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ENGR5004 / Week 5
Statics new problem 2 (session 14)TutorialOld spec ENGR2162:145 min

ENGR216 Statics "new problem 2" (set in session 11, solved in session 14) official answers

The structure in Fig. 1 is made of columns AB and CD rigidly connected to beam BC. All three members have length LL. A is pinned and D rests on a roller (vertical reaction only). Column AB carries a triangular distributed load acting horizontally to the right, growing linearly from zero at A to w0w_0 at B, i.e. w(x)=w0x/Lw(x) = w_0 x/L with xx measured along AB from A.

All members have the I cross section of Fig. 2: flange width a=OP=ZW=120a = OP = ZW = 120 mm, web height SV=120SV = 120 mm, flange thickness b=OT=UZ=20b = OT = UZ = 20 mm and web thickness c=SR=10c = SR = 10 mm (overall depth 160 mm). Use E=200E = 200 GPa and w0=30w_0 = 30 kN/m.

Fig. 1: portal frame ABCD (all members length L), pin at A, roller at D, triangular load on AB from 0 at A to w0 at B, acting to the right.
Fig. 1: portal frame ABCD (all members length L), pin at A, roller at D, triangular load on AB from 0 at A to w0 at B, acting to the right.
Fig. 2: I cross section; flanges OP and ZW of width a and thickness b, web SR of thickness c and height SV = a.
Fig. 2: I cross section; flanges OP and ZW of width a and thickness b, web SR of thickness c and height SV = a.
Fig. 3: the same frame with a uniform load w1 on beam BC.
Fig. 3: the same frame with a uniform load w1 on beam BC.
Formulas you may need
  • Resultant of a distributed load P=∫0Lw(x) dxP = \int_0^{L} w(x)\,dx and its position xP=∫0Lx w(x) dxPx_P = \dfrac{\int_0^{L} x\,w(x)\,dx}{P} (on the formula sheet)
  • dVdx=−w(x)\dfrac{dV}{dx} = -w(x), dMdx=V(x)\dfrac{dM}{dx} = V(x) (on the formula sheet)
  • Combined axial force and bending σx=NA−MyI\sigma_x = \dfrac{N}{A} - \dfrac{My}{I} (on the formula sheet)
  • Shear stress in a beam τ=VQIt\tau = \dfrac{VQ}{It}, Q=∑yˉiAiQ = \sum \bar y_i A_i above the cut (on the formula sheet)
  • Rectangle I=bh312I = \dfrac{bh^3}{12} and the parallel axis theorem I=Iˉ+Ad2I = \bar I + Ad^2 (on the formula sheet)
  • Static equilibrium; pin gives two forces, roller one (learn this)
  1. (1)
    Draw the free-body diagram of the loaded structure and calculate all reactions.
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  2. (2)
    Calculate the normal load N(x)N(x), shearing load V(x)V(x) and bending moment M(x)M(x) along all three members.
  3. (3)
    Draw N(x)N(x), V(x)V(x) and M(x)M(x) in members AB, BC and CD.
    2:2
  4. (4)
    Draw free-body diagrams of members AB, BC and CD, clearly indicating all external and internal loads on each member.
    2:2
  5. (5)
    Calculate all internal and external loads acting on members AB, BC and CD.
    2:2
  6. (6)
    Calculate the maximum member length LL for which the maximum normal stress in the cross sections of member AB does not exceed 200 MPa.
  7. (7)
    The structure is then subjected to a uniform distributed load w1=20w_1 = 20 kN/m on member BC (Fig. 3). Verify that the cross section is sufficient for the maximum shearing stress in the cross sections of member BC not to exceed 100 MPa.
    2:2