Question

D 4m - am am 2m 6m im im

with load of one unit

moving from a to c

Answer 1

• In case of moving load to calculate the vertical reaction and moment we use influence line diagram.
• mullar Breslau principles is used to draw the influence line diagram.
• the vertical reaction at location 1 and location 2 will be the shear force.
• ​​​​​​1st calculate the degree of static indeterminacy of beam

is Degree of stattic indeterminacy of beam = total no of unknown reaction=(no of equilibrium equation [ + no ofertoa hinge (internal] = 4 - [2 +2] "A .Si se (soĀT no of unknown - A. SRAIRB, RC, & MAT Calibrium Case = 3 (1920 4 5m =] Internal hinge=2 → As the degree of static indeterminacy ů zero beam is determinate & Influcence line diagram will be linear. - - Influnce line diagram for verticle reaction at location I' by Mullar Breslau 'primeiple's Give unitális placement at I 0.5- 1.5m kam om A * 25*5 q B o c Vertical reaction at' I' manimum in lunit

. For verticle reaction at I alb C 2 IT Am % A - si sq 20 B for moving load of I unit maximum verticle reactically will be maximum ordinate of influence line diagram. from similar triogle you can calculate ordinak at sa 7 + = 2 X=(3) } > Manimum ventile reaction at section II Now. for maximum moment at section I draw influence line diagram (ILD) by providing hinge at I'd giving noration o= ť al below. 1.5 - A so I si se B Manimum ordinat for 1 unit moving load will be moment at I. - 0=1=1 / 3 n = 115 unit moment Hence marimum moment at I = 105 unit

provide hinge at II y for maximum moment at section I - and draw the ILD as fallenes di t utti 0,+ O2 = 1 - 3 qm for got am ordinate at loction II - ab [ where a= &m s -3 b= am L=6m = 2X4 - - 1.333 ] > from similer triangle you can also calculate ordinate at sq which be will be -ve se- 1,33 * Sz= 1.33 so maximum moment at section II = 1.33 unit ! -
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