Zginanie belek z udziałem sił tnących, MBM PWR, Obrona (przydatne materiały), Dodatkowe materiały


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Wytrzymałość Materiałów 4

Zginanie belek z udziałem sił tnących

Wzór Żurawskiego (str. 208-210)

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z

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RAz RAx P RB

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A x

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x dx B

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a

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l

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M

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dM

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T

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M+dM N+dN

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z dx x

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z

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M

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1 -y z1 -y

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τz -z Az

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N dA τz 1 bz

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Rys. 17 Określenie naprężeń stycznych w zginanej belce

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Pix = -N + τzbzdx + N + dN = 0 0x01 graphic
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Różniczkując N po x mamy:

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dN/dx = - (dM/dx)·(Sy/Jy) (c)

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M T M+dM

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M0 = M- M- dM + Tdx = 0 0 x

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T=dM /dx (d) dx T+dT

podstawiając (c) i (d) do (b) mamy:

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gdzie 0x01 graphic
(13)

Przykład 7 (str. 210-211)

Określić rozkład naprężeń stycznych przy zginaniu belki o przekroju prostokątnym o wymiarach b×H (rys.18), oraz dobrać wartość H= 2h z warunku τmax = 50 MPa. Wartość siły tnącej w rozpatrywanym przekroju wynosi T= 104N, a stosunek 2h/b = 4.

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dA=dydz1 z Az

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z1 2h z h1 -y 0x01 graphic
; H=2h

Az = b(h-z)

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b

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Rys.18 Określenie naprężeń stycznych

Rozwiązanie

Moment statyczny pola Az względem osi obojętnej z

Sy = Azh1 = b(h2-z2)/2 Symax = bh2/2 = bH2/8

Moment bezwładności pola przekroju względem osi y

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(14)

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(c)

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Podstawiając do (c) T = 104 N, b = H/4,

τmax = 50MPa otrzymujemy:

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Odpowiedz: H = 34.6mm, b = H/4 = 34.6/4 = 8.6mm

Naprężenia średnie: 0x01 graphic
MPa

τmax /τśr = 1.5

Stan czystego ścinania (str. 78-86)

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γ τ s = γ ·1

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a) z τ2 b)

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τ3

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τ τ

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1 τ1

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o 1 0

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x τ

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1 τ4

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Rys.19 Parametry czystego ścinania

Warunki równowagi prostopadłościanu o wymiarach 1×1×1

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Px = τ2×1×1τ4×1×1=0 τ2 = τ4

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Py = τ3×1×1τ1×1×1=0 τ1 = τ3 τi = τ

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Mo = τ2×1×1×1τ3×1×1×1=0 τ2 = τ3

Energia odkształcenia w stanie czystego ścinania kostki

o wymiarach 1cm3 V1 = L(praca) = τ · s / 2 = τγ / 2.

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τ

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tgα = τ/γ =G τ = G×γ (15)

τpl τspr

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α

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γ

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Rys. 20 Wykres naprężeń τ w funkcji γ

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Skręcanie prętów (str.140-145)

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Ms dr

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* r

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s dA Rz

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γr l α r

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dα

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τmax dA r

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dr

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Rys.21 Skręcanie rd* dP

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τr

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dP = dAτr ; dMs = rdP = rdAτr ; τr /τmax = r/Rz ; (τmax /Rz )= τr / r

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gdzie (16)

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(17)

z rys.21 s = γr l = *r; τr = Gγr ; γr = τr /G; * = γr l /r

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Przykład 8 (str. 147-148)

Określić przebieg momentów skręcających w wale przedstawionym na rysunku 22.

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MA MS MB

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a b

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Rys.22

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MA1 = MS MS *1 *1

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MA2 = MB0x01 graphic
l = a+b *2

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MB

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*1= MS a /(GJ0); *2= MB l /(GJ0)

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*1+ *2 = 0; MS a + MB l =0; MB = MS a l

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MA = MA1 + MA2 = MS MS a l = MS b/l

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MA MS

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MB

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Rys.22a Wykres momentów skręcających

Przykład 9 (str.145)

Dla wału z rysunku 22 dobrać średnice tak, aby współczynnik bezpieczeństwa ne = 2. Dane τe = 125 MPa

τdop = τe /ne = 125/2 = 62.5 MPa; Ms = 104 Nm;0x01 graphic
a = 1.2m,

b = 1.5m; Rz = R

Rozwiązanie

MA = MS (b/l) = 104(1.5/2.7) = 5556 Nm

MB = - MS (a/l) = -104(1.2/2.7) = -4444 Nm

Mmax = MA = 5556Nm

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d = 2R = 76.8mm

22 WM

23WM

24WM

Rw

25WM

26WM

27WM

-y



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