4. Figure 4 sbows three struts BC.CD and CE The struts are pin-jginted together

Question

4. Figure 4 sbows three struts BC.CD and CE The struts are pin-jginted together at C and struts CD and CE are pin-jointed to a rigid support atDand Eas shown. Strut B is pin-jointed to end B of the beam AB. The beam is pin-jointed to the rigid support at and a vertical load Dis applied to the beam at the position shown. The maximum lowable magnitude of P is considered to be determined cither by buckling of one of the struts, od by the maximum allowable the beam AB. The beam and struts have the square cross-Se Young's modulus values as given in Table 1. .12oApa bending stress of 120 MPa being exceoded in cross-section dimensioas and Calculate the theoretical eritical load for each of the three struts marks) (16 marks) 25 Marks Calculate the maximum allowable magnitude of P PI 0.4 igjid Sufport 0.4 0.4 rigid support 0.2 0.4 0.8 Figure 4 (dimensions in metres) Table 1 Cross-section dimensions/mm Young's modulus, E/GPa Member AB (beam) BC (strut) CD (strut) CE (strut) 40 x 40 8 x 8 10 x 10 9 x 9 200 65 65 200

Explanation / Answer

cross sectional dimension of BC = 8x8 mm

moment of inertia of strut = bd3/12=8*83/12=341.33 mm4

effective length of strut = 0.4m=400 mm

modulus of elasticity = 65 GPa=65*103 N/mm2

Buckling load = pi2EI/L2 = (pi2*65*103*341.33)/(4002)=1368.6 N=1.37 kN

cross sectional dimension of CD = 10x10 mm

moment of inertia of strut = bd3/12=10*103/12=833.33 mm4

effective length of strut = 0.5657m=565.7 mm

modulus of elasticity = 65 GPa=65*103 N/mm2

Buckling load = pi2EI/L2 = (pi2*65*103*833.33)/(565.72)=1670.6 N=1.67 kN

cross sectional dimension of CE = 9x9 mm

moment of inertia of strut = bd3/12=9*9/12=546.75 mm4

effective length of strut =0.8944m=894.4 mm

modulus of elasticity = 200 GPa=200*103 N/mm2

Buckling load = pi2EI/L2 = (pi2*200*103*546.75)/(894.42)=1349.1 N=1.35 kN

cross sectional dimension of beam AB = 40*40

section modulus of section=40*402/6=10666.67 mm3

allowable stress=120 MPa=120 N/mm2

maximum bending moment the beam can resist=120*10666.67*10-6=1.28 kNm

maximum moment in beam due to load P = P*0.4*0.6/1=0.24P

0.24P=1.28

P=5.333 kN (allowable load from beam bending point of view)

buckling capapcity of BC=1.37 kN

axial load in BC due to P load in beam = 0.4P

0.4P=1.37

P=3.425 kN(allowable load from buckling of BC point of view)

Let us determine the axial loads in struts CD and CE when axial load in BC is 3.425 kN

From equilibrium of joint C, we get

angle of CD with horizontal=45o

angle of CE with horizontal = tan-1(0.4/0.8)=26.565o

FCDcos45=FCEcos26.565...(i)

FCDsin45 + FCEsin26.565 = 3.425...(ii)

solving (i) and (ii) we get

FCD=3.23 kN , FCE=2.55 kN

these loads exceed buckling capacities of CD and CE. therefore,load should be decreased such that load in CD=1.67 kN

We need to determine loads in struts CE and BC

when load in CD=1.67kN, load in CE=1.67*cos45/cos26.565=1.32 kN

load in strut BC = (1.67*sin45)+(1.32*sin26.565) = 1.771 kN

therefore, allowable load = 1.771/0.4=4.427 kN

Therefore, final allowable load = 4.427 kN

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