Bending Deflection – Differential Equation Method AE1108-II: Aerospace Mechanics of Materials

Dr. Calvin Rans Dr. Sofia Teixeira De Freitas

Aerospace Structures & Materials Faculty of Aerospace Engineering Recap

• So far, for symmetric beams, we have: • Looked at internal shear force and bending moment distributions • Determined normal stress distribution due to bending moments • Determined shear stress distribution due to shear force • Need to determine deflections and slopes of beams under load • Important in many design applications • Essential in the analysis of statically indeterminate beams

2 Deformation of a Beam Assumptions Shear deformation

V + V + Moment deformation V

M M

M +

Negligible (for long beams)

Bending Deformation = Shear Deformation + Moment Deformation Deformation of a Beam Assumptions • For long beams (length much greater than beam depth), shear deformation is negligible • This is the case for most engineering structures • Will consider moment deformation only in this course • Recap of sign convention +w

z N.A. x +M+ +M

+V +V + y y v Deformation of a Beam Visualizing Bending Deformation Elastic curve: plot of the deflection of the neutral axis of a beam

How does this beam deform?

We can gain insight into the deformation by looking at the bending moment diagram + - z M M

+ -

M M And by considering boundary conditions at supports Qualitatively can determine elastic curve! Moment-Curvature Relationship

z dz

m 1 z dz m2 (-ve M) (+ve v)

v(z) = vertical deflection at z dv  (z) = slope at z =  dz Moment-Curvature Relationship z dz

For small d: ds R d

1 d or   Rds curvature

dz cos  1 For small : ds dz cos when  is small Recall M E  d dv 2 1 d  dz dv I R     Rdz dz dz2 dv2 M EI  EIv dz2 (negative sign a result of sign convention) Deflection by Method of Integration

dv2 MEI Lets consider a prismatic beam dz2 (ie: EI = constant)  dv 1 Mdz   dz EI  Indefinite integrals result in constants of integration that  can be determined from boundary conditions of the 1 vMdz  problem Constant of integration EI 1 ie: zdz z2  C  2 Determining Constants of Integration Support Conditions

v = 0

v = 0

v = 0  = 0 Determining Constants of Integration Continuity Conditions

z Deformed abP shape C Discontinuity at z = a A B

z ≤ a z ≥ a Pab L Pb Pa M  z M Lz AC L CB L M 0

vavaAC  CB  

ACaa CB  Determining Constants of Integration Symmetry Conditions

• Symmetry implies reflection of P deformation across symmetry C A B plane • v is equal •  is opposite • Continuity implies equal deformation at symmetry plane • v is equal •  is equal

C  0 Procedure for Analysis Deflection by Integration • Draw a FBD including reaction forces • Determine V and M relations for the beam • Integrate Moment-displacement differential equation • Select appropriate support, symmetry, and continuity conditions to solve for constants of integration • Calculate desired deflection (v) and slopes (θ) Example 1a Problem Statement q

Determine the deflection and slope EI at point B in a prismatic beam due to the distributed load q ABL

Solution

1) FBD & Equilibrium  FR0 z q  z  MA  FRqLRqL0 yy    Rz 2 cw LqL Ry MMqLM0     AA22 A

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 13 q Example 1a EI Solution ABL 2) Determine M and V @ z qL z q q

qL2 V  2 M M AB qL V

2  -qL /2  FqLqzVVqLz0    

222 ccw qL z L z  MMqLzqzMqLzz 0        22 22

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 14 q Example 1a EI Solution ABL 3) Boundary Conditions At z = 0: v = 0, v′ = 0

Two boundary conditions dv2 Thus can solve by integrating: MEI dz2 dv2 1 M dz2 EI

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 15 q Example 1a EI Solution ABL 4) Solve Differential Equation qz22 qL d222 v1  qz qL MqLz   2 qLz 22 dz EI 22  Boundary Condition dv1  qz322 qLz qL z 0 BC: At z = 0, θ = 0  C1 dz EI 62 2 => C1 = 0

 4322 1 qz qLz qL z 0 BC: At z = 0, v = 0 vC 2 EI 24 6 4 => C2 = 0

2 qz dv qz 22 vzLzL2246  zLzL 33  24EI dz6 EI

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 16 q Example 1a EI Solution ABL 5) Calculate slopes and deflections Determine deflection and slope at B:

2 qz qz 22 vzLzL2246  zLzL33 24EI 6EI

4 qL qL3 vvBL(z )  8EI BL(z ) 6EI

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 17 Relating Deformation to Loading Shear Force-Moment Diagram Relationships • Recall from Statics (refer to Hibbler Ch. 6.2 for refresher, but be careful of coordinate system)

+w dV w dz 4 dv w(z)  dM 4 v +M+ +M  V dz EI dz dv3 V(z) +V +V  v dz3 EI Moment-Curvature dv2 M(z) Relationship (Eq. 10.1)  v dv2 dz2 EI MEI dz2 z q Example 1b EI

We can also solve Example 1 in an ABL alternative way: 4 Watch negative sign! dv wz() v EIv q = -w(z) dz4 EI  dv3 V() z  3  v EIv qz C1 = V(z) dz EI dv2 Mz()  2  qz 2  v EIv C z C = M(z) dz EI 2 12

 32 qz z EIv C  C z C = -θ(z)EI 62123 We have 4 unknown constants of integration,  43 2 qz z z thus need 4 BCs EIv C  C  C z C 241234 6 2

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 19 z q Example 1b EI

We can also solve this problem an ABL alternative way: Boundary Condition: At z = L, V = 0 EIv qz C1 = V(z)

C1 CqL1 qz2 EIv qLz C = M(z) At z = L, M = 0 2 2 22 qL2 qL C2 CqL2   22 32 qz qL2 qL EIv z  z  C = -θ(z)EI At z = 0, θ = 0 62 2 3 C3 0

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 20 z q Example 1b EI

We can also solve this problem an ABL alternative way: Boundary Condition: qz42 qL qL EIv z32  z 0 C At x = 0, v = 0 24 6 4 4 C4 0

2 qz qx 22 vzLzL2246 vzLzL   33  24EI 6EI

Same result as before!

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 21 Example 2 qL z q

Determine deflection and EI slope at B: AB EIv q = -w(z) L

 We will apply Approach 2

EIv qz C1 = V(z)  Exact same differential qz2 EIv C z C = M(z) equations as before!! 2 12

 32 What makes the qz z EIv C  C z C = -θ(z)EI problem different? 62123

 43 2 Boundary Conditions! qz z z EIv C  C  C z C 241234 6 2

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 22 qL z q Example 2 EI

Determine deflection and ABL slope at B: Boundary Condition: At z = L, V = qL EIv qz C1 = V(z)

qL CqL1 2 q q EI EI ABL ABL 2qL qL qL V V M M

-qL2/2 -3qL2/2

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 23 qL z q Example 2 EI

Determine deflection and ABL slope at B: Boundary Condition: At z = L, V = qL EIv qz C1 = V(z)

C1 CqL1 2 qz2 EIv 2 qLz C = M(z) At z = L, M = 0 2 2 22 qL2 3 qL C2 CqL2 2  22 32 qz2 3 qL EIv qLz  z  C = -θ(x)EI At z = 0, θ = 0 623 C3 0

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 24 qL z q Example 2 EI

Determine deflection and ABL slope at B: Boundary Condition: qz42 qL3 qL EIv z32  z 0 C At z = 0, v = 0 24 3 4 4 C4 0

2 qz qz 22 vzLzL22818 vzLzL 69 24EI 6EI

4 11qL 2qL3 vvBzL()  v ' 24EI BzL() 3EI

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 25 Example 2

qL xz q xz q EI EI

ABL ABL

qz2 qz2 vzLzL22818 vzLzL2246 24EI 24EI 0 qL4 8EI

11qL4 4 qLqL4  24EI 2EI 2EI 0 L

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 26 z P Example 3 EI B A C Determine deflection at C in terms of EI: L L/2 P/2 To save time, reactions are provided 3P/2

Since reaction forces act at B (discontinuity), we must split the differential equation into parts for AB and BC

We can easily see by inspection that: P EIv V  (0 < z < L) 2

EIv VP (L < z < 3L/2)

Integrate to find M

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 27 z P Example 3 EI B A C Determine deflection at C: L L/2 P/2 Moments: 3P/2 P M EIv  z  C (0 ≤ z ≤ L) 2 1

(L ≤ z ≤ 3L/2) M EIv  Pz  C2 Moment BC’s:

At z = 0, M = 0 C1 0 Integrate to find θ 3PL At z = 3L/2, M = 0 C 2 2

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 28 z P Example 3 EI B A C Determine deflection at C: L L/2 P/2 Slopes: 3P/2 P EIv z2 C (0 ≤ z ≤ L) 4 3 PPL3 EIv z2  z  C (L ≤ z ≤ 3L/2) 224 Slope Continuity Condition: 2 PL 2 At z = L, θAB = θBC  CPLC   4 34 Integrate to find v

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 29 z P Example 3 EI B A C Determine deflection at C: L L/2 P/2 Deflections: 3P/2 P EIv z3 C z C (0 ≤ z ≤ L) 12 35

PPL323 EIv z  z  C z C (L ≤ z ≤ 3L/2) 64 46 From last slide PL2 Deflection BC’s: CPLC2 4 34 At z = 0, v = 0 C5 0 2 3 PL2 5PL PL At z = L, v = 0 C C4  C6  3 12 6 4

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 30 z P Example 3 EI B A C Determine deflection at C: L L/2 Deflections: P/2 Pz 3P/2 vLz22  (0 ≤ z ≤ L) 12EI P vLLzLzz310932 23 2 (L ≤ z ≤ 3L/2) 12EI PLPL33  0.5 1212EIEI 3LPL3 0 0.5 1 1.5L vC() v ( )  0.5 28EI  1  1.5  2

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 31 Example 4 z P

What about beams with a EI non loaded free end? ABC L L

P

Will it work itself out? A C B Curved part Straight part

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 32 z P

Example 4 EI -PL ABC What about beams with a L L non loaded free end? To save time, reactions are provided P

Moments: P V M EIv  P z  L (0 ≤ z ≤ L)

M MEIv 0 (L ≤ z ≤ 2L)

-PL

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 33 z P

Example 4 EI -PL ABC What about beams with a L L non loaded free end? P Slopes: P EIv z2  PLz C (0 ≤ z ≤ L) 2 1

(L ≤ z ≤ 2L) EIv C2 Slope BC’s:

At z = 0, θ = 0 C1 0

Slope CC’s: PL2 At z = L, θ = θ C2 AC CB 2

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 34 z P

Example 4 EI -PL ABC What about beams with a L L non loaded free end? P Displacements: PPL EIv z32  z  C (0 ≤ z ≤ L) 62 3 PL2 EIv z C (L ≤ z ≤ 2L) 2 4 Displacement BC’s:

At z = 0, v = 0 C3 0

Displacement CC’s: PL3 At z = L, v = v C4 AC CB 6

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 35 z P

Example 4 EI -PL ABC What about beams with a L L non loaded free end? P Displacements: Pz2 vzL 3 (0 ≤ z ≤ L) 6EI PL2  L vz (L ≤ z ≤ 2L) 23EI  0 Formula for a straight line!  0.5 No curvature, it does work out!  1 0 1L 2 L

Aerospace Mechanics of Materials (AE1108-II) – Example Problem 36 For next time