Contoh Soal dan penyelesaian Statis Tak Tentu Teknik Sipil
Contoh Soal Statis Tak Tentu Conroh Soal Penyelesaian: A. Angka kekakuan dan ditribusi faktor Joint A= KAB = 0.75 EI/6.5= 0.115 ; Df AB= 0 (Karena tumpuan jepit) Joint B= KBA = 0.75 EI/60.5 = 0.115 ; Df BA = 0.115/0.329 = 0.35 Joint B= KBC = 1.5 EI/7 = 0.214 ; Df BC = 0.214/0.329 = 0.65 (karena joint B ada dua arah kanan dan kiri maka dijumlahkan) Joint B Total = KBA+KBC = 0.329 ; Df Total = 0.35+0.65 = 1.00 (OK) Joint C= KCB = 1.5EI/7 = 0.214 : Df CB = 1 *Tulisan tangan* B. Momen Primer *Tulisan tangan* C. Tabel Cross D. Free Body E. Analisis Pengaruh Bidang D Analisis penggambaran Bidang Momen MA = 214.956 kg (2m) = RVA (2)-214.956=168.606(2)-214.956=122.256 kg.m (3.5m) = RVA(3.5)-214.956-150(1.5)=168.606(3.5)-214.956-150(1.5)=150.165 kg.m (6.5m) = RVA (6.5)-214.956-150(4.5)-200(3)=168.606(6.5)-214.956-150(4.5)-200(3)= -394.017 kg.m MB = -394.015 kg (1m) = 406.288(1)-394.015-100(1)(0.5) = -37.727 kg.m (2m) = 406.288(2)-394.015-100(2)(1) = 218.561 kg.m (3m) = 406.288(3)-394...