Question

A beam has a cross section as shown. If the shear force is 12 kip

a) Calculate the shear stress at 1 in intervals along the depth of the section

b) Plot the shear stress distribution

4 in. 6 in. 2 in. + -4 in.

Answer 1

- Shear stress distribution is Soin shear force (s) = le kip ein k C 73 ZL Lin 3 in Lamar So lin 3 in symmetrical about Neutral axis. We need te find valus shear stress at 3 intervals. for depth of each interval. of 2 ie. i I in Hin Cross Section. shear stress distribution D at uppen most fibre, Shen stress 2= SAG Ib whear s= shear force Az area above fibre Ja distance of c.lr. of area above fibre from Neutral ani's I = m.o.I. of cross section bz width of Cross section at fibre - for to Aread above fobre LAJ=0 :(21=0 = Now, ?a= shean stress at e in from neutral axis. A = (4x1) = 4 in e (+12) 2.5 in < I. (4*) - C243) 61.33 in 4 ba 4 in Tq = SAY IK 4X9.5 z Ib 61.33 4 - 0.5 Kst kipline Ane

Now, shear stress ta = stress at from neutral avis sin febre width bal ba i in. for which hu, Tax ba= Z₂ b3 : 0.5 x 4 = Tax 1 - Zz-e kiplina Now, Tmax shear stress at Neutral axis , hin KG => Area (A) - (483) - Lare) 3 in 12-4 lin = 8 in2 ain ý - Augi - Azyz AIAZ (413)(92) – CQ42)(%) 1413)-cere) ý - 1.75 in tb) I in, = width Emax = SAT Ib = shear stress z lex8x1.75 61,33 x ) An Tmax a 2.74 2.74 kiplin? shown in in 1st fegure is stress distribution - Shear Section, for given Cross

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