Example3-1 An aluminum rod is to withstand an applied force of F=200kN (0.2MN).To assure a sufficient safety, the maximum allowable stress on the road is limited to 170MPa.The rod is at least 3.8m long,but deformed elastically no more than 6mm.The modulus of elasticity of aluminum alloy is 69GPa. Solution 6 Strain: =0.00158 3800 Stress: o=Ee=69.,000×0.00158≈109MPa PDF文件使用"pdfFactory Pro”试用版本创建ww,fineprint.com,cn
Example3-1 An aluminum rod is to withstand an applied force of F=200kN (0.2MN). To assure a sufficient safety, the maximum allowable stress on the road is limited to 170MPa. The rod is at least 3.8m long, but deformed elastically no more than 6mm. The modulus of elasticity of aluminum alloy is 69GPa. Solution Strain： 0.00158 3800 6 e = = Stress： s = Ee = 69,000 ´ 0.00158 » 109 MPa PDF 文件使用 "pdfFactory Pro" 试用版本创建 ÿwww.fineprint.com.cn
Example3-1 An aluminum rod is to withstand an applied force of F=200kN(0.2MN).To assure a sufficient safety, the maximum allowable stress on the road is limited to 170MPa.The rod is at least 3.8m long,but deformed elastically no more than 6mm.The modulus of elasticity of aluminum alloy is 69GPa. Cross-sectional area: 0.2MN 0.00182m2=1820mm2 109MN/m2 Diameter: πd2 =1820mm d 48mm PDF文件使用"pdfFactory Pro" 试用版本创建ww.fineprint.con.cn
Example3-1 An aluminum rod is to withstand an applied force of F=200kN (0.2MN).To assure a sufficient safety, the maximum allowable stress on the road is limited to 170MPa. The rod is at least 3.8m long, but deformed elastically no more than 6mm. The modulus of elasticity of aluminum alloy is 69GPa. Cross-sectional area： 2 2 0 2 0.00182 1820 109 / 0.2 m mm MN m F MN A = = = = s Diameter： 2 2 0 1820 48 4 d A mm d mm p = = = PDF 文件使用 "pdfFactory Pro" 试用版本创建 ÿwww.fineprint.com.cn
3.2.2 Shear strength Shear stress. Shear strain:Y= A H Shear modulus:G=t/y PDF文件使用"pdfFactory Pro'”试用版本创建，fineprint.com,cn
Shear stress： A0 F t = Shear strain： H x g = Shear modulus：G=t/g 3.2.2 Shear strength A0 H F F x PDF 文件使用 "pdfFactory Pro" 试用版本创建 êwww.fineprint.com.cn ê
3.2.3 Flexural or bending strength 200 rupture 150 100 50 0.1250.2503750.5 Offset (mm) PDF文件使用"pdfFactory Pro”试用版本创建w.fineprint.com.cn
3.2.3 Flexural or bending strength S t r e s s ( s ) M P a 0.125 0.25 0.375 0.5 200 150 100 50 Offset (mm) F F d w h L rupture PDF 文件使用 "pdfFactory Pro" 试用版本创建 êwww.fineprint.com.cn
h 3FL Flexural strength 2wh CF Flexural modulus 4whδ PDF文件使用"pdfFactory Pro'”试用版本创建，fineprint.com,cn
2 2 3 wh FL Flexural strength ＝ Flexural modulus ＝ d 3 3 4wh L F F F d w h L PDF 文件使用 "pdfFactory Pro" 试用版本创建 êwww.fineprint.com.cn ê