[{"data":1,"prerenderedAt":61},["ShallowReactive",2],{"q-106180-arch_structure-25":3},{"question":4,"neighbors":26,"sameNumber":45,"prev":34,"next":37},{"id":5,"classKey":6,"session":7,"year":8,"subject":9,"number":10,"groupStem":11,"ownStem":12,"stem":12,"options":13,"answer":18,"answerKind":20,"answerS":11,"answerNote":11,"images":21,"explanation":22,"withheld":11,"deepChars":23,"freq":24,"indexable":25},"arch-106180-801-0102-025","architect","106180",106,"建築結構",25,null,"有一矩形梁斷面，梁寬 30 cm，梁高 60 cm。已知材料之彈性係數為 20,000 kgf\u002Fcm2。如欲使得該斷面於梁頂與梁底處分別產生 0.003 的壓應變與張應變，則該斷面所須施加的彎矩力應為多少 kgf-cm？",{"A":14,"B":15,"C":16,"D":17},"960,000","1,080,000","1,200,000","1,360,000",[19],"B","single",[],"本題考點：均質線彈性材料矩形梁的撓曲應力公式 σ = M \u002F S（S 為斷面模數），配合虎克定律 σ = E × ε，由題目指定的最外緣應變反推所需彎矩。\n【正解理由】題目要求梁頂產生壓應變、梁底產生張應變，且兩者大小都是 0.003，代表應變沿梁深呈直線分布、在斷面中央歸零，中性軸就在梁高的一半處，屬於純彎矩作用下的線彈性撓曲。解題分兩段：第一段用虎克定律把最外緣應變換成最外緣應力，彈性係數 20000 kgf\u002Fcm² 乘以 0.003，得最外緣應力 60 kgf\u002Fcm²；第二段用撓曲公式，把最外緣應力乘上矩形斷面的斷面模數。斷面模數等於梁寬乘以梁高平方再除以 6，梁寬 30 cm、梁高 60 cm 代入得 18000 cm³。最外緣應力乘斷面模數，得所需彎矩 1080000 kgf-cm，正好等於選項 (B) 的數值。由於應力與應變、彎矩與應力在彈性範圍內都是線性關係，彎矩比這個值小，外緣應變就達不到 0.003；比這個值大，外緣應變就超過 0.003，所以題目所求只有一個值，故選 B。\n【逐項排除】\n(A) 960,000 kgf-cm：比正確彎矩小。在線彈性範圍內，外緣應力與彎矩成正比，施加此彎矩時外緣應力低於 60 kgf\u002Fcm²，梁頂、梁底的應變都達不到 0.003，不符題目所問；要讓外緣應變恰為 0.003，彎矩應為 1,080,000 kgf-cm。\n(B) 1,080,000 kgf-cm：等於 E × ε × b × h² \u002F 6 的計算結果，此時外緣應力恰為 60 kgf\u002Fcm²，對應的梁頂壓應變與梁底張應變恰為 0.003，符合題目所問。\n(C) 1,200,000 kgf-cm：比正確彎矩大，外緣應力會高於 60 kgf\u002Fcm²，梁頂、梁底的應變都會超過 0.003，不符題目所問；正確彎矩應為 1,080,000 kgf-cm。\n(D) 1,360,000 kgf-cm：比正確彎矩更大，與正確值的差距在四個選項中最遠，外緣應變超過 0.003 更多，不符題目所問；正確彎矩應為 1,080,000 kgf-cm。\n【演算步驟】\n(1) 虎克定律求最外緣應力：σ = E × ε = 20000 kgf\u002Fcm² × 0.003 = 60 kgf\u002Fcm²。\n(2) 矩形斷面模數：S = b × h × h \u002F 6 = 30 cm × 60 cm × 60 cm \u002F 6 = 18000 cm³。\n(3) 撓曲公式求彎矩：M = σ × S = 60 kgf\u002Fcm² × 18000 cm³ = 1080000 kgf-cm。\n(4) 以慣性矩驗算：I = b × h × h × h \u002F 12 = 30 cm × 60 cm × 60 cm × 60 cm \u002F 12 = 540000 cm⁴；中性軸到外緣距離 c = h \u002F 2 = 60 cm \u002F 2 = 30 cm；M = σ × I \u002F c = 60 kgf\u002Fcm² × 540000 cm⁴ \u002F 30 cm = 1080000 kgf-cm，與步驟 3 一致。\n【記憶點】頂底應變等值反向就是中性軸在正中央：先用 σ = E × ε 換成外緣應力，再乘矩形斷面模數 bh²\u002F6 就是彎矩。",772,1,true,[27,30,33,36,39,42],{"number":28,"session":7,"stem":29},22,"型鋼會因製造過程產生殘餘應力（residual stress），下列敘述何者錯誤？",{"number":31,"session":7,"stem":32},23,"如圖所示桁架，桿件 ED 的內力為何？",{"number":34,"session":7,"stem":35},24,"下圖桿件之自由端 a 點受 2P 力作用，a 點垂直向下伸長量為 δ av = ，則…",{"number":37,"session":7,"stem":38},26,"下圖簡支梁之跨度為 L。已知材料之彈性係數為 E，梁斷面之慣性矩為 I。今於跨度中央…",{"number":40,"session":7,"stem":41},27,"下圖之懸臂梁，端點 B 承受一集中載重 P，則 B 點之垂直位移為： PL2",{"number":43,"session":7,"stem":44},28,"桁架結構受力如下圖，桿件 EC 之軸向力為多少 kN？",[46,49,52,55,58],{"number":10,"session":47,"stem":48},"114180","如圖所示剛構架，各構件 EI 值皆相等，基礎支承點 D 垂直下陷將使構件產生彎矩。若…",{"number":10,"session":50,"stem":51},"113190","下圖剛架受地震力，最少出現幾個塑鉸時會發生崩塌？",{"number":10,"session":53,"stem":54},"112190","下圖所示結構中，支承 A 與 B 的反力大小為何？ 100 kN 45 kN C 4…",{"number":10,"session":56,"stem":57},"111180","三層構架如圖所示，假設質量僅分布於梁構件，那些變更會延長其基本振動週期？①增加柱斷面…",{"number":10,"session":59,"stem":60},"110180","建築物在抵抗側向地震力作用時，容許結構桿件產生破壞以損耗地震能量。依目前耐震設計原則…",1791863212012]