Homework esame Theory of Structures

Theory of Structures + Stability of Structures

February 4^th, 2021

(the solution should be delivered within February 4^th, 2021, h. 13:00)

Theory of Structures

Problem T1

The problem is referred to a parabolic arch, whose geometry is depicted in the figure. The material is homogeneous (E = 30 GPa, ν = 0.2) and the keystone section is square (side length 150 cm). The variation of the cross-section properties with respect to the arch inclination is A = A₀/cos α, I = I₀/cos α, where A₀ and I₀ are the keystone cross-section area and inertia, respectively. The external load is represented by a concentrated force (P = 150 kN). The student should solve the problem by making use of symbolic variables and, finally, should obtain the numerical values of the reaction forces and the diagrams of shear, axial force and bending moment.

Problem T2

The above figure represents a multi-cellular section with uniform thickness 50 mm (all the measurements are in mm). The students should obtain the stress field for a shear force V = 200 kN, applied in the upper left corner of the middle line. Moreover, the position of the shear center should be computed. G = 18 GPa.

THEORY OF STRUCTURES

PROBLEM T1

Homogeneous MaterialE=30GPa, ν=0.2

Keystone Square Section (A₀, I₀)

1. Redundant Structure:
   • Add a redundant action in B substituting the hinge with a roller
2. Case 0: External Load

Equilibrium

• ∑H=0 : H_A0=0
• ∑V=0 : V_A0+V_B0=P → V_A0=³⁄₄P
• ∑M_A=0 : V_B0ℓ-P^e⁄_u=0 → V_B0=P^u
• Check ∑M_B=0 : ³⁄₄Pe-P³⁄₄e=0 ✓

Focus on a part:

N₀=-(³⁄₄P+P)sinα z^e⁄₄

T₀=-(³⁄₄P+P)cosα z^e⁄₄

M_z={½Pz    z^e⁄₄        [³⁄₄P + P(z-^e⁄₄)]    z≥^e⁄₄

5) Reaction Forces

Once we have computed the HORIZONTAL THRUST x̄ we can compute REACTION FORCES and then internal forces:

EQUILIBRIUM

• ΣH = 0 : H_A = x̄ = 16.1,8 kN
• ΣV = 0 : V_A + V_B = P => V_A = P - V_B = 112.5 kN
• ΣM_A = 0 : V_B e - P ℓ e/l = 0 => V_B = P_ℓ/u = 37.5 kN

6) Internal Actions.

N = N₀ + N₁ x̄

T = T₀ + T₁ x̄

M = M₀ + M₁ x̄

con x̄ = -16,1,8 kN

N =

1. 0 < z < e
2. e < z < ℓ

N = -3/4 P sinα + x̄ cosα

N = P sinα + x̄ cosα

T =

1. 0 < z < e
2. e < z < ℓ

T = -3/4 P cosα - x̄ sinα

T = P_ℓ/u cosα - x̄ sinα

M =

1. 0 < z < e
2. e < z < ℓ

M = 3/4 Pz + y₃

M = [P_e/u + P_e/u] z + y₃

2) CENTRE OF SHEAR and PURE SHEAR

We starting considering the case of pure shear, shear force applied at centre of shear (that for sure is along y due to symmetry)

To solve this problem we introduce Doorwasy formula which requires the definition of some local abscissa in each laminate and to do this we introduce some fictitious opening.

SHEAR FLUX: q* = ^Tx/_Iy S_y^*

• S_g
• S₈
• S₁₁
• S₁₀
• S₇
• S₆
• S₄
• S₃
• S₂
• S_a

* We start computing STATIC MOMENTS S_y^*(s)

1. 0 ≤ S₁ ≤ 2000 mm
• S_y^*(S₁) = 50 x S₁ x (1000 - ^S₁/₂)
• S_y^*(S₁ = 0) = 0
• S_y^*(S₁ = 2000 mm) = 0
2. 0 ≤ S₂ ≤ 350 mm
• S_y^*(S₂) = 50 x S₂ x 1000
• S_y^*(0) = 0
• S_y^*(350) = 1.75 x 10⁷ mm³
3. 0 ≤ S₃ ≤ 400 mm
• S_y^*(S₃) = S_y^*(S₂ = 350) + 50 x S₃ (600 + 600 - ^S₃/₂)
• S_y^*(0) = 1.75 x 10⁷ mm³
• S_y^*(400) = 1.95 x 10⁷ + 2.60 x 10⁷ = 3.33 x 10⁷ mm³
4. 0 ≤ S_u ≤ 350 mm
• S_y^*(S_u) = 50 x S_u x 600
• S_y^*(0) = 0
• S_y^*(350) = 1.05 x 10⁷ mm³

ASCISSA S₆ = ∫₀⁶ S_y (S_y) ^{-508,51 mm)} (- V I_y ) dS_y

V I_y

ABSISKA S₈: ∫₀³⁵⁰ S⁺ (350mm)(V

I_y ) dS₇ V[.1,8x10^-5o²2 o = 4,25x10⁶ N mm

A3SCS S_g:∫₀³⁰⁰ (54) Jfys) (s0mm) V dS_y - V[.5x10^-63⁰obr) =lo

4,858x10⁶ Nmm

ABCSIA S₁₀: ∫₀⁴⁰⁰ S_y (Sy)(550mm)(- V I_y )ds_y = V [9.X.0-52-29x10⁶S^c2 + 4,81x10⁵³⁰1.03 =

= 9,2524x10⁵ N.mm

ABSCSSA Né: ∫₃₁¹²⁰⁰ (5x) (556mm)(- V I_y ) ds_y = 1,oY [.. 1,47x10⁶3² 1 + 1,381x10³5₁₁ =3o

=6,364x10⁹ N.mm

Now we com go back and espve tw ehte4 of S equations.C: s = 24.8x10⁵mm² + g_{28.40x10⁵mm²q92.2x10⁵mm²++f +x10⁵mm, x 7,6112x10⁷mm = 2x0⁰⁵ N.Yc}

I⁰: 30q₁ - 7q₂ + 261,26 ^N_mm = O

III⁰: 62 q₂- 9q₁ - q₃ - 26q₄ = 493,34 ^N_mm = O

IV²⁰: 30q₃-79q₂ + 261,26^N_mm = O

V⁰: 58,24-9₄ -29₂ + 26b,3^N_mm = O

Pure Shear

0.2434 0.7516 0.8045

0.1087 0.8045 0.7516 0.2434

0.0630 0.2434

0.3125 0.3903

0.7385 1.0244 0.7385

0.3805 0.2903 0.3207 0.3125

0.7739 0.7739 0.7739 0.7739

0.8111

Pure Torsion

0.4321 0.4321

q1 q2 q3

0.4321 0.3122 0.4321

0.0414

0.7029 0.7029

q4

0.7029 0.7029

0.7288

Total

1.548 1.184

0.875 0.685

q1 q2 q3

0.875 0.685

0.3805 0.3895

0.3207 0.3309

0.8797 0.9638

q4

0.8797 0.9638

0.0786 0.0722

0.0776 0.0776