Biểu đồ nội lực của các dạng khung chịu lực. cơ sở để giải các bài toán trong sức bền vật liệu, cơ kết cấu... Trình bày ngắn gọn, dễ hiểu, dễ học. Đây là tài liệu dùng cho sinh viên các ngành kĩ thuật có học môn sức bền vật liệumột trong những môn học tương đối khó
TABLE 13-1 STATICALLY DETERMINATE RECTANGULAR SINGLE-BAY FRAMES OF CONSTANT CROSS SECTION The direction of the reaction forces are shown in the figures of the configurations The signs of the moments are shown in the moment diagrams A bending moment is indicated as positive when it causes tension on the inner side of the frame and compression on the outer side Opposing moments are negative The formulas in the table give the magnitudes of these quantities The horizontal and vertical coordinate axes are x and y, respectively v jk is the displacement of point j in the k direction θ j is the slope at j Configuration Moment Diagram Important Values HA = R A = R B = 12 W v Bx = W h L2 8E I Mmax = 14 W L HA = W at point K RA = RB = W h L W h2 (3L + 2h) 6E I W h2 vC y = vC x = (L + h) 3E I Mmax = W h at point D v Bx = HA = W v Bx = RA = RB = W h2 3E I (3L + 2h) Mmax = W h HA = v Bx = RA = RB = M0 h L 2E I Mmax = M0 676 M0 L at point C TABLE 13-1 Statically Determinate Rectangular Frames TABLE 13-1 (continued) STATICALLY DETERMINATE RECTANGULAR SINGLE-BAY FRAMES OF CONSTANT CROSS SECTION Configuration Moment Diagram Important Values HA = θK = RA = RB = M0 L 12E I Mmax = 12 M0 HA = vbx = M0 L at point K RA = RB = p1 L p1 h L 12E I Mmax = p1 L at x = 12 L H A = p1 h v Bx = RA = RB = p1 h (6L + 5h) 24E I Mmax = 2 p1 h at point D H A = p1 h v Bx = p1 h 2L RA = RB = p1 h 2L p1 h (18L + 11h) 24E I Mmax = p1 h TABLE 13-1 Statically Determinate Rectangular Frames at point D 677 TABLE 13-1 (continued) STATICALLY DETERMINATE RECTANGULAR SINGLE-BAY FRAMES OF CONSTANT CROSS SECTION Configuration Moment Diagram Important Values HA = W RA = MA = W h2 (3L + 4h) 3E I WhL (L + h) v Dy = − 2E I Mmax = W h at points B, C v Dx = 10 HA = RA = W MA = W L WhL (L + 2h) v Dx = − 2E I W L2 v Dy = (L + 3h) 3E I Mmax = W L 11 HA = W RA = MA = W h W Lh 2E I 2E I Wh W Lh vC y = vC x = 3E I 2E I Mmax = W h at point A v Dx = − W h3 v Dy = 12 H A = R A = M A = M0 M0 h (L + 3h) v Dx = EI M0 L v Dy = − (L + 2h) 2E I M0 θD = (L + 2h) Mmax = M0 EI 13 HA = MA = R A = p1 L 2 p1 L p1 L h (L + 3h) 6E I p1 L = (L + 4h) 8E I v Dx = − v Dy Mmax = 678 2 p1 L TABLE 13-1 Statically Determinate Rectangular Frames TABLE 13-1 (continued) STATICALLY DETERMINATE RECTANGULAR SINGLE-BAY FRAMES OF CONSTANT CROSS SECTION Configuration 14 Moment Diagram Important Values HA = vC x = RA = W MA = W L W Lh 2E I W L2 (L + 3h) vC y = 3E I WL θC = (L + 2h) 2E I Mmax = W L 15 HA = W vC x = RA = MA = W h W h3 3E I W h2 L vC y = 2E I Mmax = W h at point A 16 HA = RA = M A = M0 h2 M0 2E I M0 L (L + 2h) vC y = 2E I M0 θC = (L + h) EL Mmax = M0 vC x = 17 HA = MA = vC x = vC y R A = p1 L 2 p1 L p1 h L 4E I p1 L (L + 4h) = 8E I Mmax = 2 p1 L TABLE 13-1 Statically Determinate Rectangular Frames 679 TABLE 13-1 (continued) STATICALLY DETERMINATE RECTANGULAR SINGLE-BAY FRAMES OF CONSTANT CROSS SECTION Configuration 18 Moment Diagram Important Values Free-end relative displacement v = v Ax − v Bx = W a2 (2a + 3L) 3E I Mmax = W a 19 Free-end relative displacement v = v Ax − v Bx = M0 a (a + L) EI Mmax = M0 20 Free-end relative displacement v = v Ax − v Bx = Mmax = 680 p1 a (a + 2L) 4E I 2 p1 a TABLE 13-1 Statically Determinate Rectangular Frames TABLE 13-2 STATICALLY INDETERMINATE RECTANGULAR FRAMES The directions of the reaction forces are shown in the figures of the configurations The signs of moments are shown in the moment diagrams A bending moment is indicated as positive when it causes tension on the inner side of the member and compression on the outer side Opposing moments are negative The formulas in the table give the magnitudes of the forces and moments e = h/L Definitions β = Ih (horizontal beam)/Iv (vertical members) Configuration Moment Diagram Important Values L −a a RB = W L L 3W a L − a H A = HB = 2h L 2βe + 3W a L − a MC = M D = 2L 2βe + W a(L − a) 4βe + MK = 2L 2βe + RA = RB = W RA = W h L H A = H B = 12 W MC = M D = 12 W h TABLE 13-2 Statically Indeterminate Rectangular Frames 681 TABLE 13-2 (continued) Configuration STATICALLY INDETERMINATE RECTANGULAR FRAMES Moment Diagram Important Values RA = RB = W h−a L HA = aβ(2h − a) W h + a − (h − a) 2h h(2hβ + 3L) HB = aβ(2h − a) W (h − a) 1+ 2h h(2hβ + 3L) MC = 12 W (h − a) + aβ(2h − a) h(2hβ + 3L) M D = 12 W (h − a) − aβ(2h − a) h(2hβ + 3L) MK = W (h − a) 2h × h + a − (h − a) RA = RB = p1 L H A = HB = p1 L 4e(2βe + 3) MC = M D = p1 L 4(2βe + 3) MK = 682 aβ(2h − a) h(2hβ + 3L) p1 L 2βe + 2βe + p1 h 2L p1 h 11βe + 18 HA = 2βe + p1 h 5βe + HB = 2βe + RA = RB = MC = p1 h 5βe + 2βe + MD = p1 h βe + 2βe + TABLE 13-2 Statically Indeterminate Rectangular Frames TABLE 13-2 (continued) STATICALLY INDETERMINATE RECTANGULAR FRAMES Configuration Moment Diagram Important Values R A = R B = 12 W 3W L 8h(βe + 2) WL M A = MB = 8(βe + 2) WL MC = M D = 4(βe + 2) W L βe + MK = βe + H A = HB = RA = RB = p1 L H A = HB = p1 L 4h(βe + 2) M A = MB = p1 L 12(βe + 2) MC = M D = p1 L 6(βe + 2) MK = p1 L (3βe + 2) 24(βe + 2) R A = R B = p1 h βe2 6βe + HA = 4βe + p1 h 8βe + 17 − 2(βe + 2) 6βe + HB = p1 h MA = βe + p1 h 4βe + + 6βe + 6(βe + 2) MB = βe + p1 h 4βe + − 6βe + 6(βe + 2) 4βe + − 6βe + 2(βe + 2) MC = p1 h 2 βe + 6βe + 6(βe + 2) M D = p1 h βe + 6βe + 6(βe + 2) TABLE 13-2 Statically Indeterminate Rectangular Frames 683 TABLE 13-2 (continued) Configuration STATICALLY INDETERMINATE RECTANGULAR FRAMES Moment Diagram Important Values RA = W a[L (2βe + 3) − a ] 2L (βe + 1) RB = W − RA H A = HB = W a(L − a ) 2h L (βe + 1) W a(L − a ) 2L (βe + 1) a M D = [W (L − a) − MC ] L p1 L 4βe + RA = βe + p1 L 4βe + RB = βe + MC = 10 H A = HB = MC = 11 p1 L 8(βe + 1) L − a2 L 3βe + RA = Wa L RB = 2a L + a W (L − a) 1− L L 3βe + H A = HB = 684 p1 L 8h(βe + 1) 1+ 3W a L − a h L 3βe + MA = W a L − a2 L 3βe + MC = 2W a L − a L 3βe + MD = 2a L + a W a(L − a) 1− L L 3βe + TABLE 13-2 Statically Indeterminate Rectangular Frames TABLE 13-2 (continued) STATICALLY INDETERMINATE RECTANGULAR FRAMES Configuration 12 Moment Diagram Important Values RA = RB = HA = β 3W a(h − a)2 3βe + hL h − a 3aβe + 2(h + a) Wa 1+ h 3βe + h2 − β 3(h − a)2 hL 3βe + HB = W − H A MA = W a(h − a) 3aβe + 2(h + a) 3βe + h2 MC = β 3W a(h − a)2 hL 3βe + M D = H A (h − a) − M A 13 RB = βe + p1 L 3βe + RA = 3βe p1 L 3βe H A = HB = 14 +5 +4 p1 L 4h(3βe + 4) MA = p1 L 4(3βe + 4) MC = p1 L 2(3βe + 4) RA = RB = βe2 p1 h 3βe + +5 +4 βe + H B = 32 p1 h 3βe + βe + M A = 14 p1 h 3βe + βe MC = 14 p1 h 3βe + HA = 3βe p1 h 3βe TABLE 13-2 Statically Indeterminate Rectangular Frames 685 TABLE 13-2 (continued) Configuration 15 STATICALLY INDETERMINATE RECTANGULAR FRAMES Moment Diagram Important Values RA = W a2 + 1) 2L (βe × [βe(3L − a) + 2(3L − 2a)] RB = W − RA H A = HB = 3W a L − a 2h L βe + W a2 L − a 2L βe + W a(L − a) MB = 2L βe(2L − a) + 2(L − a) × βe + MA = MC = W a2 L − a L βe + MD = RB a − MB 16 RA = 3βe + p1 L βe + RB = 5βe + p1 L βe + H A = HB = p1 L 8h(βe + 1) p1 L 24(βe + 1) 3βe + M B = 24 p1 L βe + MA = MC = 686 p1 L 12(βe + 1) TABLE 13-2 Statically Indeterminate Rectangular Frames TABLE 13-3 NONRECTANGULAR SINGLE-BAY FRAMES The direction of the reaction forces are shown in the figures of the configurations The signs of moments are shown in the moment diagrams A bending moment is indicated as positive when it causes tension on the inner side of the member and compression on the outer side Opposing moments are negative The formulas in the table give the magnitudes of these quantities Symmetrical Gable Frames k= γ = Configuration I1 a I2 h 3(1 − kφ) 2(1 + kφ ) φ= λ= f h α = + 3φ + φ + 6(1 + k) + kφ k η = 12[2 + 2k − γ (1 − kφ)] Moment Diagram Important Values H A = HB = W L(3 + 2φ) 2αh R A = R B = 12 W M E = MC = H B h M D = 14 W L − H B h(1 + φ) HB = W α + 3φ + k H A = W − HB Wh L M E = h(W − H B ) RA = RB = MC = H B h M D = H B h(1 + φ) − 12 W h TABLE 13-3 Nonrectangular Single-Bay Frames 687 TABLE 13-3 (continued) Configuration NONRECTANGULAR SINGLE-BAY FRAMES Moment Diagram Important Values H A = HB = p1 L (8 + 5φ) 8αh RA = RB = p1 L M E = MC = H B h M D = 18 p1 L − H B h(1 + φ) W = p1 ( f + h) HB = 8φ p1 h 12 + + 30φ 4α k + 20φ + 5φ + k H A = W − HB RA = RB = p1 (h + f )2 2L M E = H A h − 12 p1 h MC = −H B h M D = − 14 p1 (h + f )2 + H B h(1 + φ) H A = HB = W Lk (3γ + λφ) ηh R A = R B = 12 W M E = MC = W Lk (3 + 2γ φ) η M A = M B = −M E + H A h M D = −M E + 14 W L − H B f 688 TABLE 13-3 Nonrectangular Single-Bay Frames TABLE 13-3 (continued) NONRECTANGULAR SINGLE-BAY FRAMES Configuration Moment Diagram Important Values HB = 2W (λ − 3γ ) η H A = W − HB RA = RB = 3W h 2(3 + k)L M E = 4W h − 2γ 2η + 16(3 + k) −3 + 2γ 2η MC = 4W H + 16(3 + k) M A = h(W − H B ) − M E M B = −MC + H B h M D = HB f − 2W h (3 − 2γ ) η S = + 54 γ φ T = 2γ + 58 λφ H A = HB = RA = RB = M A = MB = M E = MC = MD = − p1 L T k ηh p1 L p1 L k η (T − S) p1 L Sk η p1 L Sk + p1 L −H B f η For the left half of a girder, Mx = (−M E + 14 p1 L x) × 1− TABLE 13-3 Nonrectangular Single-Bay Frames 2x L + MD 2x L 689 TABLE 13-3 (continued) Configuration NONRECTANGULAR SINGLE-BAY FRAMES Moment Diagram Important Values S = − 4γ − kφ(4 + 52 γ φ) T = 2λ + kφ(4γ + 54 λφ) − 6γ Sf h + (2 − 32 γ ) h+ f h+ f f 4h + (12 − kφ) Q= h+ f h+ f R= W = p1 ( f + h) HB = W (T f + 34 λh − 2γ h) η(h + f ) H A = W − HB RA = RB = Wh 32(3 + k)L × 4Q + 16(3 + k) ME = W h MC = −W h f φ h+ f Q R + η 16(3 + k) Q R − η 16(3 + k) M A = −M E − H B h + 12 W h h+2f h+ f M D = − 12 (M E − MC ) + H B f − Wf2 4(h + f ) M B = −MC + H B h 690 TABLE 13-3 Nonrectangular Single-Bay Frames TABLE 13-3 (continued) NONRECTANGULAR SINGLE-BAY FRAMES Symmetrical Arched Frames I1 L I2 h 1.5 − kφ β= + 0.8kφ k= Configuration φ= f h γ = + 1.5k + 0.8kφ α = 8[1 + k(1.5 + 2φ + 0.8φ )] η = 12(2 + k) − 4β(3 − 2kφ) Moment Diagram Important Values H A = HB = W Lk + 5φ αh R A = R B = 12 W M E = MC = H A h M D = 14 W L − H A (h + f ) 10 (1 + 1.5k + kφ) α HB = W e H A = W − HB e= Wh L M E = h(W − H B ) RA = RB = MC = H B h 11 H A = HB = p1 L k (1 + 45 φ) αh RA = RB = p1 L M E = MC = H A h M D = 18 p1 L − H A ( f + h) TABLE 13-3 Nonrectangular Single-Bay Frames 691 TABLE 13-3 (continued) Configuration 12 NONRECTANGULAR SINGLE-BAY FRAMES Moment Diagram Important Values e = 4(1 + 1.5k + kφ)/α p1 h (1 + αe) 2α H A = p1 h − H B HB = RA = RB = p1 h 2L M E = 12 p1 h − H B h MC = H B h 13 H A = HB = W Lk 6β + 5γ φ ηh R A = R B = 12 W M E = MC = W Lk + 5βφ η M A = M B = −M E + H A h M D = 14 W L − M E − H A f 14 HB = 2W (2γ − 3β) η H A = W − HB RA = RB = 3W h (6 + k)L Wh (6 − 4β) η 3W h + 2(6 + k) Wh MC = (6 − 4β) η 3W h + 2(6 + k) ME = M A = h(W − H B ) − M E M B = −MC + H B h 692 TABLE 13-3 Nonrectangular Single-Bay Frames TABLE 13-3 (continued) NONRECTANGULAR SINGLE-BAY FRAMES Configuration Moment Diagram 15 Important Values H A = HB = p1 L k (5β + 4γ φ) 5ηh RA = RB = p1 L M E = MC = p1 L k (5 + 4βφ) 5η M A = M B = −M E + H A h M D = 18 p1 L − M E − H B f 16 HB = p1 h (3γ − 4β) 2η H A = p1 h − H B RA = RB = ME = p1 h (6 + k)L p1 h (4 − 3β) 2η + MC = − + p1 h 2(6 + k) p1 h (4 − 3β) 2η p1 h 2(6 + k) M A = −M E − H B h + 12 p1 h M B = −MC + H B h TABLE 13-3 Nonrectangular Single-Bay Frames 693 TABLE 13-3 (continued) NONRECTANGULAR SINGLE-BAY FRAMES Symmetrical Polygonal Frames I3 a I3 d 2a a k2 = B0 = (k1 + 1) + C0 = + + 3k2 I1 e I2 e h h a B0 b h + C0 N0 = C1 = (2 + 3k2 ) C2 = + (2 + 3k2 ) h a a d b C3 = + (2 + k2 ) R = C2 − k1 N1 = k k − R L a d d h N2 = 3k1 + β + C3 β = 3k1 + + k3 = 2(k1 + 1) + (1 + C2 ) L L a b K = 2k1 + C1 a k1 = Configuration 17 Moment Diagram Important Values W cC0 + ( 34 W d)k2 2N0 X H A = HB = h X= R A = R B = 12 W M E = M D = 12 W c − X a M F = MC = X h M K = 14 W d + M E 18 p1 dcC0 + 12 p1 d k2 2N0 X H A = HB = h X= RA = RB = ME = p1 d M D = 12 p1 dc −X a M F = MC = X h M K = 18 p1 d + M E 694 TABLE 13-3 Nonrectangular Single-Bay Frames TABLE 13-3 (continued) NONRECTANGULAR SINGLE-BAY FRAMES Configuration Moment Diagram Important Values 19 W a(B0 + C0 ) 2N0 X H A = W − HB = h Wa = RB = L a = Wa − X h a = X h c Wa − X = 1− L c = Wa − X L a p1 a 2(B0 + C0 ) + k1 h = 8N0 X H A = p1 a − H B = h p1 a = RB = 2L a = p1 a − X h a = X h = R B (L − c) − X X= HB RA MF MC ME MD 20 X HB RA MF MC ME MD = X − RB c TABLE 13-3 Nonrectangular Single-Bay Frames 695 TABLE 13-3 (continued) Configuration 21 NONRECTANGULAR SINGLE-BAY FRAMES Moment Diagram Important Values 3b W dk2 4a 3h W dk2 B2 = W cC2 + 4a B1 k3 − B2 R X1 = 2N1 B2 k4 − B1 R X2 = 2N1 H A = H B = (X + X ) a B1 = W cC1 + R A = R B = 12 W M A = MB = X M F = MC = X M E = M D = 12 W c − − h X2 a b X1 a M K = 14 W d + M E 22 p1 d b k2 2a p1 d h k2 p1 dcC2 + 2a B1 k3 − B2 R 2N1 B2 k4 − B1 R 2N1 H B = (X + X ) a B1 = p1 dcC1 + B2 = X1 = X2 = HA = RA = RB = p1 d M A = MB = X M F = MC = X ME = MD = − p1 dc − ab X h X2 a M K = 18 p1 d + M E 696 TABLE 13-3 Nonrectangular Single-Bay Frames