stress p375 Bending-Stress Page 1
2 Bending-Stress Page 2
3 Bending-Stress Page 3 Bending Stress example Reactions: Force = 0.193*9.81 = N Measure the torque of the hinge mm Spring: Moment (Torque) 260*2.943 = Nmm (0.765 Nm) 300g = 0.3*9.81 = N Predict the load on the beam...
4 Bending-Stress Page Nmm Reaction Force = 765 / 170 = 4.5 N Force at centre = 4.5 * 2 = 9 N In kg = 9/9.81 = kg Hinge = 386g Extra load = = 531 g Experiment: 18*50 = 900 g (Missed!) Double Check: Mass = *50 = 1,286 g Force = 1.286*9.81 = N Bending Moment = 0.5* *170 = 1072 Nmm Could it be friction? Masses were not in the centre. Double Check 2: Mass = 386+8*50 = 786 g Force = 0.786*9.81 = N Bending Moment = 0.5*7.711*485/2 = 935 Nmm Nmm Could it be friction? Masses were not in the centre.
5 Q2 Simply Supported Beam Wednesday, 4 May :48 PM Q2: If distance a=3.7m, b=2.1m and force W=5.1kN, find the maximum bending moment. Sag=(+), Hogg=(-) If we have not done MEM30005A Forces, there is a shortcut way to find the max bending moment M. From < _Bending_Moment/images/beam_simple.jpg> Engineers often use beam bending tables. In this question the same TYPE of loading must be found in the bending table. Looking for a simply supported beam where the load is offset from the centre. (Last row of table) Solution: a=3.7m, b=2.1m and force W=5.1kN, L=a+b = = 5.8 m Max BM = Wab/L = 5100*3700*2100/5800 = E6 Nmm From < Q2: Distance a=5m, b=1.6m: Dimensions c=67mm, d=190mm: Force W=28kN. Find maximum bending stress. Solution: a=5m, b=1.6m and force W=28kN, L=a+b = = 6.6 m Max M = Wab/L = 28000*5000*1600/6600 = E7 Nmm Now find Bending Stress: = My/I We need to find I; I = bh 3 /12 = 67*190^3 / 12 = E7 mm 4 From < _Bending_Stress/images/beam_simple_rectangular.jpg> Bending-Stress Page 5
6 y = 190/2 = 95 mm (half depth) = My/I = E7 * 95 / E7 = MPa Bending-Stress Page 6
7 Cantilever Wednesday, 4 May :09 PM Q9: Force a=7.3kn and force b=8.9kn. Find the maximum bending moment. Sag=(+), Hogg=(-) Looking for a cantilevered beam where there are 2 loads. Since the max BM occurs at the wall for BOTH forces, then we can add them up separately. From < _Bending_Moment/images/cant01.jpg> Solution: M = WL = a*2 + b*6 = 7300* *6000 = Nmm From < Q7: Force a=5kn and force b=3.3kn. The beam is c=106mm wide by d=223mm deep. Find the highest stress.. Solution: M = 5000* *6000 = Nmm Now find Bending Stress: = My/I Bending-Stress Page 7
8 Now find Bending Stress: = My/I We need to find I; I = bh 3 /12 = 106*223^3 / 12 = E7 mm 4 From < _stress.jpg> y = 223/2 = mm (half depth) = My/I = * / E7 = MPa Bending-Stress Page 8
9 Bicep question Wednesday, 4 May :26 PM Q12: Freddy's forearm is 378mm long, his bicep attaches 45.9mm from the elbow. Assuming a bone diameter of 34mm, what stress would 25kg cause? From < _Bending_Stress/images/bicep_stress.jpg> This is really just a simply supported beam - upside-down! Max M = Wab/L Bending-Stress Page 9
10 Bending-Stress Page 10 Bending Stress: Shaft Diameter Friday, 13 May :11 AM A classic problem is to find the diameter of a shaft. The problem is, we can t find the stress or even Ixx until we know the diameter. A typical "engineeringy" way to do this is to guess a shaft size, calculate it, then adjust size, recalculate, etc. But a simple bending problem can be solved by algebra. Solution: Set up the problem based on diameter, d. (Do it in Nmm) I = d 4 /64 y = d/2 M = a*b = 5900*110 = Nmm = My/I = M * d/(2* d 4 /64) = 64*M /(2* d 3 ) = 32*M/(pi*d 3 ) = / d 3 So now find diameter; d = ( E6 / ) (1/3) = ( E6/270)^(1/3) = mm
11 Bending-Stress Page 11 Worked example. Tuesday, 2 August :39 PM Calculate the maximum tensile stress for the beam shown below. It is a simply supported beam made of wood that has a spec gravity of 0.7. Beam span is 8m, find the stress due to it's own weight. Process: 1. Find Centroid and Ixx 2. Find volume and mass 3. Find M 4. Find Stress
12 Bending-Stress Page 12 Worked ex. 2 Tuesday, 2 August :39 PM Spec gravity of 0.7. Beam span is 8m 1. Find Centroid and Ixx Centroid: Yc = (Ay) / (A) = 6E6/3E4 = 200 mm Parallel Axis Theorem: (p372) (The I effect of each element = It's own Ic + Area * d2 from neutral plane.) I = Ic + Ad 2 mm2 mm mm3 mm mm4 mm4 mm4 Elemt Area y Ay d Ad2 Ic I Total Ixx = yc = 200
13 Bending-Stress Page 13 Worked ex. 3 Tuesday, 2 August :39 PM Spec gravity of 0.7. Beam span is 8m Yc = 200mm I = 250E6 mm4 Find volume V=A*L = 30E3*8E3 = 240E6 mm 3 Find Mass: m= *V = 700*240E6/1E9 = 168 kg Reactions = 9.81*168/2 = N Distributed load = 9.81* 168/8 = N/m N N BM max = 0.5*4*824 = 1648 Nm
14 Bending-Stress Page 14 Worked ex. 4 Tuesday, 2 August :39 PM Spec gravity of 0.7. Beam span is 8m Yc = 200mm I = 250E6 mm4 BM max = 1648 Nm = 1648*1000 = 1648E3 Nmm Find Stress = My / I = 1648E3 * 200 / 250E6 = MPa Wood is rated (bending tensile stress - fibre stress) like this... F4 soft pine F5 F7 structural pine F14 F17 hardwood (like eucalypts, not balsa) F4 = 4 MPa (rated fibre stress) Compare to steel: Steel could handle a stress of... (a) 2 MPa (b) 20 MPa (c) 200 MPa (d) 2 GPa (e) 20 Gpa
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