Question

PLEASE SEE TEXT ANSWERS AT THE BOTTOM. I HAVE WORKED OUT PART A. PLEASE COMPLETE REST OF Q.

A simple wooden beam is constructed by bonding four 12.5 x 75 mm planks together with an adhesive as shown in Fig 1.1. load P Adhesive joint Fig. 01.1 - Cross-section through Constructed Beam The beam is Em long and must carry a load P as shown in Fig. 01.2. load P 2.5m 2.5m Fig. 01.2 - Loading of the Beam (a) Sketch shear force and bending moment diagrams for the beam under loading and hence confirm that the maximum bending moment (...)is 1.25P)Nm and the maximum shear Stress (Q.) is 051) 15 marks cond Page 2 of 9 MEE2001/2017 01. contd (b) If the yield strength of the wood is 36 MPa and the maximum allowable shear stress in the adhesive chosen for this application is 0.5 MPa calculate: (1) the maximum bending stress in the beam and the load (P) that the wood will be able to carry before failure [7 marks 1) the maximum shear stress in the bonded joint and hence determine the adhesive selected is suitable for this application. [7 marks (C) In a second beam the adhesive is replaced by a line of nails running along each of the 4 bonded interfaces. If the nails are evenly pitched along the length of the team and each nal can carry a maximum shear force of 296 N, calculate the minimum number of nais necessary to construct the beam [6 marks

ANSWERS: B) 2.59 kN, 0.473 MPa

C) 399

Answer 1

Given that >> P: 1 B RANEO 2.5m 2.5m k RAY= P/2 RB = 1/2 from the equilibrium Conditions EMA O (RB) (5)=(P) (215) RB 0.50 & Fy=0 RAY & RB=P RAY = 0.5P Efn=0 > RARE O - maximum Shear force Sfman Sfo=SFA Pla N manimom Bending moment Mman = mc = (1/2) (25) 1.250 Nim Bending stress manimum is (om man MY INA

k 75 mm ㅔ 12.5mm Z 75mm A loo mm y = 50 ohm T 12.5m 12.5m KK 50mm > kl 125mm 12.5 زی = 50 mm INA = 75X1003 50x75 4492187.5mmy 12 12 maximum bending stress. 5o = Momany IMA of = (1.25P) (103) (50) U492187.5 0.01391 (p) (meal yield strength is. Syt = ob f.s - fszl ES Golba1 CP) = 36 P=2587.5 N Pa = 2.587 kN) - The allowable load on the beam.

(in) shear stress in the bonded Joink (b=25mm] (LA) 95 (Sfman) (QA) A A INA (BA) lower Part mm QA = A5) (11.5) (50-125) " BA = (2) (12.5) = u1015.62 mm3. - 25 mm Ay, :12) 1.2935 x 103 x 41015.62 Stoman = Plz = 2.587/ (4492189.5)(25). = 1,2935kN 2 ०.५१२५ mPom eman) = 05 mpa > (A) 0 4724 mea Joint Joint hence, The adhesive selected for the Joint bonding is suitable for the application (6) -> Shear force ce Corried by each hail Sf nail = 296 N. shear for how at 'A' is q A = (SFA) (QA) INA

QA = 1.2935X103 x 41015.62 (1492187.5) QA = 11.81 N/mm two rows .) of nails. لام Shear frow QA = =) (SF) nail S (gap blw = 3(2)(296) S nails ), ① = =) 4.81 = (2) (296) S. s =500126 mm. I gap b/w each hail. Tobal no. of nails neavined / grow = or 5000 50. 126 99.748 nails. ha (u)C99.748) Total no. of nails neavired (for u onows) - 399.99 n=399