试验方法: 卡毕( Charpy)型 伊佐德(Lzod)型 原理: 卡毕型 伊佐德型 摆锤损失的能量就是材料 冲击强度(IS)的度量。 通常把抗试样冲强度引述 为断裂能量/断裂面积, 试样 量纲KJ/m2。 W=Wol(l-cos a)-Wol(1-coS B) △E=mgh1mgh2
试验方法： 卡毕（Charpy）型 伊佐德（Izod）型 原理： 摆锤损失的能量就是材料 冲击强度（IS）的度量。 通常把抗试样冲强度引述 为断裂能量 /断裂面积， 量纲 KJ /m2 。 β α l W0 W=W0 l(1-cos α)- W0 l(1-cos β ) ΔE=mgh1 -mgh2
表4-1-5一些常见聚合物缺口Ld冲击强度(24C) 材料名称 冲击强度 材料名称 ×10-2KJ/m2 KJ/ 聚苯乙烯 1.3~2.1 聚丙烯 2.65~10.6 冲击强度 ABS 5.8~63 聚碳酸酯 63~689 硬聚氯七烯 15.9 酚醛塑料(普通 1.3~19 聚氯乙烯共聚 15.9~106 酚醛塑料(布填料 5.3~15.9 物 PMM 2.1~26 酚醛塑料(玻璃纤 维填料) 5.3~15.9 醋酸纤维素 5.3~297 聚四氟乙烯 10.6~21.2 乙基纤维素 18.5~31.8 聚苯醚 26.5 尼龙-66 5.3~159 聚苯醚(25%玻璃 纤维) 74~76 尼龙-6 53~15.9 聚砜 68~26.5 聚甲醛 10.6~15.9 环氧树脂 1.0~26.2 低密度聚乙烯 >84.8 环氧树脂(玻璃纤 维填料) 53~159 高密度聚乙烯 2.65~10.6 聚酰亚胺
表4-1-5一些常见聚合物缺口Izod冲击强度（24C） 材料名称 材料名称 冲击强度 冲击强度 10 - 2 KJ/m2 10 - 2 KJ/m2 聚苯乙烯 1.3 ~ 2.1 聚丙烯 2.65 ~10.6 ABS 5.8 ~ 63 聚碳酸酯 63 ~ 68.9 硬聚氯乙烯 2.1 ~ 15.9 酚醛塑料（普通） 1.3 ~ 1.9 聚氯乙烯共聚 物 15.9 ~ 106 酚醛塑料（布填料 ） 5.3 ~ 15.9 PMMA 2.1 ~ 2.6 酚醛塑料（玻璃纤 维填料） 5.3 ~ 15.9 醋酸纤维素 5.3 ~ 29.7 聚四氟乙烯 10.6 ~ 21.2 乙基纤维素 18.5 ~ 31.8 聚苯醚 26.5 尼龙 – 66 5.3 ~ 15.9 聚苯醚（25%玻璃 纤维） 7.4 ~7.6 尼龙 – 6 5.3 ~ 15.9 聚砜 6.8 ~26.5 聚甲醛 10.6 ~ 15.9 环氧树脂 1.0 ~ 26.2 低密度聚乙烯 >84.8 环氧树脂（玻璃纤 维填料） 53 ~ 159 高密度聚乙烯 2.65 ~ 10.6 聚酰亚胺 4.7
(4)抗扭强度( (torsional strengt 材料抵抗扭曲的能力。 τb=Mh/W ≈ 扭角
（4）抗扭强度(torsional strength) 材料抵抗扭曲的能力。 b =Mb /W
EXAMPLE PROBLEM 7.3 From the tensile stress-strain behavior for the brass specimen shown in Figure7. 12. determine the following (a) The modulus of elasticity (b) The yield strength at a strain offset of 0.002 (c)The maximum load that can be sustained by a cylindrical specimen having an original diameter of 12.8mm (d) The change in length of a specimen originally 250 mm long that issubjected to a tensile stress of 345 MPa
EXAMPLE PROBLEM 7.3 From the tensile stress–strain behavior for the brass specimen shown in Figure7.12, determine the following: (a) The modulus of elasticity. (b) The yield strength at a strain offset of 0.002. (c) The maximum load that can be sustained by a cylindrical specimen having an original diameter of 12.8 mm (d) The change in length of a specimen originally 250 mm long that issubjected to a tensile stress of 345 MPa
SOLUTION (a The modulus of elasticity is the slope of the elastic or initial linear portion of the stress-strain curve E=slope U A∈ In as much as the line segment passes through the origin, it is convenient to take both o and e as zero. If o2 is arbitraril taken as 150 MPa. then E2 will have a value of 0.0016 Therefore E=_10=0NP2=938GP2(36×10p
SOLUTION (a) The modulus of elasticity is the slope of the elastic or initial linear portion of the stress–strain curve. In as much as the line segment passes through the origin, it is convenient to take both 1 and 1 as zero. If 2 is arbitrarily taken as 150 MPa, then 2 will have a value of 0.0016. Therefore