Question

A beam with simple supports as shown below has external loading of three point loads and two different uniformly distributed loads. For this beam: a. Calculate reactions at points C and D b. Derive the equations (only), as a function of x, of both Shear Force and Bending Moment between points C and D only c. Construct complete Shear Force (V) and Bending Moment (M) diagrams for the entire beam, and graph them on the lines shown below. Make sure to label maxima and minima, and show all calculations used in deriving the graphs. 40 KN 40 kN/m 45 kN 12 kN/m TH 1 m 3m Dodo 0.5 m 0.5 m 10 KN

Answer 1

Free body diagrame Yok yokulm YSKH - k Iakom to tik XF L 70 E Rb u lokn O Re * But Im lokno Im 3m For equilibrium & E Mc=o, Rpx 3 + 45x1 + 12x1x}= 40x3.5 - loxu + 40x4x2 Rox3 =420-51 - Ro= 36 T Ro= 123 kH] E fy=o, Rot Ro= 45+ 12x1440 x4 +40-10 Rc+123 - 247 RC = 124kH 124KH Take eign m= Bending moment va Shear force

o from left at a Consider distance a x lection between -o A and a 4SKH 12km/ Alle - Vx = -45-12x56 At azo, VA = -45kn dal, VP = -45-12 R V8=-57 km k an Me= -450-12x + 2 / 2 Ma=-450-6x? At xao, MA = okilim Uzl, mpz - 57 kilm Similar Again consider I 45KM - section nokrilm have to return 12 kNm 11 gou duke A Ķime 61-1)-> 124 1 V23-45-12x1 +124-40 (1) At x=1, Ve= 67 KN x=4 Vb = -45-12 + R4- 40x3 VDE -53 KM we know that moment Shear is force is zeno thon the bending maximum. Su x=0 - - 40 (x-1)= 124-45-12= 67 x= 67 + = 2.675 x= 2.675m Mx = -45a - 12 x1 (2-1+ 2) + 124 (2-) = 40 (X-1) Cal At x= 1, Mc=-51 krim X= 2.675 Mincer = 5.1128krim X=4, mo= - 30 krim

Similar again consider 6-0 45 1 12 Art alt .k 3m ) im 12 1234 KX-1) - yox 3 - 40 x-u) + 123- x = -45-12x1 +24 E At x=4, Vo = 70KN 2-4.5 Ve= Soka me -45 den 1241 (2+4+3+5) +124(9–4+3) _40X3 (3+4+} +123 (9-4) - no (kauja Camery At x=4 Mb = -30kn X=4.5, Mea OKM Again considor a section @-@ YT 12 Toledo q meet Im R4 3m 123 k ant) - 40 Tosh -al-LAB Vx= -45 - 12x1 +24- 40x3.5-406x-4.5) +123 - 40 At <= 4.5, VE = LOKM 25 UF = -10 KN mas -4506- 12x1 (2+1+1) + 129 (x4) — 40% (8-1J* Lamp +123(2-4) -40 (2–4.5)

At ME= okrim x=4.5 a= 5 MF= oktim Shear force Diagram & 67KM 70kh sokn ماما LokH чSky 53 KN JAKH 2.675m Bending moment Digram & J. 1125KNM 3o kң S1 RHM