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1. introduksi

1. introduksi. Diagram alir perancangan ( konsep dasar ). Gunakan satuan SI. 2. TEGANGAN DAN REGANGAN (KONSEP DASAR). Tegangan normal σ T egangan geser τ. Tegangan. Tegangan normal σ. Batang panjang dan luas penampang. Batang diberi gaya F sehingga bertambah panjang dengan.

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1. introduksi

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  1. 1. introduksi Diagram alirperancangan (konsepdasar)

  2. Gunakansatuan SI

  3. 2. TEGANGAN DAN REGANGAN (KONSEP DASAR) Tegangan normal σ Tegangangeserτ Tegangan Tegangan normal σ Batangpanjang dan luaspenampang BatangdiberigayaFsehinggabertambahpanjangdengan Gaya dalamdigunakanuntukmenahan agar batanghanyabertambahpanjangdengan ∆l Gaya dalamdidistribusikansecaramerata di seluruhluaspanampang ab dandiberinamategangan normal σ Tegangan normal mempunyaiorientasi normal (tegaklurus) terhadapluaspenampangnya DR.Ir Soeharsono FTI Universitas Trisakti

  4. Tegangan normal: ; Pa] Regangan normal: ϵ ; strain] Tegangangeserτ Tegangangeser c a b Gaya dalam Gambar 2-2 Tegangangeserτ: tegangangeserτempunyaiorientasigeserterhadapluaspenampangnya regangangeserϒ:

  5. Ujitarik, diagram Tegangan-regangan normal ; Pa] ; strain] Gambar 2-3 Besituang, keramiks, bronsze, dll material brittle ductile Baja, aluminium, dll

  6. Diagram tegangan-regangan normal material brittle Pada material brittle, deformasiakibatbebansangatkecildanmendekatinol (Gambar 2.4 Y1 U: ultimate Hukum Hook - Daerah OP Diagram tegangan-regangan normal material ductile E: modulus elastisitasdaribahan

  7. Diagram tegangan-regangangeser (Gambar 2-7) Pl: propotional limit U : ultimate : ultimate shear strength prpotional limit shear strength O-pl G: modulus geserdaribahan Gambar 2-7 μ: angkaperbandingan poison bahan

  8. Tegangantarik- tekan • Tegangan bending • Tegangankontak Tegangan normal σ Tegangantarik- tekan ; Pa] Tegangan bending

  9. Momeninersialuasan : Momentahanandariluasan

  10. Tegangankontak (tekananbidang)

  11. Tegangangeserτ • Geseran • puntiran Geseran Puntiran

  12. Dr Ir Soeharsono FTI Universitas Trisakti

  13. Konsentasitegangan

  14. A bar 3 m long is made of two bars, one of copper having E = 105 GN/m2 and the other of steel having E = 210 GN/m2. Each bar is 25 mm broad and 12.5 mm thick. This compound bar is stretched by a load of 50 kN. Find the increase in length of the compound bar and the stress produced in the steel and copper. The length of copper as well as of steel bar is 3 m each.

  15. A central steel rod 18 mm diameter passes through a copper tube 24 mm inside and 40 mm outside diameter, as shown in. It is provided with nuts and washers at each end. The nuts are tightened until a stress of 10 MPa is set up in the steel. The whole assembly is then placed in a lathe and turned along half the length of the tube removing the copper to a depth of 1.5 mm. Calculate the stress now existing in the steel. Take Es= 2Ec

  16. BIAXIAL STRESSES Tebal =t Gambar 7-5a

  17. FAILURE THEORIES Maximum-Normal-Stress Theory

  18. Maximum-Shear-Stress Theory Distortion-Energy Theory

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